Category: Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves

Very Short Answer Questions

Question 1.
What is the average wavelength of X-rays? (A.P. Mar. ’16 )
Answer:
The wavelength range of X-rays is from 10^-8m(10nm) to 10^-13 m (10^-4 nm).
Average wavelength of X – rays = \(\frac{10+0.0001}{2}\) = 5.00005nm.

Question 2.
Give anyone use of infrared rays. (T.S. Mar. ’19)
Answer:

1. Infrared radiation plays an important role in maintaining the Earth warm.
2. Infrared lamps are used in physical therapy.
3. Infrared detectors are used in Earth Satellites.
4. These are used in taking photographs during conditions of fog, smoke, etc.

Question 3.
If the wavelength of electromagnetic radiation is doubled, what happens to the energy of photon ? (T.S. Mar. ’16)
Answer:
Photon energy (E) = hv = \(\frac{\mathrm{hc}}{\lambda}\)
E ∝ \(\frac{1}{\lambda}\)
Given λ₁ = λ, λ₂ = 2λ, E₁ = E
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{\lambda_2}{\lambda_1}\)
\(\frac{E}{\mathrm{E}_2}\) = \(\frac{2 \lambda}{\lambda}\)
E₂ = E/2
∴ The energy of photon reduces to half of its initial value.

Question 4.
What is the principle of production of electromagnetic waves ?
Answer:
If the charge is accelerated both the magnetic field and electric field will change with space and time, then electromagnetic waves are produced.

Question 5.
What is the ratio of speed of infrared rays and ultraviolet rays in vacuum ?
Answer:
The ratio of speed of infrared rays and ultraviolet rays in vacuum is 1 : 1.
All electromagnetic waves travel with same speed 3 × 10⁸ m /s in vacuum.

Question 6.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave ?
Answer:
E₀ = CB₀
Where E₀ = Amplitude of electric field.
B₀ = Amplitude of magnetic field.
C = velocity of light.

Question 7.
What are the applications of microwaves ? (T.S. Mar. ’15)
Answer:

1. Microwaves are used in Radars.
2. Microwaves are used for cooking purposes.
3. A radar using microwave can help in detecting the speed of automobile while in motion.

Question 8.
Microwaves are used in Radars, why ? (Mar. ’14)
Answer:
As microwaves are of smaller wavelengths, hence they can be transmitted as a beam signal in a particular direction. Microwaves do not bend around the comers of any obstacle coming in their path.

Question 9.
Give two uses of infrared rays. (A.P. Mar. ’19)
Answer:

1. Infrared rays are used for producing dehydrated fruits.
2. They are used in the secret writings on the ancient walls.
3. They are used in green houses to keep the plants warm.

Question 10.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates?
Answer:
i = charging current for a capacitor = 0.6 A
i = id = \(\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
∴ i = i = i_d = 0.6A
∴ Displacement current (i_d) = 0.6 A.

Short Answer Questions

Question 1.
What does an electromagnetic wave consists of ? On what factors does its velocity in vacuum depend ?
Answer:
Maxwell concluded that the variation in electric and magnetic field vectors perpendicular to each other leads to the production of electromagnetic waves in space. They can travel in space even without any material medium. These waves are called electromagnetic waves.

According to Maxwell, electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angles to each other as well as at right angles to the direction of wave propagation. Thus electomagnetic waves have transverse nature.

Electric field E_x = E₀ Sin (kz – ωt)
Magnetic field B_y = B₀ sin (kz – ωt)
Where K is propagation constant (K = \(\frac{2 \pi}{\lambda}\))
The velocity of electromagnetic waves C = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)

Velocity of E.M waves depends on

1. Permeability in free space (μ₀).
2. Permittivity in free space (ε₀).

Velocity of e.m waves is 3 × 10⁸ m / s.

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth ?
Answer:
Green house effect: Temperature of the earth increases due to the radiation emitted by the earth is trapped by atmospheric gases like CO₂, CH₄, N₂, Chlorofluoro carbons etc is called green house effect.

1. Radiation from the sun enters the atmosphere and heat the objects on the earth. These heated objects emit infrared rays.
2. These rays are reflected back to Earth’s surface and trapped in the Earth’s atmosphere. Due to this temperature of the earth increases.
3. The layers of carbon dioxide (CO₂) and low lying clouds prevent infrared rays to escape Earth’s atmosphere.
4. Since day-by-day the amount of carbondioxide in the atmosphere increases, more infrared rays are entrapped in the atmosphere.
5. Hence the temperature of the Earth’s surface increases day by day.

Long Answer Questions

Question 1.
Give the brief history of discovery of knowledge of electromagnetic waves.
Answer:

1. Faraday from his experimental study of electromagnetic induction magnetic field changing with time, gives rise to an electric field.
2. Maxwell in 1865 from his theoritical study concluded that, an electric field changing with time gives rise to magnetic field.
3. It is a consequence of the displacement current being a source of magnetic field.
4. It means a change in electric (or) magnetic field with time produces the other field.
5. Maxwell concluded that the variation in electric and magnetic field vectors perpendicular to each other leads to the production of electromagnetic waves in space.
6. These electromagnetic waves travel in space without any material medium.
7. Both electric and magnetic fields vary with time and space and have the same frequency.
   
8. The electric field vector \(\overrightarrow{\mathrm{E}}\) and magnetic field vector \(\overrightarrow{\mathrm{B}}\) are vibrating along y and z axis and propagation of electromagnetic waves along x – axis.
9. Maxwell found that the electromagnetic waves travel in vacuum with a speed is given by
   
   Where μ₀ = 4π × 10^-7 H/m = permeability in free space.
   ε₀ = 8.85 × 10^-12 c² N^-1 m^-2 = permittivity in free space.
10. The velocity of electromagnetic waves in a medium is given by v = \(\frac{1}{\sqrt{\mu \varepsilon}}\)
11. Maxwell also concluded that electromagnetic waves are transverse in nature.
12. In 1888 Hertz demonstrated experimentally the production and detection of E.M waves using spark oscillator.
13. In 1895 Jagadish Chandra Bose was able to produce E.M waves of wavelength 5m.m to 25 m.m.
14. 1899 Marconi was the first to establish a wireless communication at a distance of about 50 km.

Question 2.
State six characteristics of electromagnetic waves. What is Greenhouse effect ?
Answer:
Characteristics (or) properties of electromagnetic waves :

1. Electromagnetic waves do not require any material medium for their propagation. They propagate in vacuum as well as in a medium.
2. Speed of E.M. waves in free space (or) vacuum is given by
   C = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\) = 3 × 10⁸ m/s.
3. Speed of E.M waves in a medium is given by
   v = \(\frac{1}{\sqrt{\mu \varepsilon}}\)
4. Electromagnetic waves are transverse in nature.
   Electric field \(\overrightarrow{\mathrm{E}}\) and magnetic field \(\overrightarrow{\mathrm{B}}\) which constitute the E.M waves an mutually perpendicular to each other as well as perpendicular to the direction of propagation of the wave.
5. Electromagnetic waves are self sustaining electric and magnetic field oscillations in space.
6. Electromagnetic waves transport energy.
   Poynting vector (\(\overrightarrow{\mathrm{P}}\)) = \(\frac{1}{\mu_0}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}})\)
7. Relation between electric field vector E and magnetic field vector g in vacuum is given by
   C = \(\frac{\mathrm{E}_0}{\mathrm{~B}_0}\)
8. Electromagnetic waves are not deflected by magnetic and electric fields.
9. Electromagnetic waves can be reflected, refracted, interferenced, diffracted and polarised.
10. Electromagnetic wave follow the superposition principle.
11. Average electric energy density of E.M wave is given by
    U_av = U_E + U_B
    U_av = \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2\) + \(\frac{1}{2} \cdot \frac{\mathrm{B}^2}{\mu_0}\)
    U_av = 2U_E = 2U_B
12. Intensity of an E.M waves depends on its average energy density.
    I = \(\frac{1}{2} \varepsilon_0 \mathrm{C} \mathrm{E}_0^2\)
13. E.M. waves carry momentum and exert radiation pressure is given by
    P = \(\frac{\mathrm{F}}{\mathrm{A}}\) = \(\frac{1}{\mathrm{~A}} \frac{\mathrm{dp}}{\mathrm{dt}}\) = \(\frac{\text { Intensity (I) }}{C}\)

Green house effect:
The temperature of the Earth increases due to radiation emitted by the Earth.is trapped by atmospheric gases like CO₂, CH₄, N₂O, chlorofluoro carbons etc is called green house effect.

Textual Exercises

Question 1.
The figure shows a capacitor made of two circular plates each of radius 12cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
a) Calculate the capacitance and the rate of charge of potential difference between the plates.

b) Obtain the displacement current across the plates.
c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor ? Explain.
Answer:
Given ε₀ = 8.85 × 10^-12 C² N^-1 m^-2
Here, R = 12cm = 0.12m, d = 5.0mm = 5 × 10^-3m, I = 0.15A
Area, A = πR² = 3.14 × (0.12)²m²

a) We know that capacity of a parallel plate capacitor is given by

b) Displacement current is equal to conduction current i.e., 0.15 A.
c) Yes, Kirchhoffs first rule is valid at each plate of the capacitor provided. We take the current to be the sum of the conduction and displacement currents.

Question 2.
A parallel plate capacitor in the figure made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^-1.
a) What is the rms value of the conduction current ?
b) Is the conduction current equal to the displacement current ?
c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answer:
a) I_rms = \(\frac{E_{\mathrm{fms}}}{X_c}\) = \(\frac{E_{\mathrm{rms}}}{\frac{1}{\omega C}}\) = E_rms × ωC
∴ I_rms = 230 × 300 × 100 × 10^-12 = 6.9 × 10^-6A = 6.9µA

b) Yes, I = I_d where I is steady d.c or a.c. This is shown below

c) We know, B = \(\frac{\mu_0}{2 \pi} \times \frac{\pi}{R^2} \times I_d\)
The formula is valid even if I_d is oscillating. As I_d = I, therefore
B = \(\frac{\mu_0 r \mathrm{I}}{2 \pi \mathrm{R}^2}\)
If I = I₀, the maximum value of current, then
Amplitude of B = max. value of B = \(\frac{\mu_0 \mathrm{rI}_0}{2 \pi \mathrm{R}^2}\) = \(\frac{\mu_0 \mathrm{r} \sqrt{2} \mathrm{I}_{\mathrm{rms}}}{2 \pi \mathrm{R}^2}\)
= \(\frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \times 3.14 \times(0.06)^2}\) = 1.63 × 10^-11T.

Question 3.
What physical quantity is the same for X-rays of wavelength 10^-10m, red light of wavelength 6800 A and radiowaves of wavelength 500m?
Answer:
The speed in vacuum is same for all the given wavelengths, which is 3 × 10⁸ m/s.

Question 4.
A plane electromagnetic wave travels In vacuum aloñg z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
In electromagnetic wave, the electric field vector \(\overrightarrow{\mathrm{E}}\) and magnetic field vector \(\overrightarrow{\mathrm{B}}\) show their variations perpendicular to the direction of propagation of wave as well as perpendicular to each other. As the electromagnetic wave is travelling along z – direction, hence \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) show their variation in x – y plane.
wave length λ = \(\frac{\mathrm{c}}{\mathrm{v}}\) = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{30 \times 10^6 \mathrm{~s}^{-1}}\) = 10m

Question 5.
A radio can tune into any station in the 7.5 MHz to 12MHz band. What is-the corresponding wavelength band?
Answer:
λ₁ = \(\frac{3 \times 10^8}{7.5 \times 10^6}\) = 40m
λ₂ = \(\frac{3 \times 10^8}{12 \times 10^6}\) = 25m
Thus wavelength band is 40m to 25m.

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator ?
Answer:
The frequency of electromagnetic wave is same as that of oscillating cliarged particle about its equilibrium position; which is 10⁹Hz.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave ?
Answer:
Here, B₀ = 510nT = 510 × 10^-9T
E₀ = CB₀ = 3 × 10⁸ × 510 × 10^-9 = 153 NC^-1.

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz.
(a) Determine. B₀, ω, k, and λ.
(b) Find expressions for E and B.
Answer:
a)
B₀ = \(\frac{E_0}{\mathrm{c}}\) = \(\frac{120}{3 \times 10^8}\) = 4 × 10^-7 T
ω = 2πv = 2 × 3.14 × (50 × 10⁶) = 3.14 × 10⁸ rad/s
K = \(\frac{\omega}{\mathrm{C}}\) = \(\frac{3.14 \times 10^8}{3 \times 10^8}\) = 1.05 rad/m
λ = \(\frac{C}{V}\) = \(\frac{3 \times 10^8}{50 \times 10^6}\) = 6.00 m

b) Expression for \(\overrightarrow{\mathrm{E}}\) is E = E₀ sin (kx – ωt)
= (120 N/c) Sin [(1.05 rad/m) x – (3.14 × 108 rad /s)t] \(\hat{\mathrm{j}}\)
Expression for \(\overrightarrow{\mathrm{B}}\) is B = B₀ sin (kx – ωt)
= (4 × 10^-7 T) sin [(1.05 rad/m) x – (3.14 × 10⁸ rad/s)t] \(\hat{\mathrm{k}}\).

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum, in what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation ?
Answer:
Energy of a photon of frequency v is given by E = hv joules = \(\frac{\mathrm{hv}}{1.6 \times 10^{-19} \mathrm{ev}}\)
Where h = 6.6 × 10^-34 J. The energy of photon of different parts of electromagnetic spectrum in joules and eV are shown in table below, along with their sources of origin.

Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m^-1.
a) What is the wavelength of the wave ?
b) What is the amplitude of the oscillating magnetic field ?
c) Show that the average energy density of the E field equals the average energy density of the B field. [C = 3 × 10⁸ ms^-1].
Answer:
Here, v = 2.0 × 10¹⁰Hz, E₀ = 48 Vm^-1, C = 3 × 10⁸ m/s

a) wavelength of the wave, λ = \(\frac{C}{v}\) = \(\frac{3 \times 10^8}{2.0 \times 10^{10}}\) = 1.5 × 10^-2 m

b) Amplitude of oscillating magnetic field,
B₀ = \(\frac{E_0}{C}\) = \(\frac{48}{3 \times 10^8}\) = 1.6 × 10^-7T

c) For average energy density
U_E = \(\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2\) …… (1)
We know that \(\frac{E_0}{B_0}\) = C
Putting in Eq (1)
U_E = \(\frac{1}{4} \varepsilon_0 \cdot C^2 \mathrm{~B}_0^2\) …. (2)
Speed of Electro magnetic waves, C = \(\frac{1}{\sqrt{\mu_0 \mathrm{E}_0}}\)
Putting in Eq (2) We get.
U_E = \(\frac{1}{4} \varepsilon_0 B_0^2 \cdot \frac{1}{\mu_0 \varepsilon_0}\)
U_E = \(\frac{1}{4} \cdot \frac{\mathrm{B}_{\mathrm{O}}^2}{\mu_0}=\frac{\mathrm{Bo}^2}{2 \mu_0}\) = μ_B
Thus, the average energy density of E field equals the average energy density of B field.

Additional Exercises

Question 1.
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + {5.4 × 10⁶ rad /s} t]} i.
a) What is the direction of propagation ?
b) What is the wavelength λ ?
c) What is the frequency v ?
d) What is the amplitude of the magnetic field part of the wave ?
e) Write an expression for the magnetic field part of the wave.
Answer:
a) From the given question it is clear that direction of motion of e.m. wave is along negative y direction i.e along – \(\hat{\mathrm{j}}\)

b) Comparing the given question with equation E = E₀ cos (ky + ωt).
We have, K = 1.8 rad/m, ω = 5.4 × 10⁸ rad/s, E₀ = 3.1 N/C
λ = \(\frac{2 \pi}{\mathrm{k}}\) = \(\frac{2 \times(22 / 7)}{1.8}\) = 3.492 m ≈ 3.5m

c) V = \(\frac{\omega}{2 \pi}\) = \(\frac{5.4 \times 10^8}{2 \times\left(\frac{22}{7}\right)}\) = 85.9 × 10⁶ ≈ 86MHz

d) B₀ = \(\frac{\mathrm{E}_0}{\mathrm{C}}\) = \(\frac{3.1}{3 \times 10^8}\) = 1.03 × 10^-8T ≈ 10.3nT

e) B = B0 cos (ky + ωt) \(\hat{\mathbf{k}}\) = (10.3nT) cos [(1.8 rad/m/y + (5.4 × 10⁸ rad / s)t] \(\hat{\mathbf{k}}\)

Question 2.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
a) at a distance of lm from the bulb ?
b) at a distance of 10 m ?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:
a) Intensity,

b) I = \(\frac{100 \times(5 / 100)}{4 \pi(10)^2}\) = 4 × 10^-3 W/m²

Question 3.
Use the formula λ_m T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you ?
Answer:
We know, every body at given temperature T₁ emits radiations of all wavelengths in certain range. For a black body, the wavelength corresponding to maximum intensity of radiation at a given temperature.
λ_m T = 0.29cmk or T = \(\frac{0.29}{\lambda_{\mathrm{m}}}\)
For λ_m = 10^-6m = 10^-4cm, T = \(\frac{0.29}{10^{-4}}\) = 2900 k.

Temperature for other wavelengths can be similarly found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of e.m spectrum. Thus to obtain visible radiation, say, λ_m = 5 × 10^-5cm, the source should have a temperature
T = \(\frac{0.29}{5 \times 10^{-5}}\) = 6000 k
It is to be noted that, a body at lower temperature will also produce this wavelength but not with maximum intensity.

Question 4.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
d) 5890 A – 5896 A [double lines of sodium]
e) 14.4 keV [energy of a particular transition in ⁵⁷Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy).
Answer:
a) This wavelength corresponds to radiowaves.

b) This frequency also corresponds to radiowaves.

c) Given T = 2.7 K As λ_m T = 0.29cm °k
∴ λ_m = \(\frac{0.29}{\mathrm{~T}}\) = \(\frac{0.29}{2.7}\) ≈ 0.11cm
This wavelength corresponds to microwave region of the electromagnetic spectrum.

d) This wavelength lies in the visible region of the electromagnetic spectrum.

e) Here, Energy E = 14.4KeV = 14.4 × 10³ × 1.6 × 10^-19J
As E = hv
∴ v = \(\frac{E}{h}\) = \(\frac{14.4 \times 10^3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}\) ≈ 3 × 10¹¹ MHz
This frequency lies in the X-ray region of electromagnetic spectrum.

Question 5.
Answer the following questions :
a) Long distance radio broadcasts use short-wave bands. Why ?
b) It is neccessary to use satellites for long distance TV transmission. Why ?
c) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the earth. Why ?
d) The small ozone layer on top of the stratosphere is crucial for human survival. Why ?
e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now ?
f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction ?
Answer:
a) It is so because ionosphere reflects the waves in these bands.

b) Yes, television signals being of high frequency are not reflected by ionosphere, Therefore to reflect them satellites are needed. That is why, satellites are used for long distance TV transmission.

c) Optical and radiowaves can penetrate the atmosphere where as X-rays being of much smaller wavelength are absorbed by the atmosphere. That is why we can work with optical and radio telescopes on earth’s surface but X-ray astronomical telescopes must be used on the satellite orbiting above the earth’s atmosphere.

d) The small ozone layer present on the top of the stratosphere absorbs most of ultraviolet radiations from the sun which are dangerous and cause genetic damage to living cells, prevents them from reaching the earth’s surface and thus helps in the survival of life.

e) The temperature of earth would be lower because the green house effect of atmosphere would be absent.

f) The clouds by a global nuclear war would perhaps cover most parts of sky preventing solar light from reaching many parts of globe. This would cause a winter.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 10th Lesson Alternating Current Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 10th Lesson Alternating Current

Very Short Answer Questions

Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the secondary if the primary has 10 turns. (A.P. Mar.’19; T.S. Mar. ’16 )
Answer:
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) = \(\frac{N_s}{N_p}\)
V_p = 200V, V_s = 2000V, N_p = 10
N_s = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}} \times \mathrm{N}_{\mathrm{p}}\) = \(\frac{2000}{200} \times 10\)
N_s = 100.

Question 2.
What type of transformer is used in a 6V bed lamp ?
Answer:
Step down transformer is used in 6V bed lamp.

Question 3.
What is the phenomenon involved in the working of a transformer ? (T.S. Mar.19; A.P. Mar. 16, Mar. 14)
Answer:
Transformer works on the principle of mutual induction.

Question 4.
What is transformer ratio ?
Answer:
The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
Transformer ratio = \(\frac{V_s}{V_p}\) = \(\frac{N_s}{N_p}\).

Question 5.
Write the expression for the reactance of

1. an inductor and
2. a capacitor.

Answer:

1. Inductive reactance (X_L) = ωL
2. Capacitive reactance (X_C) = \(\frac{1}{\omega C}\)

Question 6.
What is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor.
Answer:

1. In pure resistor A.C. e.in.f and current are in phase with each other.
2. In pure inductor, current lags behind the e.m.f. by an angle of \(\frac{\pi}{2}\) (or) 90°.
3. In pure capacitor, current leads the e.m.f by an angle \(\frac{\pi}{2}\) (or) 90°.

Question 7.
Define power factor. On which factors does power factor depend ?
Answer:
The ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
Power factor (cosϕ) = \(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{rms}}}\). (∴ P_rms = V_rmsI_rms)
Power factor depends on r.m.s voltage, r.m.s current and average power (P).

Question 8.
What is meant by wattless component of current ?
Answer:
Average power (P) = V_rms(I_rms sinϕ)cos \(\frac{\pi}{2}\)
The average power consumed in the circuit due to I_rms sin ϕ) component of current is zero. This component of current is known as wattless current. (I_rms sin ϕ) is the wattless component of current.

Question 9.
When does a LCR series circuit have minimum impedance?
Answer:
In LCR series circuit, Impedance (Z) = \(\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}\)
At a particular frequency, ωL = \(\frac{1}{\omega C}\)
The impedance is minimum (Z = R)
This frequency is called resonant frequency

Question 10.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
Answer:
In LCR series circuit power factor (cos ϕ) = 1
∴ Phase difference between voltage and current is zero. (ϕ = 0)

Short Answer Questions

Question 1.
Obtain an expression for the current through an inductor when an AC emf is applied.
Answer:
Circuit consists of pure inductor of inductance L. Let an ac e.m.f V = V_m sin ωt is applied to it. Let i be the instantaneous current.
Back e.m.f developed across the inductor = -L\(\frac{\mathrm{di}}{\mathrm{dt}}\)

Total e.m.f = V_m sin ωt – L\(\frac{\mathrm{di}}{\mathrm{dt}}\) ——- (1)
According to ohms law this must be equal to iR = 0

This is the expression for the instantaneous current through inductor. Here current lags behind the e.m.f by \(\frac{\pi}{2}\) radian (or) 90°.

Question 2.
Obtain an expression for the current in a capacitor when an AC emf is applied.
Answer:
Circuit consists of pure capacitor of capacitance C. Let an A.C emf V = V_m sin ωt is applied to it. Let i and q be the instantaneous values of current and charge.
Potential difference across capacitor = –\(\frac{\mathrm{q}}{\mathrm{C}}\).
Total emf = V_m sin ωt –\(\frac{\mathrm{q}}{\mathrm{C}}\)

According to Ohms law this must be equal to iR = 0
V_m sin ωt –\(\frac{\mathrm{q}}{\mathrm{C}}\) = 0

i₀ is peak value of current. Here the current leads the applied e.m.f. by \(\frac{\pi}{2}\) or (or) 90°

Question 3.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Transformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
Principle : It works on the principle of mutual induction between two coils.
Working : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.
This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.

Theory: Let N₁ and N₂ be the number of turns in the primary and secondary. Let V_P and Vs be the emf s across the primary and secondary.

Efficiency of transformer:
It is the ratio of output power to the input power.

Long Answer Questions

Question 1.
Obtain on expression for impedance and current in series LCR circuit. Deduce an expresssion for the resonating frequency of an LCR series resonating circuit
Answer:
Circuit consists of resistor of resistance R, inductor of inductance L and capacitor of capacitance C connected in series with an a.c Voltage V = V_m sin ωt.
Let i be the current and q be the charge at any instant t.

Back e.m.f across inductor is -L\(\frac{\mathrm{di}}{\mathrm{dt}}\)
and across capacitor is \(\frac{-\mathrm{q}}{\mathrm{C}}\)
Total e.m.f = V_m sin ωt – L\(\frac{\mathrm{di}}{\mathrm{dt}}\) – \(\frac{\mathrm{q}}{\mathrm{C}}\)
According to Ohms law, this must be equal to iR
V_m sinωt – L\(\frac{\mathrm{di}}{\mathrm{dt}}\) – \(\frac{q}{C}\) = iR
L\(\frac{\mathrm{di}}{\mathrm{dt}}\) + iR + \(\frac{q}{C}\) = V_m sin ωt ——- (1)
The current i any instant in the circuit is
i = i_m sin(ωt – ϕ) if ωL > \(\frac{1}{\omega \mathrm{C}}\) which is possible at high frequencies.
i = i_m sin (ωt + ϕ) if ωL which is possible at low frequencies.
The maximum current (i_m) is given by

Let ϕ be the phase difference between current and e.m.f

Resonant frequency(f₀):
At this frequency, the impedance of LCR circuit is minimum and is equal to R. At this frequency current is maximum This frequency is called resonant frequency (f₀).

In frequency response curve, at resonant frequency (f₀), current is maximum. This series resonant circuit is called acceptor circuit.

Problems

Question 1.
An ideal inductor (no internal resistance for the coil) or 20 mH is connected in series with an AC ammeter to an AC source whose emf is given by e = 20\(\sqrt{2}\) sin(200t + π/3)V, where t is in seconds. Find the reading of the ammeter ?
Solution:
Given that L = 20 mH = 20 × 10^-3H

Question 2.
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance
are given by i = \(\sqrt{2}\) sin (100t – π/4)A and υ = 40 sin (100t) V. Calculate the resistance ?
Solution:
i = \(\sqrt{2}\) sin (100t – π/4)A
(∵ i = i₀ sin(ωt – ϕ))
υ = 40 sin (100t)V (∵ V = V₀sin(ωt))
i₀ = \(\sqrt{2}\), V₀ = 40, ω = 100, ϕ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0} \cos \phi\) = \(\frac{40}{\sqrt{2}} \cos \frac{\pi}{4}\)
R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\)
R = 20Ω

Question 3.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
Solution:
V_C = 20V, V_R = 35V, V_L = 20V
V = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}^2-\mathrm{V}_{\mathrm{C}}^2\right)}\)
V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\)
V = \(\sqrt{35^2}\)
V = 35 Volt.

Question 4.
An AC circuit contains a resistance R, an inductance L and a capacitance C connected in series across an alternator of constant voltage and variable frequency. At resonant frequency, it is found that the inductive reactance, the capacitive reactance and the resistance are equal and the current in the circuit
is i₀. Find the current in the circuit at a frequency twice that of the resonant frequency.
Solution:
At Resonance R = ω₀L = \(\frac{1}{\omega_0 C}\)

Question 5.
A series resonant circuit contains L₁, R₁ and C₁. The resonant frequency is f. Another series resonant circuit contains L₂, R₂ and C₂. The resonant frequency is also f. If these two circuits are connected in series, calculate the resonant frequency.
Solution:
Given that resonant frequency (f)
= \(\frac{1}{2 \pi \sqrt{\mathrm{L}_1 C_1}}\) = \(\frac{1}{2 \pi \sqrt{\mathrm{L}_2 \mathrm{C}_2}}\)
L₁C₁ = L₂C₂
L₁ = \(\frac{\mathrm{L}_2 \mathrm{C}_2}{\mathrm{C}_1}\) —— (1)
When these two circuits are connected in series
The total inductance L = L₁ + L₂
Total capacitance is given by
\(\frac{1}{\mathrm{C}}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) (or) C = \(\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\)
The resonant frequency of combined circuit is given by

Question 6.
In a series LCR circuit R = 200Ω and the voltage and the frequency of the mains supply is 200 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 45°. On taking out the inductor from the circuit the current leads the voltage by 45°. Calculate the power dissipated in the LCR circuit.
Solution:
R = 200Ω, V = 200V, f = 50Hz
ϕ = 45°, At resonance ωL = \(\frac{1}{\omega \mathrm{C}}\)
Power dissipated (P) = Vi = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) (∵ i = \(\frac{V}{R}\))
P = \(\frac{(200)^2}{200}\) = 200ω

Question 7.
The primary of a transformer with primary to secondary turns ratio of 1: 2, is connected to an alternator of voltage 200 V. A current of 4A is flowing though the primary coil. Assuming that the transformer has no losses, find the secondary voltage and current are respectively.
Solution:
Given that \(\frac{\mathbf{N}_1}{\mathbf{N}_2}\) = \(\frac{1}{2}\)
E₁ = 200V, i₁ = 4A

i)
Transformer ratio = \(\frac{E_2}{E_1}\) = \(\frac{\mathrm{N}_2}{\mathrm{~N}_1}\)
\(\frac{E_2}{200}\) = \(\frac{2}{1}\)
E₂ = 400V

ii)
\(\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = \(\frac{i_1}{i_2}\)
\(\frac{2}{1}\) = \(\frac{4}{i_2}\)
i₂ = 2A

Textual Exercises

Question 1.
A 100Ω resistor is connected to a 220 V. 50 Hz ac supply.
(a) What is the rms value of current in the circuit ?
(b) What is the net power consumed over a full cycle ?
Solution:
Given resistance R = 100Ω
V_rms = 220V
Frequency f = 50Hz

a) Current in the circuit
I_rms = \(\frac{V_{r m s}}{R}\) = \(\frac{220}{100}\) = 2.2A

b) Net power consumed in full cycle
P = V_rms × I_rms
= 220 × 2.2
= 474 W

Question 2.
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage ?
(b) The rms value of current in an ac circuit is 10A. What is the peak current ?
Solution:
a) Given, peak value of voltage V₀ = 300V
The rms value of current I_rms = 10A
The rms value of voltage
V_rms = \(\frac{\mathrm{V}_0}{\sqrt{2}}\) = \(\frac{300}{\sqrt{2}}\) = 212.1V

b) Using the formula
I_rms = \(\frac{\mathrm{I}_0}{\sqrt{2}}\)
The peak value of current
I₀ = \(\sqrt{2} \times \mathrm{I}_{\mathrm{rms}}\)
= \(\sqrt{2}\) × 10 = 1.414 × 10
I₀ = 14.14 A

Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of current in the circuit.
Solution:
Given inductance
L = 44mH = 44 × 10^-3H
V_rms = 220V
Frequency of Inductor f = 50Hz
Inductive resistance X_L = 2πfL
= 2 × 3.14 × 50 × 44 × 10^-3
= 13.83Ω
The rms value of current in the circuit
I_rms = \(\frac{V_{\text {rms }}}{X_L}\) = \(\frac{220}{13.83}\)15.9A

Question 4.
A 60 μF capacitor is connected to a 110V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
Given, capacitance of the capacitor C
= 60μF
= 60 × 10^-6F
V_rms = 110
Frequency of AC supply f = 60Hz
Capacitive reactance
X_C = \(\frac{1}{2 \pi \mathrm{fC}}\) = \(\frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}}\)
= 44.23Ω
The rms value of current in the circuit
V_rms = \(\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{X}_{\mathrm{C}}}\) = \(\frac{110}{44.23}\) = 2.49A

Question 5.
In Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Solution:
In Exercise 3 Average power P = V_rms, I_rms. cos ϕ

As we know that the phase difference between current and voltage in case of inductor is 90°.
P = V_rms. I_rms. cos 90° = 0
In Exercise 4 Average power
P = V_rms. I_rms. cos ϕ
We know that the phase difference between current and voltage in case of capacitor is 90°
P = V_rms. I_rms. cos 90° = 0

Question 6.
Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0H. C = 32 μF and R = 10Ω. What is the Q-value of this circuit ?
Solution:
Given, L = 2H, C = 32μF, R = 10Ω
Resonant angular frequency

Question 7.
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ?
Solution:
Capacitance of capacitor C = 30μF = 30 × 10^-6F
Inductance L = 27mH = 27 × 10^-3H
For free oscillations, the angular frequency should be resonant frequency. Resonant angular frequency of oscillation of the circuit

Question 8.
Suppose the initial charge on the capacitor in Exercise 7 is 6 mC. What is the total energy stored in the circuit initially ? What is the total energy at later time ?
Answer:
Given, charge on the capacitor Q = 6mC = 6 × 10^-3
C = 30μF = 30 × 10^-6F
Energy stored in the circuit
E = \(\frac{Q^2}{2 C}\) = \(\frac{\left(6 \times 10^{-3}\right)^2}{2 \times 30 \times 10^{-2}}\) = \(\frac{36}{60}\) = 0.6J
After some time, the energy is shared between C and L, but the total energy remains constant.
So, we assume that there is no loss of energy.

Question 9.
A series LCR circuit with R = 20Ω, L = 1.5 H and C = 35µF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle ?
Given resistance R = 200Ω
Inductance L = 1.5H, capacitance C = 35µF = 35 × 10^-6F and voltage V_rms = 200V
When, the frequency of the supply equal to the natural frequency of the circuit, this is the condition of resonance. At the condition of resonance, Impedance Z = R = 20Ω
The rms value of current in the circuit
I_rms = \(\frac{V_{r m s}}{Z}\) = \(\frac{200}{20}\) = 10A
ϕ = 0°
Power transferred to the circuit in one complete cycle.
P = I_rms V_rms cos ϕ
= 10 × 200 × cos0° = 2000 ω = 2Kω

Question 10.
A radio can tune over the frequency range o a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

(Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.)
Solution:
Given, minimum frequency
f₁ = 800KHz = 8 × 10⁵HZ
Inductance
L = 200μH = 200 × 10^-5H = 2 × 10^-4H.
Maximum frequency
f₂ = 1200 KHz = 12 × 10⁵Hz for tuning, the natural frequency is equal to the frequency of oscillations that means it is the case of resonance.
Frequency of oscillations f = \(\frac{1}{2 \pi \sqrt{L C}}\)
For capacitance C₁, f₁ = \(\frac{1}{2 \pi \sqrt{L C_i}}\)
C₁ = \(\frac{1}{4 \pi^2 \mathrm{f}_1^2 \mathrm{~L}}\)
= \(\frac{1}{4 \times 3.14 \times 3.14 \times\left(8 \times 10^5\right)^2 \times 2 \times 10^{-4}}\)
= 197.7 × 10^-12F
= 197.7PF
for capacitance C₂, f₂ = \(\frac{1}{2 \pi \sqrt{L C_2}}\)
C₂ = \(\frac{1}{4 \pi^2 f_1^2 \mathrm{C}}\)
= \(\frac{1}{4 \times 3.14 \times 3.14 \times\left(12 \times 10^5\right)^2 \times 2 \times 10^{-4}}\)
= 87.8 × 10^-12F = 87.8pF
Thus, the range of capacitor is 87.8pF to 197.7pF.

Question 11.
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40Ω.

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Solution:
Given, the rms value of voltage
V_rms = 230V
Inductance L = 5H
Capacitance C = 80μf = 80 × 10^-6F
Resistance R = 40Ω

a) For resonance frequency of circuit
ω_r = \(\frac{1}{\sqrt{\mathrm{LC}}}\) = \(\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}\) = 50 rad/s
Source frequency at resonance, then
υ₀ = \(\frac{\omega_0}{2 \pi}\) = \(\frac{50}{2 \times 3.14}\) = 7.76 Hz

b) At the resonant frequency, X_L = X_C
So, impedance of the circuit Z = R
∴ Impedance Z = 40Ω
The rms value of current in the circuit
I_rms = \(\frac{V_{r m s}}{Z}\) = \(\frac{230}{40}\) = 5.75A
Amplitude of current I₀ = I_rms\(\sqrt{2}\)
= 5.75 × \(\sqrt{2}\)
= 8.13A

c) The rms potential drop across L
V_L = I_rms X_L
= I_rms ω_rL = 5.75 × 50 × 5 = 1437.5V
The rms potential drop across R
V_R = I_rms = R = 5.75 × 40 = 230V
The rms potential drop across C,

Potential drop across LC combinations
= I_rms(X_L – X_C)
= I_rms(X_L – X_L) = 0 (∵ X_L – X_C)

Additional Exercises

Question 1.
An LC circuit contains a 20 mH inductor and a 50μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially ? Is it conserved during LC oscillations ?
(b) What is the natural frequency of the circuit ?
(c) At What time is the energy stored
(i) Completely electrical (i.e., stored in the capacitor) ?
(ii) completely magnetic (i.e., stored in the inductor) ?
(d) At what times is the total energy shared equally between the inductor and the capacitor ?
(e) As a resistor is inserted in the circuit, how much energy is eventually dissipated as heat ?
Solution:
Given Inductance L = 20mH = 20 × 10^-3H
Capacitance of capacitor
C = 50μf = 50 × 10^-6F
Initial charge on the capacitor,
Q_i = 10mc = 10 × 10³C

a) The total energy stored across the capacitor initially,

Yes, this energy is conserved during LC oscillations.

b) To get the natural frequency or resonant frequency,
f_r = \(\frac{1}{2 \pi \sqrt{L C}}\)
= \(\frac{1}{2 \times 3.14 \times \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}\)
= \(\frac{7 \times 10^3}{44}\) = 159.2 Hz.
The natural frequency of the circuit
ω = 2πV = 2π × 159.2
= 999.78 ≈ 1000 = 10³ rad/s

c)

i) Let at any instant the energy stored is completely electrical the charge on the capacitor Q = Q₀ cos ωt
Q = Q₀ cos \(\frac{2 \pi}{T} \cdot t\) —– (1)
Q is maximum as it is equal to Q₀, only if cos \(\frac{2 \pi}{T}\) t = ±1 = cosnπt or t
= \(\frac{\mathrm{nT}}{2}\), where n = 1, 2, 3. ..
t = 0, T/2, T, 3T/2, ………..
Thus, the energy stored is completely electrical at t = 0, T/2, T, 3/2……….

ii) Let at any instant, the energy stored is completely magnetic as when the electrical energy across the capacitor is zero.
q = 0
Q = Q₀cos \(\frac{2 \pi t}{T}\) = 0 (from eq(i))
∴ cos \(\frac{2 \pi}{T} \cdot t\) = 0 = \(\frac{\mathrm{n} \pi}{2}\) or t = \(\frac{\mathrm{nT}}{4}\) = cos.
It happens if t = T/4, 3T/4, \(\frac{\mathrm{5T}}{4}\),………..
Thus, the energy stored is completely magnetic at
t = T/4, 3T/4, 5T/4………..

d) Equal sharing of energy between inductor and capacitor means the energy stored in capacitor = \(\frac{1}{2}\) × Maximum energy

or cos(2n + 1)π/4 = cos \(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\)
\(\frac{(2 n+1) \pi}{4}\) = \(\frac{2 \pi t}{T}\)
t = T/8(2n + 1) (n = 0, 1, 2, 3, ……..)
Hence the energy will be shared half on capacitor and half on inductor.
t = \(\frac{T}{8}\), \(\frac{3 \mathrm{~T}}{8}\), \(\frac{5 \mathrm{~T}}{8}\), ………..

e) As a resistor is inserted in the circuit, all of the energy loss during heating. Energy loss = 1J. The oscillations becomes damped and becomes disappear after sometime as the total energy loss in the form of heat.

Question 2.
A coil of inductance 0.50 H and resistance 100Ω is connected to a 240 V, 50Hz ac supply.
a) What is the maximum current in the coil ?
b) What is the time lag between the voltage maximum and the current maximum ?
Solution:
Given, Inductance L = 0.50H
Resistance R = 100Ω
The rms value of voltage
V_rms = 240V, f = 50Hz

a) Impedance of circuit

the rms value of current
I_rms = \(\frac{\mathrm{V}_{\mathrm{rms}}}{\mathbf{Z}}\) = \(frac{240}{31400.15}\) = 0.00764
the maximum value of current in the circuit
I₀ = \(\sqrt{2}\) I_rms = 1.414 × 0.00764 = 1.824A

b) Using the formula of time lag,
t = \(\frac{\phi}{\omega}\)
tan ϕ = \(\frac{X_L}{R}\) = \(\frac{\omega \mathrm{L}}{\mathrm{R}}\) = \(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\) = \(\frac{2 \times 3.14 \times 50 \times 0.50}{100}\)
ϕ = tan^-6(1.571) = 57.5

Time lag t = \(\frac{\phi}{\omega}\) = \(\frac{57.5 \pi}{180 \times 2 \pi \mathrm{f}}\)
= \(\frac{57.5}{180 \times 2 \times 50}\)
= 3.19 × 10^-3S
Thus, the time lag between the voltage maximum and the current maximum is 3.19 × 10^-3S

Question 3.
Obtain the answers (a) to (b) in Exercise 13 if the circuit is connected to a high frequency supply (240V, 10 kHz). Hence, explain the statement that at very high frequency an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Answer:
Given frequency f = 10kHz = 10⁴Hz
the rms value of voltage V_rms = 240v
from Exercise 13
Resistance R = 100Ω
Inductance L = 0.5H

Question 4.
A 100μF capacitor in series with a 40Ω resistance is connected to a 110V. 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Answer:
Given capacitance of capacitor
C = 100µF= 100 × 10^-6F
Resistance R = 40Ω
the rms value of voltage V_rms = 110V
Frequency f = 60Hz

(a) Impedance Z

(b) Time lag (t) = \(\frac{\phi}{\omega}\)

Thus, the time lag between the voltage maximum and the current maximum is 1.55 × 10^-3S.

Question 5.
Obtain the answers to (a) and (b) in Exercise 15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Answer:
Given, the rms value of voltage,
V_rms = 110V
The frequency of capacitor
f = 12kHz = 12000 Hz.
Capacitance of conductor C = 10^-4F.
Resistance R = 40Ω
Capacitive Resistance
X_C = \(\frac{1}{2 \pi \mathrm{fC}}\) = \(\frac{1}{2 \times 3.14 \times 12000 \times 10^{-4}}\)
= 0.133Ω
The rms value of current

The maximum value of current,
I₀ = \(\sqrt{2}\) I_rms
= 1.414 × 2.75
= 3.9A
Here, the value of X_C is very small, so term containing C is negligible.
tan ϕ = \(\frac{1}{\omega \mathrm{CR}}\)
= \(\frac{1}{2 \times 3.14 \times 12000 \times 10^{-4} \times 40}\)
= \(\frac{1}{96 \pi}\)
It is very very small.
In DC circuits, ω = 0
X_C = \(\frac{1}{\omega C}\) = ∞
So, it behaves like an open circuit.

Question 6.
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.
Solution:
As they are corrected in the parallel combination
\(\frac{1}{Z}\) = \(\frac{1}{R}\) + (\(\frac{1}{X_L}\) + \(\frac{1}{X_C}\))
As the reactance (X_C – X_L) is perpendicular to the ohmic resistance R, therefore we can write as

That means \(\frac{1}{Z}\) = minimum and thus Z = maximum. As Z is maximum, current will be minimum, current through inductor

Question 7.
A circuit containing a 80 mH inductor and a 60μF capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’]
Answer:
Given Inductance
L = 80mH = 80 × 10^-3H
Capacitance of capacitor
C = 60μF = 60 × 10^-6F .
The rms value of voltage
V_rms = 230V
Frequency f = 50Hz; Resistance R = 0

a) Impedance of circuit

As ZCO, means X_L < X_C, emf lags by current by 90°
The rms value of current
I_rms = \(\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}\) = \(\frac{230}{27.95}\) = 8.29A
The maximum value of current
I₀ = \(\sqrt{2} I_{\text {rms }}\) = 1.414 × 8.29 = 11.64A

b) Potentiál drop acrõss L
V_rmsL = I_rms × X_L = 8.29 × 2 × 3.14 × 50 × 80 × 10^-3
= 208.25 V
Potential drop across C
V_rmsC = I_rms × X_C
= 8.29 × \(\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}\)
= 440.02 V
∴ Applied rms voltage = V_C – V_L
= 440 – 208.25 = 231.75V

c) Average power transferred to the inductor
P = I_rms V_rms. cosϕ
As the phase difference is 90°, So P = 0

d) Average power transferred to the capacitor.
P = I_rms V_rms. cosϕ
As the phase difference is 90°, So P = 0

e) As there is no resistance in the circuit, so average power is equal to sum of average power due to inductor and capacitor. That means the average power consumed is zero.

Question 8.
Suppose the circuit in Exercise 18 has a resistance of 15Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Solution:
Given, the rms value of voltage V = 230V
V_rms = 230V
Resistance R = 15Ω
Frequency f = 50Hz
Average power across inductor and capacitor is zero as the phase difference between current and voltage is 90°.
Total power absorbed = power absorbed in resistor, P_av

Total power absorbed = 790.6W

Question 9.
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23Ω is connected to a 230 V variable frequency supply.
a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency ? what is the current amplitude at these frequencies ?
d) What is the Q-factor of the given circuit ?
Answer:
Inductance L = 0.12H,
Capacitance C = 480nF = 480 × 10^-9F
Resistance R = 23Ω
The rms value of voltage
I_rms = \(\frac{V_{r m s}}{Z}\) = \(\frac{230}{23}\) = 10A
The maximum value of current I₀ = \(\sqrt{2}\)
I_rms = 1.414 × 10 = 14.14A.
At natural frequency, the current amplitude is maximum.
ω = \(\frac{1}{\sqrt{\mathrm{LC}}} \frac{1}{2 \pi \sqrt{0.12 \times 480 \times 10^{-9}}}\) = 4166.6
= 4167 rad/s

b) Average power is maximum for resonance.
P_av(max) = I²_rms R = 10 × 10 × 23 = 2300W.

c) Power transferred to the circuit is half the power at resonant frequency
Δω = \(\frac{\mathrm{R}}{2 \mathrm{~L}}\) = \(\frac{23}{2 \times 0.12}\) = 95.83 rad/s
ΔV = \(\frac{\Delta \mathrm{W}}{2 \pi}\) = 15.2Hz
The frequencies at which power transferred is half .
V = V₀ ± ΔV = 663.48 ± 15.26
So, frequencies are 448.3Hz and 678.2Hz, the maximum current
I = \(\frac{\mathrm{I}_0}{\sqrt{2}}\) = \(\frac{14.14}{\sqrt{2}}\) = 10A

d) Q-factor = \(\frac{\omega_{\mathrm{r}}}{\mathrm{R}}\) = \(\frac{4166.7 \times 0.12}{23}\) = 21.74

Question 10.
Obtain the resonant frequency and Q-factor of series LCR circuit with L = 3.0 H, C = 27μF, and R = 10.4Ω. It is desired to improved the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Solution:
Given, inductance L = 3H
Capacitance of capacitor
C = 27μF = 27 × 10^-6F
Resistance R = 7.4Ω
the resonant frequency of circuit
ω_r = \(\frac{1}{\sqrt{\mathrm{LC}}}\) = \(\frac{1}{\sqrt{3 \times 27 \times 10^{-6}}}\) = \(\frac{1000}{9}\)
= 111.1 rad/s.
Q-factor of a series LCR circuit
Q-factor = \(\frac{\omega_{\mathrm{r}} \mathrm{L}}{\mathrm{R}}\) = \(\frac{111.1 \times 3}{7.4}\) = 45.04
To reduce the full width at half by factor Q, we have to reduce the value of R as R/2
\(\frac{R}{2}\) = \(\frac{7.4}{2}\) = 3.7Ω

Question 11.
Answer the following questions :
a) In any ac circuit, is the applied Instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit ? Is the same true for rms voltage ?
Answer:
Yes, the applied voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit
No, it is not true for rms voltage because there is some phase differences across different elements of circuits.

b) A capacitor is used in the primary circuit of an induction coil.
Answer:
A capacitor is used in the primary circuit of an induction coil because when the circuit is broken, a large induced voltage is used up in charging the capacitor. So, the parking or any damages are avoided.

c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
Solution:
As we know that
X_C = \(\frac{1}{2 \pi f c}\), X_L = 2πfL
f = 0, X_C = ∞, X_L = 0

d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an Iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
Answer:
When the choke coil is connected to dc, there is no change in the brightness. Because f = 0, X_L = 6. So, no change in the brightness. In Ac, the choke offers impedance, so, it glows dim. As we insert an iron core the magnetic field increases and hence inductance increases.
BA = LI = ϕ
L ∝ B
So, X_L also increases and the brightness of bulb decreases.

e) Why is choke coil needed in the use of fluorescent tubes with ac mains ? Why can we not use an ordinary resistor instead of the choke coil ?
Answer:
We use the choke coil instead of resistance because the power loss across resistor is maximum while the power loss across choke is zero.
For resistor, ϕ = 0
P = I_rms. V_rms. cos ϕ
= I_rms. V_rms = maximum
For Inductor ϕ = 90°
P = I_rms. V_rms. cos 90° = 0

Question 12.
A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns, what should be the number of turns in the secondary in order to get output power at 230 V ?
Solution:
Given Primary voltage V_P = 2300V N_P = 4000 turns
Secondary voltage V_S = 230V
Using formula,
\(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}\) = \(\frac{N_S}{N_P}\)
\(\frac{230}{2300}\) = \(\frac{\mathrm{N}_{\mathrm{S}}}{4000}\)
N_S = 400
∴ thus, the number of turns in secondary are 400.

Question 13.
At a hydroelectric power plant, the water, pressure head is at a height of 300 m and the water flow available is 100 m³s^-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (q = 9.8 ms^-2).
Answer:
Given, height of water h = 300m
Rate of flow of water V = 100m³/s
efficiency η = 60%
g = 9.8m/s².
As we know that input power is required to raised the water up to height
h = 300 m

Suppose, the power output is P_out which is equal to the power available from the plant. The efficiency of generator
η = \(\frac{P_{\text {out }}}{P_{\text {in }}}\)

p_out = 176.4MW = 1764 × 10⁵W

Question 14.
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power in 0.5 Ω per km. The town gets power from the line through a 4000 – 220V step-down transformer at a substation in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage ?
(c) Characterise the step up transformer at the plant.
Answer:
Generating power of Electric plant = 800 kW at V = 220V
Distance = 15km,
Generating voltage = 440V
Resistance/length = 0.50Ω/km
Primary voltage V_P = 4000V
Secondary voltage V_S = 220V
(a) Power = I_p. V_p
800 × 1000 = I_p × 4000
I_p = 200A
Line power loss in form of heat
= (I_p)² × Resistance of line
= (I_p)² × 0.5 × 15 × 2
= (200)² × 0.5 × 15 × 2
= 60 × 10⁴W
= 600KW

(b) If there is no power loss due to leakage the plant supply should be =
800 + 600 = 1400KW

(c) Voltage drop across the line
= I_P. R
= 200 × 0.5 × 15 × 2
= 3000V
Voltage from transmission
= 3000 + 4000 = 7000V
As it is given that the power generated at 440V
So, the step-up transformer needed at the plant is 440V – 7000V

Question 15.
Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred ?
Answer:
Given primary voltage V_P = 40,000V
Let the current in primary is I_P
∴ V_PI_P = P
800 × 1000 = 40000 × I_P
I_P = 20A
a) Line power loss = (I_P)² × R
= (20)² × 2 × 0.5 × 15
= 6000W = 6KW

b) Power supply by plant
= 800 + 6 = 806KW

Voltage drop of line = I_P.R
= 20 × 2 × 0.5 × 15
= 300V
Voltage for transmission
= 40000 + 300 = 40300V
Power loss at higher voltage
= \(\frac{6}{800}\) × 100 = 0.74%
Power loss at lower voltage
= \(\frac{600}{1400}\) × 100 = 42.8%
Hence, the power loss is minimum at higher voltage. So, the high voltage transmission is preferred.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction

Very Short Answer Questions

Question 1.
What did the experiments of Faraday and Henry show?
Answer:
The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry.

Question 2.
Define magnetic flux.
Answer:
Magnetic flux is defined as the number of magnetic lines of force crossing through the surface.
ϕ_B = \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = BA cos θ
C.G.S unit → Maxwell
S.I. unit → Weber (wb)
Magnetic flux is a scalar.

Question 3.
State Faraday’s law of electromagnetic induction.
Answer:
“Magnitude of induced e.m.f is directly proportional to the rate of change of magnetic flux”
ε ∝ \(-\frac{d \phi}{d t}\)

Question 4.
State Lenz’s law.
Answer:
The direction of induced e.m.f (or) current is such that it opposes the cause which produce it. Lenz’s law is in accordance with law of conservation of energy.

Question 5.
What happens to the mechanical energy (of motion) .when a conductor is moved in a uniform magnetic field ?
Answer:
Motional e.m.f is produced to the motion of the conductor,in a magnetic field.
Motion e.m.f (ε) = Blυ

Question 6.
What are Eddy currents ? (T.S. Mar. ’19; A.P. Mar. ’15)
Answer:
Eddy currents (or) Foucault currents : The induced circulating currents produced in a conductor itself due to change in magnetic flux linked with the conductor are called Eddy currents.
Due to Eddy currents, the energy is dissipated in the form of heat energy.

Question 7.
Define ‘inductance’.
Answer:
Inductance is a coefficient of electromagnetic induction and is an intrinsic property of a material just like capacitance.
Inductance is an important scalar quantity which depends upon the geometry (i.e, dimensions) of a coil.

Question 8.
What do you understand by ‘self inductance’ ?
Answer:
Self inductance of a coil is defined as the induced e.m.f produced in the coil through which the rate of change of current is unity.
ε = \(-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\) ; ε = -L, If \(\frac{\mathrm{dI}}{\mathrm{dt}}\) = 1 A/s.

Short Answer Questions

Question 1.
Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion.
Answer:
Consider a conductor PQ of length l moving freely in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) with uniform velocity u on a rectangular conductor ABCD. Let any arbitrary charge q in the conductor also move in the field with same velocity.

Magnitude of Lorentz force on this charge (F) = Bqυ —– (1)
Workdone in moving the charge from P to Q is given by
W = Force × displacement
W = Bqυ × l —– (2) (∵ Direction of force on the charge as per Fleming’s left hand rule)
Electromotive force (ε) = \(\frac{W}{Q}\)
ε = \(\frac{\mathrm{Bqv} l}{\mathrm{q}}\) ⇒ ε = Blυ —- (3)

Question 2.
Describe the ways in which Eddy currents are used to advantage. (A.P. Mar. ’16, AP & T.S. Mar. ’15)
Answer:
Eddy currents are used to advantage in

1. Magnetic braking in trains : A strong magnetic field is applied across the metallic drum rotating with the axle of the electric train. Thus large eddy currents are produced in the metallic drum. These currents oppose the motion of the drum and hence the axle of the train which ultimately makes the train come to rest.
2. Induction Motor: Eddy currents are used to rotate the short circuited rotor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current.
3. Electromagnetic damping : Certain galvanometers have a fixed core made of non magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.
4. Induction furnace : Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil. The eddy currents generated in the metals produce high temperatures sufficient to melt it.
5. Analogue energy meters : Concept of eddy currents is used in energy meters to record
   the consumption of electricity. Aluminium disc used in these meters get induced due to varying magnetic field. It rotates due to eddy currents produced in it.

Question 3.
Obtain an expression for the mutual inductance of two long co-axial solenoids. (A.P. Mar. ’19)
Answer:
Consider two solenoids as shown in figure. The length of primary coil be l and are a of cross section A. Let N₁ and N₂ are the total number of turns in the primary and secondary solenoids. Let n₁ and n₂ be the number of turns per unit length

(n₁ = \(\frac{\mathrm{N}_1}{l}\) and n₂ = \(\frac{\mathrm{N}_2}{l}\)). Current in the primary coil is i.
∴ Magnetic field inside the primary (B) = μ₀n₁ I = μ₀\(\frac{\mathrm{N}_1}{l}\) I —– (1)
Magnetic flux through each turn of primary
ϕ_B = \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = μ₀\(\frac{\mathrm{N}_1}{l} \mathrm{I} \times \mathrm{A}\) —— (2)

The same magnetic flux is linked with the secondary coil.
∴ Total magnetic flux linked with secondary = μ₀\(\frac{\mathrm{N}_1 \mathrm{i}}{l} \times \mathrm{A} \times \mathrm{N}_2\) ——– (3)
If M be mutual inductance of the two coils, the total flux linked with the secondary is Mi

Question 4.
Obtain an expression for the magnetic energy stored a solenoid in terms of the magnetic field, area and length of the solenoid.
Answer:
When the current flows through the inductor of inductance L, an e.m.f is induced in it is given by
ε = -L\(\frac{\mathrm{dI}}{\mathrm{dt}}\) —– (1)

(-ve sign shows that e.m.f opposes the passage of current)
Let an infinite small charge dq be driven through the inductor. So the work done by the external voltage is given by

Total work done to maintain the maximum current (I₀) is given by

This work done is stored in the form of potential energy in the magnetic field (U)

In case of solenoid B = μ₀nI₀ (or) I₀ = \(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\) and L = μ₀n²Al

Long Answer Questions

Question 1.
Outline the path-breaking experiments of Faraday and Henry and highlight the contributions of these experiments to our understanding of electromagnetism.
Answer:
Faraday’s and Henry’s experiments :

Experiment 1: A magnet induces current due to relative motion

1. The apparatus consists of a coil with a galvanometer G and a bar magnet.
2. When the bar magnet (NS) was at rest, the galvanometer shows no deflection.
3. When North pole of the bar magnet moved towards the coil, galvanometer shows the deflection in one direction indicating the flow of current in the coil.
   
4. When North pole of the bar magnet moved away from the coil, galvanometer again showed the deflection but now in the opposite direction.
5. The deflection of the galvanometer was large when the bar magnet was moved faster towards (or) away from the coil.
6. When south pole of the magnet was brought near the coil (or) moved away from the coil, the deflections in the galvanometer are opposite to that observed with the north pole for similar movements.

Conclusion:

1. Whenever there is a relative motion between a coil and a magnet, induced current flows through the coil.
2. Large induced e.m.f. (or) current is produced in the coil if the relative motion between magnet and the coil is large.

Experiment 2 : Current induces current due to relative motion of coils :

1. The bar magnet is replaced by a secondary coil C₂ connected to a battery as shown in figure.
2. The steady current in the coil C₂ produces a steady magnetic field.
3. As coil C₂ is moved towards the coil C₁, the galvanometer shows a deflection. This indicates current is induced in the coil C₁.
4. When coil C₂ is moved away, the galvanometer shows a deflection again, but opposite direction.
5. The deflection lasts as long as coil C₂ is in motion.
6. When the coil C₂ is held fixed and C₁ is moved, the same effects are observed.

Conclusion : Induced e.m.f (or) current is produced, when there is a relative motion between the coils.

Experiment 3: Changing current, Induces current without relative motion:

1. Faraday showed that this relative motion is not an absolute requirement.
2. Figure shows two coils C₁ and C₂ held stationary.
3. Coil C₁ is connected to a battery through a tap key K and coil C₂ is connected to a galvanometer (G).
4. it is observed that the galvanometer shows a momentary deflection when the tap key K is pressed.
5. The pointer in the galvanometer returns to zero immediately.
6. If the key is held pressed continuously, there is no deflection in the galvanometer.
7. When the key is released, the galvanometer shows deflection again but in the opposite direction.
8. Deflection of the galvanometer increases a lot when wooden bar is replaced by iron bar.

Question 2.
Describe the working of a AC generator with the aid of a simple diagram and necessary expressions.
Answer:
“An electrical machine used to convert mechanical energy into electrical energy is known as A.C generator/altemator”.
Principle : It works on the principle of electromagnetic induction.

Construction :

1. Armature : Armature coil (ABCD) consists of a large number of turns of insulated copper wire wound over a soft iron core.
2. Strong field magnet : A strong permanent magnet (or) an electromagnet whose poles (N and S) are cylindrical in shape used as a field magnet. The armature coil rotates between the pole pieces of the field magnet.
   
3. Slip rings: The two ends of the armature coil are connected to two brass slip rings R₁ and R₂. These rings rotate along with the armature coil.
4. Brushes : Two carbon brushes B₁ and B₂ are pressed against the slip rings. The brushes remain fixed while slip rings rotate along with the armature. These brushes are connected to the load through which the out put is obtained.

Working : When the armature coil ABCD rotates is the magnetic field provided by the strong field magnet, it cuts the magnetic line of force. The magnetic flux linked with the coil changes due to the rotation of the armature and hence induced e.m.f is set up in the coil.

The current flows out through the brush B in one direction of half of revolution and through the brush B₂ in the next half revolution in the reverse direction. This process is repeated, therefore e.m.f produced is of alternating nature.

Theory:

1. When the coil is rotated with a constant angular velocity (ω)
2. The angle between the normal to the coil and magnetic field \(\overrightarrow{\mathrm{B}}\) at any instant is given by θ = ωt —— (1)
3. The component of magnetic field normal to the plane of the coil = B cos θ = B cosωt —— (2)
4. Magnetic flux linked with the single turn of the coil = (B cos ωt) A —— (3)
   where A is the area of the coil, if the coil has n turns
5. Total magnetic flux linked with the coil (ϕ) = n(B cos ωt) A —– (4)
   According to Faraday’s law,

ε = \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(-\frac{\mathrm{d}}{\mathrm{dt}}\)(nBA cos ωt) = -nBA(-ω sin ωt)
ε = nBA ω sin ωt —— (5)
Where nBAω is the maximum value of e.m.f. (ε₀)
ε = ε₀ sin ωt —– (6) (∵ ω = 2πυ)

Instantaneous current is the circuit is given by
I = \(\frac{\varepsilon}{\mathrm{R}}\) = \(\frac{\varepsilon_0}{\mathrm{R}}\) sin ωt [∵ i = \(\frac{\varepsilon_0}{\mathrm{R}}\)]
I = I₀ sin ωt
The direction of the current changes periodically and therefore the current is called alternating current (a.c.).

Textual Exercises

Question 1.
Predict the direction of induced current in the situations described by the following (a) to (f).

Answer:
a) Here south pole is moving towards the coil, so according to Lenz’s law this end becomes s-pole. (To oppose the motion of south pole by repelling it). Hence, the direction of current is clockwise (by using clock rule) and the current flows from p to q.

b) In coil p-q at the end q ⇒ s-pole is moving towards end q, so it behaves like a south pole (by lenz’s law). The direction of current is clockwise (by clock rule) i.e. from p to q. North pole is moving away so this end will behave like south pole (To oppose its away motion by attracting it). In coil x-y, S-pole is induced (by Lenz’s law) and the direction of current is clockwise i.e. x to y.

c) As the tapping key is just, closed, the current in coil increases so, the magnetic flux and field increases. According to Maxwell’s right hand grip rule, the direction of magnetic field is left wards. Thus, the direction of induced current in the neighbouring coil is such that it try to decrease the field, thus the direction of field in the neighbouring coil should be rightwards i.e., according to Maxwell’s right hand rule the direction of induced current is anti clock wise i.e. xyz.

d) As the rheostat setting is changed, the current is changed. The direction of field due to the coil is leftwards according to Maxwell’s right hand grip rule. The direction of induced current in the left coil is such that the magnetic field produced by it in rightwards, thus the direction of current in left coil is Anticlockwise i.e from zyx.

e) As the key is just released, the current which is flowing anticlockwise goes on decreasing. Thus, the induced current developed in such a sense the magnetic field due to left coil increases (which is towards right). So, the magnetic field due to the right coil should also towards right and hence the induced current is in anticlockwise, i.e., x to yx – direction.

f) The magnetic field lines due to the current carrying wire are in the plane of the loop. Hence, no induced current is produced in the loop (because no flux lines crosses the area of loop).

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by fig. a, b.

(a) A wire of irregular shape turning into a circular shape.
(b) A circular loop being deformed into a narrow straight wire.
Answer:
a) Here, the direction of magnetic field is perpendicularly inwards to the plane of paper. If a wire of irregular shape turns into a circular shape then its area increases (∵ the circular loop has greater area than the loop of irregular shape) so that the magnetic flux linked also increases. Now, the induced current is produced in a direction such that it decreases the magnetic field (i.e.) the current will flow in such a direction so that the wire forming the loop is pulled inwards in all directions (to decreases the area) i.e., current is in anticlockwise direction, i.e., adcba

b) When a circular loop deforms into a narrow straight wire, the magnetic flux linked with it also decreases. The current induced due to change in flux will flow in such a direction that it will oppose the decrease in magnetic flux so it will flow anti clockwise i.e along a’d’c’b’a’, due to which the magnetic field produced will be out of the plane of paper.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing ?
Solution:
Given number of turns (n) = 15 per cm = 1500 per metre. .
Area of small loop A = 2 cm² = 2 × 10^-4m²
Change in current \(\frac{\mathrm{dI}}{\mathrm{dt}}\) = \(\frac{4-2}{0.1}\) = \(\frac{2}{0.1}\) = 20 A/s
Let e be the induced emf, According to Faraday’s law,
e = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(BA) or e = \(\text { A. } \frac{\mathrm{dB}}{\mathrm{dt}}\) (∵ ϕ = BA)
e = A. \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\mu_0 \mathrm{nI}\right)\) (∵ Magnetic field inside the solenoid B = μ₀nI)
or e = Aμ₀n\(\frac{\mathrm{dI}}{\mathrm{dt}}\)
e = 2 × 10^-4 × 4 × 3.14 × 10^-7 × 1500 × 20
e = 7.5 × 10⁶v (∵ μ₀ = 4π × 10^-7) Thus, the induced emf in the loop is 7.5 × 10⁶V.

Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop ? For how long does the induced voltage last in each case ?
Solution:
Given length of the loop l = 8 cm = 8 × 10^-2 m.
Width of the loop b = 2 cm = 2 × 10^-2 m.
Velocity of the loop = 1 cm/s = 0.01 m/s.
Magnitude of magnetic field (B) = 0.3

a) When velocity is normal to the longer side .
(l = 8 cm = 8 × 10^-2 m)
In this case, motional emf
e = B/υ = 0.3 × 8 × 10^-2 × 0.01 = 2.4 × 10^-4 V

b) When velocity is normal to the shorter side
(l = 2cm = 2 × 10^-2 m)
In this case, the developed emf
e = Blυ = 0.3 × 2 × 10^-2 × 0.01
e = 0.6 × 10^-4 V.

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Solution:
Length of rod l = 1 m
Angular frequency of rod ω = 400 rad/s
Magnetic field B = 0.5 T
The linear velocity of fixed end = 0
The linear velocity of other end = lω (∵ V = rω)

Average linear velocity V = \(\frac{0+l \omega}{2}\)
V = \(\frac{l \omega}{2}\) —– (i) By using the formula of motional emf,
e = Bυl = \(\frac{\mathrm{B} l \omega}{2}\) (from equation (i)), e = \(\frac{0.5 \times 1 \times 400 \times 1}{2}\), e = 100.
Thus, the emf developed between the centre and ring is 100 V

Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from ?
Solution:
Given radius of coil = 8 cm = 0.08 m, Number of turns = 20
Resistance of closed loop = 10Ω, Angular speed ω = 50 rad/s
Magnitude of magnetic field B = 3 × 10^-2 T
Induced emf produced in the coil e = NBA W sin ωt
For maximum emf, sin ωt = 1
∴ Maximum emf e₀ = NBAω

The source of power dissipated as heat in the coil is the external rotar. The current induced in the coil causes a torque which opposes the rotation of the coil, so the external agent rotar counter this torque to keep the coil rotating uniformly.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^-4 Wb m².
(a) What is the instantaneous value of the emf induced in the wire ?
(b) What is the direction of the emf ?
(c) Which end of the wire is at the higher electrical potential ?
Solution:
Given, velocity of straight wire = 5 m/s
Magnetic field of straight wire,
B = 0.30 × 10^-4 Wb/m²
length of wire l = 10m

a) Emf induced in the wire e = B/υsin θ
Here, θ = 90°
∴ sin θ = 1 .
(∴ wire is falling at right angle to earth’s horizontal magnetic field component)
= 0.3 × 10^-4 × 10 × 5 = 1.5 × 10^-3 V.

b) According to the Fleming’s right hand rule, the force is downward, then the direction of induced emf will be from west to east.

c) As the direction of induced emf or current is from west to east, the west end of the wire is at higher potential.
(∵ current always flows from a point at higher potential to a point at lower potential).

Question 8.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. (T.S. Mar. ’16, Mar. ’14)
Solution:
Change in current, dI = 5 – 0 = 5A, Time taken in current change dt = 0.1 s
Induced average emf e_av = 200 V
Induced emf in the circuit e = L. \(\frac{\mathrm{dI}}{\mathrm{dt}}\) ⇒ 200 = L\(\left(\frac{5}{0.1}\right)\) or L = \(\frac{200}{50}\) = 4H

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ?
Solution:
Given, mutual inductance of coil M = 1.5 H, Current change in coil dl = 20 – 0 = 20 A
Time taken in change dt = 0.5s, Induced emf in the coil e = M\(\frac{\mathrm{dI}}{\mathrm{dt}}\) = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\)
dϕ = M.dl = 1.5 × 20, dϕ = 30 Wb, Thus the change of flux linkage is 30 Wb.

Question 10.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Solution:
Speed of jet plane V = 1800 km/h = 1800 × \(\frac{5}{18}\) = 500 m/s
l = Distance between the ends of wings = 25 m
The magnitude of magnetic field B = 5 × 10^-4 T
Angle of dip γ = 30°.
Use the formula of motional emf
e = B_vVl, e = B sin γ Vl (B_v = B sin γ),
e = 5 × 10^-4 × sin 30° × 500 × 25, e = 3.1 V
Thus, the voltage difference developed between the ends is 3.1 V

Additional Question

Question 1.
Obtain an expression for the self inductance of a solenoid.
Answer:
i) Consider a long solenoid of length l, area of cross-section A. Let n be the number of turns per unit length (n = \(\frac{\mathrm{N}}{l}\))

ii) Let i be the current flows through it. The magnetic field inside the solenoid is given by
B = μ₀ni —— (1)

iii) Magnetic flux linked with each turn of the solenoid = B.A = μ₀niA —— (2)

iv) Total magnetic flux linked with whole solenoid
ϕ = μ₀niA × N (∵ N = nl)
ϕ = μ₀ni A × nl (Where N Total number of turns in the solenoid)
ϕ = μ₀n²iAl —— (3) Also, ϕ = Li —– (4) From eq’s (3) & (4), Li = μ₀n²iAl
L = μ₀n²Al —– (5) (or) L = μ₀\(\frac{\mathrm{N}^2}{l}\)A —– (6)

Additional Exercises

Question 1.
Suppose the loop in Textual Exercise 4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^-1. If the cut is joined and the loop has a resistance of 1.6Ω how much power is dissipated by the loop as heat ? What is the source of this power ?
Solution:
Area of loop = 8 × 2 = 16 cm²
Rate of change of magnetic field \(\frac{\mathrm{dB}}{\mathrm{dt}}\) = 0.02 T/S,
Resistance of loop R = 1.6Ω
Induced emf of loop e = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})\) (∵ ϕ = BA)
e = A. \(\frac{\mathrm{dB}}{\mathrm{dt}}\) ⇒ e = 16 × 10^-4 × 0.02 ⇒ e = 0.32 × 10^-4V
Induced current in the loop I = \(\frac{e}{R}\) = \(\frac{0.32 \times 10^{-4}}{1.6}\) = 0.2 × 10^-4A
Power of source as heat P = I²R = (0.2 × 10^-4)² × 1.6 = 6.4 × 10^-10 W ⇒ P = 6.4 × 10^-10W
The agency which changing the magnetic field with time is the source of this power.

Question 2.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^-1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T cm^-1 along the negative x-direction (that is it increases by 10^-3 T cm^-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^-3 Ts^-1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Solution:
Given, side of loop a = 12 cm
∴ Area of loop (A) = a² = (12)² = 144 cm² = 144 × 10^-4 m² (∵ Area of square = (side)²)
Velocity v = 8 cm/s = 8 × 10^-2 m/s.
Rate of change of magnetic field with distance \(\frac{\mathrm{dB}}{\mathrm{dx}}\) = 10^-3 T/cm
Rate of change of magnetic field with time \(\frac{\mathrm{dB}}{\mathrm{dt}}\) = 10^-3 T/s
Resistance of the loop R = 4.5 mΩ = 4.5 × 10^-3Ω
Rate of change of magnetic flux with respect to time
\(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}\) = \(\left(\frac{\mathrm{dB}}{\mathrm{dt}}\right) \mathrm{A}\) = 10^-3 × 144 × 10^-4
= 1.44 × 10^-5 Wb/s

Rate of change of magnetic flux due to the motion of loop
\(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{\mathrm{dB}}{\mathrm{dx}} \cdot \mathrm{A} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}\) = 10^-3 × 144 × 10^-4 × 8 = 11.52 × 10^-5 Wb/s
Both of the effects cause a decrease in magnetic flux along the positive z-direction.
Total induced emf in the loop e = 1.44 × 10^-5 + 11.52 × 10^-5
e = 12.96 × 10^-5 V
Induced current in the loop = \(\frac{\mathrm{e}}{\mathrm{R}}\) = \(\frac{12.96 \times 10^{-5}}{4.5 \times 10^{-3}}\) = 2.88 × 10^-2A.
The direction of induced current is such as to increase the flux through the loop along positive z-direction.

Question 3.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50Ω Estimate the field strength of magnet.
Solution:
Area of coil A = 2cm² = 2 × 10^-4 m², Number of turns N = 25
Total charge in the coil θ = 7.5 mC (1 mC = 10^-3 C) = 7.5 × 10^-3 C
Resistance of coil R = 0.5Ω
When the coil is removed from the field, the flux is zero ϕ_f = 0. Induced current in the coil

Thus, the strength of magnetic field is 0.75 T.

Question 4.
Following figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open ? What if K is closed ?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed ?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^-1) when K is closed ? How much power is required when K is open ?
(f) How much power is dissipated as heat in the closed circuit ? What is source of this power ?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular ?
Solution:
Given, length of the rod l = 15 cm = 15 × 10^-2, Magnetic field B = 0.50 T
Resistance of the closed – loop containing the rod, R = 9mΩ ⇒ 9 × 10^-3Ω
Velocity of rod V = 12 cm/s = 12 × 10^-2 m/s,
a) The magnitude of the motional emf
e = BVl = 0.50 × 12 × 10^-2 × 15 × 10^-2, e = 9 × 10^-3 V.
According to the Fleming’s left hand rule, the direction of Lorentz force (f = -e(V × B)) on electrons in PQ is from P to Q. So, p would acquire positive charge and Q would acquire negative charge.

b) Yes, an excess positive charge developes at P and the same amount of negative charge developes at Q as the key is open. When the key K is closed, the induced current flows and maintains the excess charge.

c) When key is open, there is no net force on the electrons because the presence of excess charge at P and Q sets up an electric field and magnetic force on the electrons is balanced by force on them due to force by the electric field. So, there is no net force on the rod.

d) When the key K is closed, current flow in the loop and the current carrying wise experience a retarding force in the magnetic field which is given by
Force = BIl = B . \(\frac{\mathrm{e}}{\mathrm{R}}\) . l = \(\frac{0.5 \times 9 \times 10^{-3} \times 15 \times 10^{-12}}{9 \times 10^{-3}}\) = 7.5 × 10^-2 N.

e) To keep the rod moving at the same speed the required power
= retarding force × velocity = 7.5 × 10^-2 × 12 × 10^-2 = 9 × 10^-3 W

f) Power dissipated in closed circuit due to flow of current = I²R
= \(\left(\frac{\mathrm{e}}{\mathrm{R}}\right)^2\) × R = \(\frac{\left(9 \times 10^{-3}\right)^2}{9 \times 10^{-3}}\) = 9 × 10^-3 W₂ The source of this power is the external agent.

g) When the field is parallel to length of rails θ = 0°, induced emf = e = BVl sin θ = 0 (∵ sin θ° = 0). In this situation, the moving rod will not cut the field lines so that flux change is zero and hence induced emf is zero.

Question 5.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Solution:
Given, length of solenoid I = 30 cm = 30 × 10^-2 m
Area of cross-section A = 25 cm² = 25 × 10^-4 m²
Number of turns N = 500, Current I₁ = 2.5A, I₂ = 0, Brief time dt = 10^-3s.
Induced emf in the solenoid e = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = \(\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}\) (∵ ϕ = BA)
Magnetic field induction B at a point well inside the long solenoid carring current I is

Question 6.
Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in the figure.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, υ = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take α = 0.1 m and assume that the loop has a large resistance.

Solution:
a) Let us assume that an elementary strip of width dx a at distance x from the wire carring current I.
Side of square = a.
The magnetic field due to current carring wire at a distance x from the wire is
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{x}\) —— (i)
Small amount of magnetic flux associated with the strip

As we know that ϕ = MI —— (iii)
where, M is mutual inductance. From equations (ii) and (iii) we get

This is mutual inductance between wire and square loop.

b) Given current I = 50 A, Velocity V = 10 m/s, x = 0.2 m and a = 0.1 m

Question 7.
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (figure). A uniform magnetic field extends over a circular region within the rim. It is given by,
B = B₀k (r ≤ a, a < R)
= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?
Solution:
Given Linear charge density

Radius of rim = R,
Mass of rim = M,
Magnetic field extends over a circular region
B = -B₀ (r ≤ a, a < R) = O (otherwise).
Let the angular velocity of the wheel be w, then induced emf e = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
E exist along circumference of radius a due to change in magnetic flux.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter

Very Short Answer Questions

Question 1.
A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field ?
Answer:
The nature of the magnetic field’ is uniform, magnetic dipole (bar magnet) experiences a net force (or torque).

Question 2.
What happens to compass needles at the Earth’s poles ?
Answer:
At the poles, earths field is exactly vertical. So, the campass needles free to rotate in a horizontal plane only, it may point out in any direction.

Question 3.
What do you understand by the ‘magnetization’ of a sample ? (A.P. Mar. ’16)
Answer:
When a magnetic sample is placed in a magnetic field, their magnetic moments are add up in the direction of magnetic field. Hence the sample get a net magnetic moment (m_net ≠ 0)
Magnetisation is defined as the net magnetic moment per unit volume i.e., M = \(\frac{m_{\text {net }}}{V}\)

Question 4.
What is the magnetic moment associated with a solenoid?
Answer:
Magnetic dipole moment in a solinoid m = NIA
Where ‘N’ is the number of turns in the loop, ‘I’ the current and A the area vector.

Question 5.
What are the units of magnetic moment, magnetic induction and magnetic filed ?
Answer:
Unit of

1. magnetic moment m is Am² or J T^-1.
2. Magnetic induction – wb m^-2 or Tesla (T)
3. magnetic field – Tesla.

Question 6.
Magnetic lines form continuous closed loops. Why ? (A.P. Mar. 16)
Answer:
Magnetic lines of force always start from north pole and forming curved path, enter south pole and travel to north pole inside the magnet. Thus lines of force are forming closed loops.

Question 7.
Define magnetic declination. (Mar. ’14)
Answer:
Magnetic Declinatin (D) : The angle between the true geographic north and the north shown by a compass needle is called magnetic declination or simply declination (D).

Question 8.
Define magnetic inclination or angle of dip. (A.P. & T.S. Mar. ’15)
Answer:
Inclination or Dip (I) : The angle which the total intensity of earth’s magnetic field makes with the horizontal at any place is called inclination (I).

Question 9.
Classify the following materials with regard to magnetism : Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. (A.P. Mar. ’19 & T.S. Mar. ’16, ’15)
Answer:
Ferromagnetic materials → Cobalt, Nickel.
Paramagnetic materials → Oxygen, Manganese
Diamagnetic materials → Bismuth, Copper

Short Answer Questions

Question 1.
Derive an expression for the axial field of a solenoid of radius “a”, containing “n” turns per unit length and carrying current “i”.
Answer:
Expression for the axial field of a solenoid :

1. Consider a solinoid of length ‘2l’ and radius ‘a’ having ‘n’ turns per unit length.
2. Let ‘I’ be the current in the solenoid.
3. We have to calculate magnetic field at any point P on the axis of solenoid, where OP = r.
   
4. Consider a small element of thickness dx of the solenoid, at a distance ‘x’ from ‘O’.
5. Number of turns in the element = ndx.
6. Magnitude of magnetic field at P due to this current element is dB = \(\frac{\mu_0 \mathrm{ndx} \cdot \mathrm{Ia}^2}{2\left[(\mathrm{r}-\mathrm{x})^2+\mathrm{a}^2\right]^{3 / 2}}\)
7. If P lies at a very large distance from 0, i.e., r > > a and r > > x, then [(r – x)² + a²]^3/2 = r³.
   ⇒ dB = \(\frac{\mu_0 \mathrm{ndx} \mid \mathrm{a}^2}{2 \mathrm{r}^3}\)
8. To get total magnetic field, integrating the above equation between the limits from X = -l to X = +l.
   
9. The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa²).
   B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\)
10. Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 2.
The force between two magnet poles separated by a distance’d’ in air is ‘F. At what distance between them does the force become doubled ?
Answer:
Force between two magnetic poles, F₁ = F;
Distance between two magnetic poles, d₁ = d
Force between two magnetic poles increased by double F₂ = 2F
Distance between two magnetic poles, d₂ = ?
From Coulombs law, \(\mathrm{F}_1 \mathrm{~d}_1^2\) = \(\mathrm{F}_2 \mathrm{~d}_2^2\)
Fd² = 2 F \(\mathrm{d}_2^2\)
⇒ \(\mathrm{d}_2^2\) = \(\frac{\mathrm{d}^2}{2}\)
∴ d₂ = \(\frac{\mathrm{d}}{\sqrt{2}}\)

Question 3.
Compare the properties of para, dia and ferromagnetic substances.
Answer:
Diamagnetic substances

a) When these materials placed in a magnetic field, they are magnetised feebly in the opposite direction to the applied external field.
b) When a rod of diamagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the perpendicular direction to the magnetic field.
c) When they kept in a non uniform magnetic field, they moves from the region of greater field strength to the region of less field strength.
d) The relative permeability is less than 1. μ_r < 1 and negative.
e) The susceptibility \((\chi)\) is low and negative.
E.g. : Copper, Silver, Water, Gold, Antimony, Bismuth, Mercury Quartz Diamond etc.

Paramagnetic Substance

a) When these materials placed in a magnetic field, they are magnetised feebly in the direction of the applied magnetic field.
b) When a rod of paramagnetic material is suspended freely in a uniform magneticfield, it comes to rest in the direction of the applied magnetic field.
c) When they kept in a non-uniform magnetic field, they moves from the region of less field strength to the region of greater field strength.
d) The relative permeability is greater than 1. μ_r > 1 and positive.
e) The susceptibility \((\chi)\) is small and positive.
E.g. : Aluminium, Magnetiam, Tungsten, Platinum, Manganese, liquid oxygen, Ferric chloride, Cupric chloride etc.

Fêrromagnetic substances

a) When these materials placed in a magnetic field, they are magnetised strongly in the direction of the applied external field.
b) When a rod of ferromagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they kept in a non-uniform magnetic field they moves from the regions of lesser (magnetic field) strength to the regions of stronger (magnetic field) strength.
d) The relative permeability is much greater than μ_r >> 1 and positive.
e) The stisceptibility \((\chi)\) is high and positive.
E.g. : Iron, Cobalt, Nickel, Gadolinium and their alloys.

Question 4.
Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.
Answer:
The magnetic field of the earth at a point on its surface can be specified by the declination D, the angle of dip or the inclination I and the horizontal component of the earth’s field H_E. These are known as the elements of the earth’s magnetic field.

Explanation:

1. The total magnetic field at P can be resolved into a horizontal component H_E and a vertical component Z_E.
2. The angle that B_E makes with H_E is the angle of dip, I.
3. Representing the vertical component by Z_E, we have
   Z_E = B_E Sin I
   H_E = B_E Cos I
   Which gives Tan I = \(\frac{\mathrm{Z}_{\mathrm{E}}}{\mathrm{H}_{\mathrm{E}}}\)

Question 5.
Define retentivity and coercivity. Draw the hysteresis curve for soft iron and steel. What do you infer from these curves ?
Answer:

1. Retentivity : The value of magnetic induction \(\overrightarrow{\mathrm{B}}\) left in the specimen, when the magnetising force (H) is reduced to zero is called Retentivity or Remanence or Residual magnetism.
2. Coercivity : To reduce the retentivity to zero, we have to apply a magnetising force in opposite direction. This value of magnetising force is called coercivity or coercive force.
3. Hysterisis eurve: The curve represents the relation between magnetic induction \(\overrightarrow{\mathrm{B}}\) (or) intensity of magnetization (I) of a ferromagnetic material with magnetizing force (or) magnetic intensity (\(\overrightarrow{\mathrm{H}}\)) is called Hystersis curve.
   
4. Hysterisis curve for soft Iron and steel is shown below.
   The hysterisis loops of soft iron and steel reveal that
   1. The retentivity of soft iron is greater than the retentivity of steel.
   2. Soft Iron is more strongly magnetised than steel.
   3. Coercivity of Soft Iron is less than coercivity of steel. It means soft Iron loses its magnetisation more rapidly than steel does.
   4. As area of I – H loop for soft Iron is smaller than the area of I – H loop for steel. Therefore hysterisis loss in case of soft Iron is smaller than the hysterisis loss in case of steel.

Question 6.
If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns ?
Answer:
For first circular coil; B₁ = B, n₁ = 1; I₁ = I; a₁ = \(\frac{\mathrm{L}}{2 \pi}\)
For second circular coil, B₂ = ? n₂ = 10; I₂ = I; a₂ = \(\frac{\mathrm{L}}{2 \pi}\)
As B = \(\frac{\mu_0 \mathrm{n} I \mathrm{a}^2}{2 \mathrm{r}}\), B ∝ n
\(\frac{\mathrm{B}_2}{\mathrm{~B}_1}\) = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\)
\(\frac{\mathrm{B}_2}{\mathrm{~B}}\) = \(\frac{10}{1}\)
∴ B₂ = 10 B

Question 7.
If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic field at the axis of the solenoid change?
Answer:
B₁ = B (say); n₁ = n; n₂ = 2n; B₂ =?
Magnetic field at the centre of a solinoid is given by B = \(\frac{\mu_0 \mathrm{nI} \mathrm{a}^2(2 l)}{2 \mathrm{r}^3}\)
⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}_1}\) = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\) ⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}}\) = \(\frac{2 \mathrm{n}}{\mathrm{n}}\)
∴ B₂ = 2 B

Long Answer Questions

Question 1.
Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop.
Answer:
Expression for the magnetic field at a point on the axis of a current carrying circular loop:

1. Consider ‘O’ is the centre of a circular coil of one turn and radius ‘a’.
2. Let P is a point at a distance r from the centre, along the axis of coil.
3. The plane of the coil is ⊥^r to the plane of paper.
4. Consider two elements AB and AB’ each of length dl which are diametrically opposite.
5. Then, the magnetic fields at P due to these two elements will be dB and dB in the direction PM and PN respectively.
6. These directions are ⊥^r to the lines joining the mid-points of the elements with the point P
7. Resolve these fields into two components parallel (dB sin θ) and perpendicular (dB cos θ) to the axis of the coil.
8. The dB cos θ components cancel one another and dB sin θ components are in the same direction and add up due to the symmetric elements of the circular coil.
9. Therefore, the total magnetic field along the axis = B = \(\int \mathrm{dB}\) sin θ of the circular coil along PC —– (1)
10. The magnetic field at ‘P’ due to current element of length ‘dl’ is
    ‘dB’ = \(\frac{\mu_0}{4 \pi} \frac{I d l \sin \phi}{\left(a^2+x^2\right)}\) = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{a}^2+\mathrm{r}^2\right)}\) —– (II) [∵ φ = 90°]
11. From equation (I) and (II), B = \(\int \frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{a}^2+\mathrm{r}^2\right)} \sin \theta\)
    From Δ^le OPE, sin θ = \(\frac{a}{\sqrt{\left(a^2+r^2\right)}}\)
    ⇒ B = \(\int \frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \mathrm{a}}{\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}}\) = \(\frac{\mu_0 \mathrm{Ia}}{4 \pi\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}} \int \mathrm{d} l\)
    But \(\int \mathrm{d} l\) = Circumference of the coil = 2πa
    ∴ B = \(\frac{\mu_0 \text { I a }}{4 \pi\left(a^2+r^2\right)^{3 / 2}}\) × 2πa = \(\frac{\mu_0 \mathrm{Ia}^2}{2\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}}\), Tesla along the direction of PC.
12. If the coil contains N turns, then B = \(\frac{\mu_0 \text { NIa }^2}{2\left(a^2+r^2\right)^{3 / 2}}\)
13. At the centre of the coil r = 0, B = \(\frac{\mu_0 \mathrm{NI} \mathrm{a}^2}{2 \mathrm{a}^3}\) = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{a}}\)
    Note: If r >> a, B = \(\frac{\mu_0 \mathrm{NI}\left(\pi \mathrm{a}^2\right)}{2 \pi \mathrm{r}^3}\) = \(\frac{\mu_0 \mathrm{NI}_{\mathrm{A}}}{2 \pi^3}\)
    A = πa² = Area of current loop.

Question 2.
Prove that a bar magnet and a solenoid produce similar fields.
Answer:
Bar magnet produce similar field of Solenoid:

1. We know that the current loop acts as a magnetic dipole. According to Ampere’s all magnetic phenomena can be explained interms of circulating currents.
2. Cutting a bar magnet is like a solenoid. We get two similar solenoids with weaker magnetic properties.
3. The magnetic field lines remain continuous, emerging from one face of solenoid and entering into other face of solenoid.
4. If we were to move a small compass needle in the neighbourhood of a bar magnet and a current carrying solenoid, we would find that the deflections of the needle are similar in both cases as shown in diagrams.
   
   
   The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet –
5. The magnetic field at point P due to bar magnet in the form of solenoid is B = \(\frac{\mu_0}{4 \pi} \frac{2 m}{r^3}\)
6. The total magnetic field, at a point P due to solenoid is given by
7. The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa²).
   ∴ B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\)
8. Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 3.
A small magnetic needle is set into oscillations in a magnetic field B obtain an expression for the time period of oscillation.
Answer:
Expression for time period of oscillation :

1. A small compass needle (magnetic dipole) of known magnetic moment m and moment of Inertia is placing in uniform magnetic field B and allowing it to oscillate in the magnetic field.
2. This arrangement is shown in Figure.
3. The torque on the needle is \(\tau\) = m × B
4. In magnitude \(\tau\) = mB sin θ.
   Here \(\tau\) is restoring torque and θ is the angle between m and B.
5. Therefore, in equilibrium
   
   Negative sign with mB sin θ implies that restoring torque is in opposition to deflecting torque.
   
6. For small values of θ in radians, we approximate sin θ ≈ θ and get
   
   
   This represents a simple harmonic motion.
7. From defination of simple harmonic motion, we have \(\) = -ω²θ ——– (II)
   From equation (1) and (II), we get ⇒ ω² = \(\frac{\mathrm{mB}}{\mathcal{J}}\)
   
8. Therefore, the time period is
   

Question 4.
A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T₁ and T₂ at two places, where the angles of dip are θ₁ and θ₂ respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.
Answer:

1. Suppose, the resultant magnetic fields is to be compared at two places A and B.
2. A bar magnet, held horizontally at A and which is set into angular oscillatins in the Earths magnetic field.
3. Let time period of a bar magnet at place ‘A’ is T₁ and angular displacement or angle of dip is θ₁.
4. As the bar magnet is free to rotate horizontally, it does not remain vertical component (B₁sin θ₁) It can have only horizontal component (B₁ cos θ₁)
5. The time period of a bar magnet in uniform magnetic field is given by T = \(2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mB}_{\mathrm{H}}}}\)
6. Now, in this case T = T₁ and B_H = B₁cosθ₁
   
7. Therefore time period of a bar magnet at place ‘A’ is given by
   T₁ = \(2 \pi \sqrt{\frac{I}{m B_1 \cos \theta_1}}\) —– (1) Where I is moment of Inertia of a barmagnet and m is magnitude of magnetic moment.
8. Similarly, the same bar magnet is placed at B and which is set into angular oscillations in the earths magnetic field.
9. Let time period of a bar magnet at place B is T₂ and angle of dip is θ₂.
10. Since horizontal component of earths field at B is B_H = B₂ cos θ₂, time period,
    T₂ = \(2 \pi \sqrt{\frac{1}{\mathrm{mB}_2 \cos \theta_2}}\) —— (2)
11. Dividing equation (1) by equation (2), we get \(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\) = \(\sqrt{\frac{\mathrm{mB}_2 \cos \theta_2}{\mathrm{mB}_1 \cos \theta_1}}\)
    Squaring on both sides, we have \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}\) = \(\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}\)
12. But B₁ = μ₀H₁ and B₂ = μ₀H₂
    \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}\) = \(\frac{\mu_0 \mathrm{H}_2 \cos \theta_2}{\mu_0 \mathrm{H}_1 \cos \theta_1}\)
13. Therefore, \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = \(\frac{\mathrm{T}_2^2 \cos \theta_2}{\mathrm{~T}_1^2 \cos \theta_1}\)
    H2 T?cosOi .
14. By knowing T₁, T₂ and θ₁, θ₂ at different places A and B, we can find the ratio of resultant magnetic fields.

Question 5.
Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility. (A.P. Mar. ’15)
Answer:

1. SusceptIbility: When a material is placed in a magnetic field, the ratio of the intensity of magnetization acquired by it to the intensity of the applied magnetic field is called its susceptibility.
   
2. The susceptibility of a material represents its ability to get magnetism.
3. Susceptibility is a dimension less quantity.
4. Relation between μ_r and \(\chi\):
   • Suppose that material is placed in a magnetic field of intensity H. Let I be the intensity of magnetisation acquired by it.
   • Then the magnetic induction with in the material is
     B = μ₀H + μ₀I ⇒ \(\frac{\mathrm{B}}{\mathrm{H}}\) = μ₀[1 + \(\frac{\mathrm{I}}{\mathrm{H}}\)]
     ⇒ μ = μ₀[1 + \(\chi\)] ⇒ \(\frac{\mu}{\mu_0}\) = 1 + \(\chi\)
     μ_r = 1 + \(\chi\) [∵ μ_r = \(\frac{\mu}{\mu_0}\)]
5. Negative susceptibility (y) of diamagnetic elements4are Bismuth (-1.66 × 10^-5) and copper (-9.8 × 10^-6).
6. Positive susceptibility of Ferromagnetic elements are Aluminium (2.3 × 10^-5) and oxygen at STP (2.1 × 10^-6).
7. Large and positive susceptibility of Ferromagnetic elements are Cobalt and Nickel.

Question 6.
Obtain Gauss Law for magnetism and explain it. (T.S. Mar. ’19)
Answer:
Gauss law for Magnetism:

1. According to Gauss’s law for magnetism, the net magnetic flux (ϕ_B) through any closed surface is always zero.
2. The law implies that the no. of magnetic field lines leaving any closed surface is always equal to the number of magnetic field lines entering it.
3. Suppose a closed surface S is held in a uniform magnetic field B. Consider a small vector area element ΔS of this surface as shown in figure.
4. Magnetic flux through this area element is defined as
   Δϕ_B = B. ΔS. Then the net flux ϕ_B, is,
   
5. If the area elements are really small, we can rewrite this equation as
   
   
6. Comparing this equation with Gauss’s law of electrostatics i.e., electric flux through a closed surface S is given by
   
   Where q is the electric charge enclosed by the surface.
7. In an electric dipole were enclosed by the surface equal and opposite charges in the dipole add upto zero. Therefore, ϕ_E would be zero.
8. The fact that ϕ_B = 0 indicates that the simplest magnetic element is a dipole or current loop.
9. The isolated magnetic poles, called magnetic monopoles are not known to exist.
10. All magnetic phenomena can be explained interms of an arrangement of magnetic dipoles and /or current loops.
11. Thus corresponding to equation (II) of Gauss’s theorem in electrostatics, we can visualize equation (I) as
    ϕ_B = \(\int_{S_{,}} B . d S\) = µ₀ (m) + µ₀ (-m) = 0 where m is strength of N-pole and -m is strength of S -pole of same magnet.
12. The net magnetic flux through any closed surface is zero.

Question 7.
What do you understand by “hysteresis” ? How does this propetry influence the choice of materials used in different appliances where electromagnets are used ?
Answer:

1. Cycle of magnetisation : When a fen or magnetic specimen is slowly magnetised, the intensity of magnetisation varies with magnetic field through a cycle is called cycle of magnetisation.
2. Hysterisis : The lagging of intensity of magnetisation (I) and magnetic induction (B) behind magnetic field intensity (H) when a magnetic specimen is subjected to a cycle of magnetisation is called hysterisis.
3. Retentivity : The value of I for which H = 0 is called retentivity or residual magnetism.
4. Coercivity: The value of magnetising force required to reduce I is zero in reverse direction of H is called coercive force or coercivity.
5. Hysterisis curve : The curve represents the relation between B or I of a ferromagnetic material with magnetising force or magnetic intensity H is known as Hysterisis curve.
6. Explanation of hysterisis loop or curve :
   • In fig, a closed curve ABCDEFA in H – I plane, called hysterisis loop is shown in fig.
   • When ferromagnetic specimen is slowly magnetised, I increases with H.
   • Part OA of the curve shows that I increases with H.
     
   • At point A, the value I becomes constant is called saturation value.
   • At B, I has some value while H is zero.
   • In fig. BO represents retentivity and OC represents coercivity.
7. Uses : The properties of hysterisis curve, i.e., saturation, retentivity, coercivity and hysterisis loss help us to choose the material for specific purpose.
   1. Permanent magnets : A permanent magnet should have both large retentivity and large coercivity. Permanent magnets are used in galvanometers, voltmeres, ammeters, etc.
   2. An electromagnet core : The electromagnet core material should have maximum induction field B even with small fields H, low hysterisis loss and high initial permeability.
   3. Transformer cores, Dynamocore, Chokes, Telephone diaphragms: The core material should have high initial permeability, low hysterisis loss and high specific resistance to reduce eddy currents. Soft iron is the best suited material.

Problems

Question 1.
What is torque acting on a plane coil of “n” turns carrying a current “i” and having an area A, when placed in a constant magnetic field B ?
Answer:

For Rectangular LOOP PQRS :
Length PR = QS = l
breadth PQ = RS = b
Current = i
Magnetic field Induction = B
Angle made by normal to plane of coil with B = θ
Forces on conductor PR and SQ, F = Bil sin θ
Force on conductor PQ and RS, F = 0
Torque on a rectangular coil, T – F × ⊥^r distnace (b) ⇒ \(\tau\) – Bil sin θ (b)
∴ \(\tau\) = BiA sinθ[∵ A = l × b]
If the loop has n turns, then \(\tau\) = B in A sin θ.

Question 2.
A coil of 20 turns has an area of 800 mm² and carries a current of 0.5A. If it is placed in a magnetic field of intensity 0.3T with its plane parallel to the field, what is the torque that it experiences ?
Answer:
n = 20; A = 800 mm² = 800 × 10^-6 m²; i = 0.5A; B = 0.3T; θ = 0°.
When the plane parallel to the field,
T = B in A cos θ = 0.3 × 0.5 × 20 × 800 × 10^-6 × cos 0°
\(\tau\) = 2.4 × 10^-3 Nm

Question 3.
In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression the magnetic moment (μ) of the electron in a Hydrogen atom in terms of its angular momentum L.
Answer:
Consider an electron of charge e, moves with constant speed v in a circular orbit of radius ‘r’ in Hydrogen atom as shown in fig.
Time period of orbiting electron,


The current constitute by revolving electron in circular motion around a nucleus, I = \(\frac{\mathrm{e}}{\mathrm{T}}\).
orbital magnetic moment, μ = IA = I (πr²)
⇒ μ = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\left(\pi \mathrm{r}^2\right)=\frac{\mathrm{evr}}{2}\)
μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}(\mathrm{mvr})\) [∵ Multiplying and dividing with ‘m’ on right side]
∴ μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}} \mathrm{~L}\) where L = mvr = angular momentum.

Question 4.
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid ?
Answer:
l = 22.5 cm = 22.5 × 10^-2 m = \(\frac{45}{2}\) × 10^-2m
N = 900; I = 0.8A ; H = ?
H = \(\frac{\mathrm{NI}}{l}\) = \(\frac{900 \times 0.8}{\left(\frac{45}{2}\right) \times 10^{-2}}\)
H = \(\frac{900}{45}\) × 0.8 × 10² × 2
∴ H = 3200Am^-1

Question 5.
A bar magnet of length 0.1 m and with a magnetic moment of 5Am² is placed in a uniform amagnetic field n of intensity 0.4T, with its axis making an angle of 60° with the field. What is the torque on the magnet ? (Mar. ’14)
Answer:
Given, 2l = 0.1m; m = 5A – m²; B = OAT; θ = 60°.
Torque, T = mB sin θ = 5 × 0.4 × sin 60° = 2 × \(\frac{\sqrt{3}}{2}\)
∴ T = 1.732 N-m

Question 6.
If the Earth’s magnetic field at the equator is about 4 × 10⁵T, What is its approximate magnetic dipole moment ?
Answer:
Given, B_E = 4 × 10^-5 T; r = 6.4 × 10⁶m; m = ?
B_E = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^3}\)
4 × 10^-5 = \(\frac{4 \pi \times 10^{-7}}{4 \pi}\) × \(\frac{\mathrm{m}}{\left(6.4 \times 10^6\right)^3}\)
m = 4 × 10² × (6.4 × 10⁶)³
∴ m = 1.05 × 10²³ Am² ≈ 1 × 10²³ Am²

Question 7.
The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10^-5T and the angle of dip is 60°. What is the magnetic field of the earth at this location ?
Answer:
Given H_E = 2.6 × 10^-5T;
D (or) δ = 60°
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos \mathrm{D}}\) = \(\frac{2.6 \times 10^{-5}}{\cos 60^{\circ}}\) = \(\frac{2.6 \times 10^{-5}}{(1 / 2)}\) = 5.2 × 10^-5T
∴ B_E = 5.2 × 10^-5T

Question 8.
A solenoid, of insulated wire, is wound on a core with relative permeability 400. If the number of turns per metre is 1000 and the solenoid carries a current of 2A, calculate H, B and the magnetisation M.
Answer:
Given, µ_r = 400, I = 2A, n = 1000
H = nI = 1000 × 2 = 2 × 10³ A/m
B = µ_rµ₀ H = 400 × 4π × 10^-7 × 2 × 10³ = 1.0 T Magnetisation M = (µ_r – 1) H = (400 – 1)H = 399 × 2 × 10³
∴ M \(\simeq\) 8 × 10⁵ A/m

Textual Exercises

Question 1.
Answer the following questions regarding earth’s magnetism :
a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
Answer:
The three independent quantities used to specify earth’s magnetic field are, Magnetic declination (θ), Magnetic dip (δ) and Horizontal component of earth’s field (H).

b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain ?
Answer:
Yes, we expect greater dip angle in Britian, because it is located close to North pole; δ = 70° in Britain.

c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground ?
Answer:
As Melbourne is situated in Southern hemisphere where north pole of earth’s magnetic field lies, therefore, magnetic lines of foece seem to come out of the ground.

d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north ro south pole ?
Answer:
At the poles, earth’s field is exactly vertical. As the compass needle is free to rotate in a horizontal plane only, it may point out in any direction.

e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² J T^-1 located at its centre. Check the order of magnitude of this number in some way.
Answer:
Here, M = 8 × 10²² J T^-1.
Let us calculate magnetic field intensity at magnetic line of short mangetic dipole for which, d = R = radius of earth 6,400 km = 6.4 × 10⁶ m.
B = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}^3}\) = 10^-7 × \(\frac{8 \times 10^{22}}{\left(6.4 \times 10^6\right)^3}\) = 0.31 × 10^-4T.
= 0.31 gauss
This value is in good approximation with observed values of earth’s magnetic field.

f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all ?
Answer:
The earth’s magnetic field is only approximately a dipole field. Therefore, local N-S poles may exist oriented in different directions. This is possible due to deposits of magnetised minerals.

Question 2.
Answer the following questions :
a) The earth’s magnetic field varies from point to point in space. Does it also change with time ? If so, on what time scale does it change appreciably ?
Answer:
Yes, earth’s field undergoes a change with time. For example, daily changes, annual changes secular changes with period of the order of 960 years and irregular changes like magnetic storms. Time scale for appreciable change is roughly a few hundred years.

b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why ?
Answer:
The earth’s core does contain iron but in the molten form only. This is not ferromagnetic and hence it cannot be treated as a source of earth’s magnetism.

c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents ?
Answer:
One of the possibilities is radioactivity in the interior of the earth. But it is not certain.

d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past ?
Answer:
Earth’s magnetic field gets recorded weakly in certain rocks during their solidification. An analysis of these rocks may reveal the history of earth’s magnetism.

e) The earth’s field departs from its dipole shape substantially at large, distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
Answer:
The earth’s magnetic field gets modified by the field produced by motion of ions in earth’s ionosphere.

f) Interstellar space has an extremely weak magnetic field of the order of 10^-12 T. Can such a weak field be of any significant consequence? Explain.
[Note : Exercise 2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.)
Answer:
When a charged particle moves in a magnetic field, it is deflected along a circular path such that BeV = \(\frac{\mathrm{mV}^2}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{mV}}{\mathrm{Be}}\)
When B is low, r is high i.e., radius of curvature of path is very large. Therefore, over the gigantic inter stellar distance, the deflection of charged particles becomes less noticeable.

Question 3.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a troque of magnitude equal to 4.5 × 10^-2 J. What is the magnitude of magnetic moment of the magnet?
Answer:
Here θ = 30°, B = 0.25 T, r = 4.5 × 10^-2 J, M =?
As r = mB sin θ ∴ m = \(\frac{r}{B \sin \theta}\) = \(\frac{4.5 \times 10^{-2}}{0.25 \sin 30^{\circ}}\) = 0.36JT^-1

Question 4.
A short bar magnet of magnetic moment m = 0.32 JT^-1 is placed in a uniform magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium ? What is the potential energy of the magnet in each case?
Answer:
Here m = 0.32JT^-1, B = 0.15T

1. In stable equilibrium, the bar magnet is aligned along the magnetic field, i.e., θ = 0°.
   Potentiâl Energy = -mB cos θ° = -032 × 0.15 × 0 = -4.8 × 10^-2J.
2. In unstable equilibrium the magnet is so oriented that magnetic moment is at 180° to the magnetic field i.e., θ = 180°.
   Potential Energy = -mB cos 180° = – 0.32 × 0.15 (-1) = 4.8 × 10^-2 J.

Question 5.
A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^-4 m² carries a
current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
Answer:
Here n = 800, a = 2.5 × 10^-4 m², I = 3.0 A
A magnetic field develop along the axis of the solenoid. Therefore the current carrying solenoid behaves like a bar magnet m = N IA = 800 × 3.0 × 2.5 × 10^-4
= 0.6 JT^-1 along the axis of solenoid.

Question 6.
If the solenoid in Exercise 5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field ?
Answer:
Here M = 0.6 JT^-1 (from Question 5)
B = 0.25 T r = ? q = 30°
As r = m B sin θ’ ∴ r = 0.6 × 0.25 sin 30° = 0.075 N.m.

Question 7.
A bar magnet of magnetic moment 1.5 JT^-1 lies aligned with the direction of a uniform magnetic field of 0.22T.
a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:
(i) normal to the field direction,
(ii) opposite to the field direction ?
Answer:
Here m = 1.5 JT^-1, B = 0.22 T, W = ?

a) Here θ₁ = 0° (along the field)
θ₂ = 90° (⊥ to the field)
As W = -mB (cos θ₂ – cos θ₁)
W =-1.5 × 0.22 (cos 90° – cos 0°) = -0.33 (0 – 1)
= 0.33 f

ii) Here θ₁ = 0°, θ₂ = 180°
W = -1.5 × 0.22 (cos 180° – cos 0°)
= – 0.33 (-1 – 1) = 0.66 J.

b) What is the torque on the magnet in cases (i) and (ii) ?
Answer:
Torque r = mB sin θ
i) Here θ = 90°, r = 1.5 × 0.22 sin 90° = 0.33 Nm
ii) Here θ = 180°, r = 1.5 × 0.22 sin 180° = 0

Question 8.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
a) What is the magnetic moment associated with the solenoid ?
Answer:
N = 2000, A = 1.6 × 10^-4 m², I = 4 amp, M = ?
As m = NIA
∴ M = 2000 × 4 × 1.6 × 10^-4 = 1.28 JT^-1

b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 T is set up at an angle of 30° with the axis of the solenoid ?
Answer:
Net force on the solenoid = 0
Torque, r = m B sin θ = 1.28 × 7.5 × 10^-2 sin 30°
= 1.28 × 7.5 × 10^-2 × \(\frac{1}{2}\)
r = 4.8 × 10^-2 Nm.

Question 9.
A circular coil of 76 turns and radius 10 cm. carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 8.0 × 10^-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^-1. What is the moment of inertia of the coil about its axis of rotation.
Answer:
Here n = 16, r = 10 cm = 0.1 m, I = 0.75A, B = 5.0 × 10^-2T
v = 2.0 s^-1, I = ?

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:
Here θ = 22° 1 + 0.35 G, R = ?
As H = R cos θ
R = H/cos θ = \(\frac{0.35}{\cos 22^{\circ}}\) = \(\frac{0.35}{0.9272}\) = 0.38 G

Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Here declination θ = 12° west, dip θ = 60°
H = 0.16 gauss = 0.16 × 10^-4 tesla. R = ?
As H = R cos θ

R = \(\frac{\mathrm{H}}{\cos \theta}\) = \(\frac{0.16 \times 10^{-4}}{\cos 60^{\circ}}\)
R = \(\frac{0.16 \times 10^{-4}}{1 / 2}\) = 0.32 × 10^-4T
The earth’s field lies in a vertical plane 12° west of geographic meridian at an angle of 60° above the horizontal.

Question 12.
A short bar magnet has a magnetic moment of 0.48 J T^-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equitorial lines (normal bisector) of the magnet.
Answer:
Here M = 0.48 J T^-1, B = ?
d = 10 cm = 0.1 m

a) On the axis of the magnet
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3}\) = 10^-7 × \(\frac{2 \times 0.48}{(0.1)^3}\)
= 0.96 × 10^T along S – N direction

b) On the equitorial line of the magnet
B = \(\frac{\mu_{\mathrm{0}}}{4 \pi} \times \frac{\mathrm{M}}{\mathrm{d}^3}\) = 10^-7 × \(\frac{0.48}{(0.1)^3}\) = 0.48 × 10^-4T, along N – S direction.

Question 13.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point- (i.e., 14 cm) from the centre of the magnet ? (At null distants, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
As null points are on the axis of the magnet, therefore
B₁ = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^3}\) = H
On the equitorial line of magnet at same distance (d), field due to the magnet is
B₂ = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}^3}\) = \(\frac{B_1}{2}\) = \(\frac{\mathrm{H}}{2}\)

∴ Total Magnetic field at this point on equitorial line is
B = B₂ + H = H + \(\frac{\mathrm{H}}{2}\) = \(\frac{3}{2} \mathrm{H}\)
B = \(\frac{3}{2}\) × 0.36 = 0.54G

Question 14.
If the bar magnet in Exercise 13 is turned around by 180°, where will the new null points be located ?
Answer:
When the bar magnet is turned through 180°, neutral points would lie on equitorial line, so that
B₂ = \(\frac{\mu_0}{4 \pi} \frac{M}{d_2^3}\) = H
On the equitorial line of magnet at same direction (d), field due to the magnet is
B₂ = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}_2^3}\) = H
In the previous question
B₁ = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}_1^3}\)
\(\mathrm{d}_2^3\) = \(\frac{d_1^3}{2}\) = \(\frac{(14)^3}{2}\)
d₂ = \(\frac{14}{2^{1 / 3}}\) = 11.1 cm

Question 15.
A short bar magnet of magnetic movement 5.25 × 10^-2 J T^-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Here M = 5.25 × 10^-2 J T^-1 r = ?

Earth’s field \(\overrightarrow{\mathrm{B}}_{\mathrm{e}}\) = 0.42 G = 0.42 × 10^-4 T
a) At a point P distant r on normal bisector, fig, field due to the magnet is
\(\overrightarrow{\mathrm{B}}_2\) = \(\frac{\mu_0}{4 \pi} \frac{M}{r^3}\) along PAI/NS.
The resultant field \(\overrightarrow{\mathrm{R}}\) will be inclined at 45° to the earth’s field along PQ only when
|\(\vec{B}_2\)| = |\(\overrightarrow{\mathrm{B}}_{\mathrm{e}}\)|
\(\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3}\) = 0.42 × 10^-4
which gives, r = 0.05 m = 5 cm

b) When the point P lies on axis of the magnet such that OP = r, field due to magnet [fig.] is

\(\overrightarrow{\mathrm{B}}_1\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{r}^3}\) along PO, Earth’s field \(\overrightarrow{\mathrm{B}}_{\mathrm{e}}\) is along \(\overrightarrow{\mathrm{PA}}\). The resultant field \(\overrightarrow{\mathrm{R}}\) will be inclined at 45° to Earth’s field [Figure.] only when
|\(\overrightarrow{\mathrm{B}}_1\)| = |\(\overrightarrow{\mathrm{B}_{\mathrm{e}}}\)|
\(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{r}^3}\) = 0.42 × 10^-4 which gives
r = 6.3 × 10^-2 m = 6.3 cm

Additional Exercises

Question 1.
Answer the following Questions :
a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled ?
Answer:
This is because at lower temperatures, the tendency to disrupt the alignment of dipoles (due to magnetising field) decreases on account of reduced random thermal motion.

b) Why is diamagnetism, in contrast, almost independent of temperature ?
Answer:
In a diamagnetic sample, each molecule is not a magnetic dipole in itself. Therefore, random thermal motion of molecules does not affect the magnetism of the specimen. This is why diamagnetism is independent of temperature.

c) If a toroid uses bismuth for its core,’ will the field in the core be (slightly) greater or (slightly) less than when the core is empty ?
Answer:
As bismuth is diamagnetic, therefore, the field in the core will be slightly less than when the core is empty.

d) Is the permeability of a ferromagnetic material independent of the magnetic field ? If not, is it more for lower or higher fields ?
Answer:
No, permeability of a ferromagnetic material is not independent of magnetic field. As is clear from the hysteresis curve, μ is greater for lower fields.

e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why ?
Answer:
Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. The proof of this important fact is based on the boundary conditions of magnetic fields (B and H) at the interface of two media.

f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?
Answer:
The magnetic permeability of a ferromagnetic material μ > > 1. That is why the field lines meet this medium normally.

Question 2.
Answer the following questions :
a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
Answer:
Since, in a ferromagnetic substance the magnetic properties are due to alignment of domains, therefore on with drawing the magnetising field the original domain formation does not take place.

b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy ?
Answer:
Carbon steel piece, because heat lost per cycle is proportional to the area of the hysteresis loop.

c) A system displaying a hysterisis loop such as a ferromagnet, is a device for storing memory ?’ Explain the meaning of this statement.
Answer:
Magnetisation of a ferromagnet is not a single – valued function of the magnetizing field. Its value for a particular field depends both on the field and also on the history of magnetisatiop. In other words, the value of magnetisation is a record or ‘memory’ of its cycle of magnetisation. If information bits can be made to correspond these cycles, the system displaying such a hysterisis loop can act as a device for storing information.

d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer ?
Answer:
Ceramics (specially treated barium iron oxides) also called ferrites.

e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer:
Surround the region by soft iron rings. Magnetic field lines will be drawn into the rings, and the enclosed space will be free of magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field.

Question 3.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable) ? (At neutral points, magnetic field due to a current – carrying cable is equal and opposite to the horizontal component of earth’s magnetic field).
Answer:
Here i = 2.5 amp
R = 0.33G = 0.33 × 10^-4 T; θ = 0°
Horizontal component of earth’s field
H = R cos θ = 0.39 × 10^-4 cos 35°
= 0.39 × 10^-4 × 0.8192
= 3.9 × 10^-5 tesla.
Vertical component of earth’s field.
H = R cos θ = 0.33 × 10^-4 cos 0°
= 0.33 × 10^-4 tesla.
Let the neutral points lie at a distance r from the cable
Strength of magnetic field on this line due to current in the cable = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

At neutral point,
\(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) = H
r = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{H}}\) = \(\frac{4 \pi \times 10^{-7} \times 2.5}{2 \pi \times 0.33 \times 10^{-4}}\) = 1.5 × 10^-2m
Hence neutral points lie on a straigt line parallel to the cable at a perpendicular distance of 1.5 cm above te plane of the paper.

Question 4.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable ?
Answer:
There, no. of wires, n = 4, i = 1.0 amp
Earth’s field R = 0.39 G = 0.39 × 10^-4 T
dip, θ = 35 declination θ = 0°
R₁ = ?, R₂ = ?

r = 4 cm each 4 × 10^-2 m
Magnetic field at 4 cm due to currents in 4 wires
B = 4 × \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) = 4 × \(\frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 4 \times 10^{-2}}\)
Horizontal component of earth’s field
H = R cos θ = 0.39 × 10^-4 cos 35°
= 0.39 × 10^-4 × 0.8192 = 3.19 × 10^-5tesla
Vertical component of Earth’s field
V = R sin θ = 0.39 × 10^-4 sin 35°
= 0.39 × 10^-4 × 0.5736 = 2.2 × 10^-5 tesla.
At point Q, 4 cm below the wire, horizontal component due to Earth’s field and field due to current are in opposite directions (fig.)
H₁ = H – B
∴ H₁ = 3.19 × 10^-5 – 2 × 10^-5
= 1.19 × 10^-5 tesla.
Hence R₁ = \(\sqrt{\mathrm{H}_1^2+\mathrm{V}^2}\)
= \(\sqrt{\left(1.19 \times 10^{-5}\right)^2+\left(2.2 \times 10^{-5}\right)^2}\)
= 2.5 × 10^-5 tesla.
At point P, 4 cm above the wire, horizontal component of Earth’s field and field due to current are in the same direction [fig.]
H₂ = H + B = 3.19 × 10^-5 + 2 × 10^-5 = 5.19 × 10^-5 T
R₂ = \(\sqrt{\mathrm{H}_2^2+\mathrm{V}^2}\) = \(\sqrt{\left(5.19 \times 10^{-5}\right)^2+\left(2.2+10^{-5}\right)^2}\) = 5.54 × 10^-5 tesla.

Question 5.
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
a) Determine the horizontal component of the earth’s magnetic field at the location.
b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer:
a) Here n = 30, r = 12 cm = 12 × 10^-2 m
i = 0.35 amp, H = ?
As is clear from fig. 11 the needle can point west to east only when H = B sin 45°
Where B = Magnetic field strength due to current in coil = \(\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{ni}}{\mathrm{r}}\)

b) When current in coil is reversed and coil is turned through 90° anticlock wise, the direction of needle will reverse (i.e., it will point from East to West). This follow from the figure.

Question 6.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field ?
Answer:
Here θ = 60°; B₁ = 1.2 × 10^-2 tesla
θ₁ = 15°; θ₂ = 60° – 15° = 45°

In equilibrium, torque due to two fields must balance i.e., r₁ = r₂
MB₁ sin θ₁ = MB₂ sin θ₂
B₂ = \(\frac{B_1 \sin \theta_1}{\sin \theta_2}\) = \(\frac{1.2 \times 10^{-2} \sin 15^{\circ}}{\sin 45^{\circ}}\)
B₂ = \(\frac{1.2 \times 10^{-2} \times 0.2588}{0.7071}\) = 4.4 × 10^-3 tesla.

Question 7.
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e = 9.11 × 10^-31 kg).
[Note : Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field bn the motion of the electron beam from the electron gun to the screen in a TV set] .
Answer:
Here energy E = 18 KeV = 18 × 1.6 × 10^-19J
B = 0.40 G = 0.40 × 10^-4 T
x = 30 cm = 0.3 m
As E = \(\frac{1}{2}\) mυ² ∴ υ = \(\sqrt{\frac{2 E}{m}}\)
In a magnetic field electron beam is deflected along a circular arc of radius r, such that

BeV = \(\frac{m v^2}{r}\) or r = \(\frac{\mathrm{mv}}{\mathrm{Be}}\)
r = \(\frac{m}{B e} \sqrt{\frac{2 E}{m}}\) = \(\frac{1}{\mathrm{Be}} \sqrt{2 \mathrm{Em}}\) = 11.3 m
If y is the deflection at the end of the path it is clear from fig.
θ = \(\frac{\mathrm{x}}{\mathrm{r}}\) = \(\frac{\mathrm{y}}{\mathrm{x} / 2}\) = 2\(\frac{y}{x}\)
or y = \(\frac{x^2}{2 r}\) = \(\frac{0.30 \times(0.30)}{2 \times 11.3}\)m = 0.004m = 4mm

Question 8.
A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 10^-23 J T^-1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie’s law)
Answer:
Here no. of dipoles n = 2 × 10²⁴
Magnetic moment of each dipole M = 1.5 × 10^-23 J T^-1.
Total dipole moment of sample = n × M = 2 × 10²⁴ × 1.5 × 10^-23 = 30
As saturation achieved is 15% therefore, effective dipole moment
M₁ = \(\frac{15}{100}\) × 30 = 4.5 J T^-1; B₁ = 0.64T, T₁ = 4.2 k,
M₂ = ?; B₂ = 0.98 T,T₂ = 2.8 k

Question 9.
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Answer:
Here r = 15 cm = 15 × 10^-2 m, N = 3500, M_r = 800
I = 1.2A, B =?
Number of turns, length, n = \(\frac{N}{2 \pi r}\) = \(\frac{3500}{2 \pi \times 15 \times 10^{-2}}\)
B = μ₀μ_rnI
= 4π × 10^-7 × 800 × \(\frac{3500 \times 1.2}{2 \pi \times 15 \times 10^{-2}}\)
= 4.48T

Question 10.
The magnetic moment vectors μ_s and μ_l associated with the intrinsic spin angular momentum S and orbital angular momentum 1, respectively, of an electron are predicted
by quantum theory (and verified experimentally to a high accuracy) to be given by:
m_s = -(e/m) S,
μ_l = -(e/2m)l
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:
Out of the two relations given only one is in accordance with classical physics. This is
\(\vec{\mu}_1\) = \(-\left(\frac{\mathrm{e}}{2 \mathrm{~m}}\right) \vec{l}\)
It follows from the definitions of μ₁ and l.
μ = iA = \(\left(\frac{-\mathrm{e}}{\mathrm{T}}\right) \pi \mathrm{r}^2\)
l = mvr = \(\mathrm{m}\left(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\right) \mathrm{r}\)
Where r is the radius of the circular orbit, which the electron of mass m and charge (-e) completes in time T.
Divide (i) by (ii) \(\frac{\mu_1}{l}\) = \(\frac{-\mathrm{e}}{\mathrm{T}} \pi \mathrm{r}^2 \times \frac{\mathrm{T}}{\mathrm{m}^2 \pi \mathrm{r}^2}=\frac{-\mathrm{e}}{2 \mathrm{~m}}\)
\(\vec{\mu}_1\) = \(\left(\frac{-\mathrm{e}}{2 \mathrm{~m}}\right) \vec{l}\)
‘Clearly \(\vec{\mu}_1\) and \(\vec{l}\) will be antiparallel (both being normal to the plane of the orbit)
In contrast \(\frac{\mu_{\mathrm{S}}}{\mathrm{S}}\) = \(\frac{\mathrm{e}}{\mathrm{m}}\). It is obtained on the basis of quantum mechanics.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism

Very Short Answer Questions

Question 1.
What is the importance of Oersted’s experiment?
Answer:
The importance of Oersted’s experiment is every current-carrying conductor produces a magnetic field around it which is perpendicular to the current-carrying conductor.

Question 2.
State Ampere’s law and Biot-Savart’s law. (T.S. Mar. ’19)
Answer:
Ampere’s law: The line integral of the intensity of magnetic induction around a closed path is equal to µ₀ times the total current enclosed in it.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\mu_0 \mathrm{i}\)

Biot – Savart’s laws: Biot – Savart’s states that the intensity of magnetic induction (dB) due to a small element is directly proportional to the

1. current (i)
2. length of the element (dl)
3. sine angle between radius vector (r) and dl and
4. Inversely proportional to the square of the point from current element.
   ∴ dB ∝ \(\frac{\mathrm{i} \mathrm{d} l \sin \theta}{\mathrm{r}^2}\)
   dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} d l \sin \theta}{r^2}\)

Question 3.
Write the expression for the magnetic induction at any point on the axis of a circular current-carrying coil. Hence, obtain an expression for the magnetic induction at the centre of the circular coil.
Answer:

1. Intensity of magnetic induction field on the axis of the circular coil B = \(\frac{\mu_0 n i r^2}{2\left(r^2+x^2\right)^{3 / 2}}\)
2. At the centre of the coil B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 4.
A circular coil of radius V having N turns carries a current “i”. What is its magnetic moment ?
Answer:
Magnetic moment (M) = N i A
M = N i (πr²) (∵ A = πr²)
∴ M = n N i r²

Question 5.
What is the force on a conductor of length L carrying a current “i” placed in a magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a conductor (F) = B i L sin θ
2. If θ = 90°, F_Max = B i L
   i.e., the direction of current and magnetic field are perpendicular to each other, then force is maximum.

Question 6.
What is the force on a charged particle of charge “q” moving with a velocity “v” in a uniform magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a charged particle (F) = B q v sin θ.
2. If θ = 90°, F_Max = B q v.

Question 7.
Distinguish between ammeter and voltmeter. (A.P. Mar. ’15)
Answer:
Ammeter

1. It is used to measure current.
2. Resistance of an ideal Ammeter is zero.
3. It is connected in series in the circuits.

Voltmeter

1. It is used to measure P.D between two points.
2. Resistance of ideal voltmeter is infinity.
3. It is connected in parallel in the circuits.

Question 8.
What is the principle of a moving coil galvanometer ?
Answer:
Moving coil galvanometer is based on the fact that when a current carrying coil is placed in a uniform magnetic field, it experiences a torque.
∴ current in the coil (i) ∝ deflecting angle (θ).

Question 9.
What is the smallest value of current that can be measured with a moving coil galvanometer ?
Answer:
Moving coil galvanometer is sensitive galvanometer, it is used to measure very small current upto 10^-9A.

Question 10.
How do you convert a moving coil galvanometer into an ammeter ? (A.P. Mar. ’19)
Answer:
A small resistance is connected in parallel to the moving coil galvanometer, then it converts to Ammeter.
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)

Question 11.
How do you convert a moving coil galvanometer into a voltmeter? (T.S. & A.P. Mar. ’16, T.S. Mar. ’15 Mar. ’14)
Answer:
A high resistance is connected in series to the moving coil galvanometer, then it converts to voltmeter.
R = \(\frac{\mathrm{v}}{\mathrm{i}_{\mathrm{g}}}-\mathrm{G}\)

Question 12.
What is the relation between the permittivity of free space e₀, the permeability of free space m₀ and the speed of light in vacuum ?
Answer:
Speed of light in vacuum (C) = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Here μ₀ = m₀ = permeability in vacuum
ε₀ = permittivity in vacuum.

Question 13.
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns about the vertical axis ?
Answer:
Torque \((\tau)\) = \(\overrightarrow{\mathrm{M}}\) × \(\overrightarrow{\mathrm{B}}\) = i\(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) (M = n i A)
where i is current, \(\overrightarrow{\mathrm{A}}\) is area vector, \(\overrightarrow{\mathrm{B}}\) is magnetic field. Area vector \(\overrightarrow{\mathrm{A}}\) acts normal to the loop, so torque \(\vec{\tau}\) cannot act along the vertical axis. The magnetic field is not set up to turn the loop around it self.

Question 14.
A current carrying circular loop is placed in a uniform external magnetic field. If the loop is free to turn, what is its orientation when it is achieves stable equilibrium ?
Answer:
The plane of the loop is perpendicular to the direction of magnetic field because the torque on the loop in this orientation is zero.

Question 15.
A wire loop of Irregular shape carrying current is placed In an external magnetic field. If the wire is flexible, what shape will the loop change to ? Why?
Answer:
For a given perimeter, a circle has maximum area among all geometrical shapes. So to maximise tie magnetic flux through it will assume a circular shape with its plane normal to the field.

Short Answer Questions

Question 1.
State and explain Biot-Savart law. (T.S. Mar. ’16, Mar. ’14)
Answer:
Consider a very small element of length dl of a conductor carrying current

Magnetic induction due to small element at a point P distance r form the element. Magnetic induction (dB) is directly proportional to

1. 1. current (i)
   2. Length of the element (dl)
   3.  sine angle between r and dl and
   4. Inversely proportional to the square of the distance from small element to point R
      
      dB ∝ \(\frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
      dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
      where μ = permeability in free space.
      \(\frac{\mu_0}{4 \pi}\) = 10^-7 Wb m^-1 A^-1

Question 2.
State and explain Ampere’s law.
Answer:
Ampere’s law : The line integral of the intensity of magnetic induction field around closed path is equal to μ₀ times the net current (i) enclosed by the path.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i
Proof: Consider a long” straight conductor carrying current i as shown in figure. Magnetic induction at a distance r from the conductor is given by

B = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) (from Biot-Savart’s law)
The value of B is same at all points on the circle.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\oint \mathrm{Bd} l \cos \theta\)
= \(\mathrm{B} \oint \mathrm{d} l\) = B × 2πr (∵ θ = 0° Angle between \(\overrightarrow{\mathrm{B}}\) & \(\overrightarrow{\mathrm{d} l}\) is zero)
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) × 2πr
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i.
This proves Ampere’s circuital laws.

Question 3.
Find the magnetic induction due to a long current carrying conductor.
Answer:
Consider a long straight conductor carrying a current i. Let P be a point at a distance r from the conductor. Let r be the radius of the circle passing through point p.

(∵ \(\oint \mathrm{d} l\) = 2πr = circumference of the circle)
Magnetic induction is same at all points on the circle. Consider a small element of length dl.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\oint \mathrm{Bd} l \cos \theta\)
Angle between B and dl is zero i.e. θ = 0
= B\(\oint \mathrm{dl}\)
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = B (2πr) ——> (1)
According to Ampere’s laws
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = µ₀i ——-> (2)
From equations (1) and (2), B (2πr) = µ₀ i
B = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

Question 4.
Derive an expression for the magnetic induction at the centre of a current carrying circular coil using Biot-Savart law.
Answer:
Consider a circular coil of radius r and carry a current i. Consider a small element ‘dl’. Let O is the centre of the coil. By using Biot – Savart’s law
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} d l \sin \theta}{\mathrm{r}^2}\)
Here angle \(\overrightarrow{\mathrm{d} t}\) and \(\overrightarrow{\mathrm{r}}\) is 90° (i.e. θ = 90°)
dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{id} l}{\mathrm{r}^2}\) — (1)

As the field due to all elements of the circular loop have the same direction.
The resultant magnetic field can be obtained by integrating equation (1)
\(\int \mathrm{dB}\) = \(\int \frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\)
B = \(\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}^2} \int \mathrm{d} l\) ∵ (\(\int\)dl = 2πr)
= \(\frac{\mu 0 \mathrm{i}}{4 \pi \mathrm{r}^2}\) × 2πr
B = \(\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}}\)
If the circular coil has n turns.
B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 5.
Derive an expression for the magnetic induction a point on the axis of a current carrying circular coil using
Answer:

Consider a circular coil of radius R and carrying a current i. Let P is a point on the axis at a distance x from the centre O. Let r be the distance of small element (dl) from P.
From Biot – savart’s law
dB = \(\frac{\mu_0}{4 \pi} \frac{\text { id } l \sin \theta}{r^2}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\) —– (1) (∵ θ = 90° Angle between \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\))
dB can be resolved into two components dB cos θ and dB sinθ. If we consider another element diametrically opposite to AB.
This also resolved into dB cosθ and dB sinθ.
The components along the axis will add up and perpendicular to the axis will cancel.
∴ Resultant magnetic induction at P is
B = \(\int \frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
= \(\frac{\mu_0 i}{4 \pi^2} \int d / \sin \theta\) (∵ sin θ = \(\frac{R}{r}\))
= \(\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}^2}\) × 2πR × \(\frac{\mathrm{R}}{\mathrm{r}}\) (∵ \(\int dl\) = 2πr)
B = \(\frac{\mu_0 \mathrm{iR}^2}{2 \mathrm{r}^3}\) —–> (3)
From figure r = \(\sqrt{\mathrm{R}^2+\mathrm{x}^2}\)
B = \(\frac{\mu_0 i R^2}{2\left(R^2+x^2\right)^{3 / 2}}\) —–> (4)
If the coil contains N turns,
B = \(\frac{\mu_0 N i R^2}{2\left(R^2+x^2\right)^{3 / 2}}\) —–> (5)

Question 6.
Obtain an expression for the magnetic dipole moment of current loop.
Answer:
We know that magnetic induction on the axial line of a circular coil is B = \(\frac{\mu_0 N \mathrm{iR}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}\)
where N = Number of turns in the coil
R = Radius of the coil
x = Distance from centre of the coil
i = Current in a coil
If x >> R, Then B = \(\frac{\mu_0 \mathrm{Ni} \mathrm{R}^2}{2 \mathrm{x}^3}\)
Multiplying and dividing with 2π
B = \(\frac{\mu_0 \mathrm{Ni} \mathrm{R}^2}{2 \mathrm{x}^3} \times \frac{2 \pi}{2 \pi}\)
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{Ni}\left(\pi \mathrm{R}^2\right)}{\mathrm{x}^3}\) (∵ A = πR²)
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{Ni} \mathrm{A}}{\mathrm{x}^3}\) —– (1)
We know that magnetic induction field on the axial line of a bar magnet
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{x}^3}\) —– (2)
Comparing the equations (1) and (2)
Magnetic moment (M) = N i A

Question 7.
Derive an expression for the magnetic dipole moment of a revolving electron. (Ã.P. Mar. ’16)
Answer:
Consider an electron revolving in a circular orbit of radius r with speed v and frequency υ.
If the electron cross a point P on the circle in every revolution, then distance travelled by electron to complete one revolution = 2πr.
No. of revolutions in one second (υ) = \(\frac{\mathrm{v}}{2 \pi r}\)
The electric current (i) = \(\frac{\text { charge }}{\text { time }}\) = charge × frequency
i = e × \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
∴ Magnetic dipole moment (M) = I A (∵ N = 1)
M = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^2\) (∵ A = πr²)
M = \(\frac{\text { evr }}{2}\)

Question 8.
Explain how crossed E and B fields serve as a velocity selector.
Answer:
When a charged particle q moving with a velocity v in presence of both electric and magnetic fields.
The force experienced due to electric field F_E = q\(\overrightarrow{\mathrm{E}}\)
The force experienced due to magnetic field F_B = q \(\left(\begin{array}{l}
\vec{v} \times \vec{B}
\end{array}\right)\)
Consider electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.
E = E\(\hat{j}\), B = B\(\hat{k}\), v = v\(\hat{i}\)
F_E = qE\(\hat{j}\), F_B = q (v\(\hat{i}\) × B\(\hat{k}\)) = – qvB\(\hat{j}\)
∴ F = F_E + F_B
F = q (E – υB) \(\hat{j}\)

Thus electric and magnetic forces are in opposite directions.
We adjust E and B such that, the forces are equal
F_E = F_B
qE’ = q υ B
υ = \(\frac{E}{B}\)
This condition can be used to select charged particles of a particular velocity. The crossed field E and B serve as a velocity selector.

Question 9.
What are the basic components of a cyclotron ? Mention its uses ?
Answer:
Cyclotron is a device used to accelerate positively charged particles like protons, α – particles, deutrons etc.

Cyclotron mainly consists of

1. Two hollow D-shaped metallic chambers D₁ and D₂
2. High frequency oscillator
3. Strong electro magnet
4. Vacuum chamber.

Uses of cyclotron :

1. It is used for producing radioactive material for medical purposes i.e. diagnostics and treatment of chronic diseases.
2. It is used to improve the quality of solids by adding ions.
3. It is used to synthesise fresh substances.
4. It is used to bombard the atoms with highly accelerated particles to study the nuclear reactions.

Long Answer Questions

Question 1.
Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive an expression for the force per unit length between two parallel current carrying conductors.
Answer:
Expression for the Force acting on a current carrying conductor :
Consider a straight conductor (wire) of length ‘l’, area of cross section ‘A’, carrying a current ’i’, ‘which is placed in a uniform magnetic field of induction B’ as shown in fig.
We know the external magnetic field exerts a force on the conductor.
The electrons in effect move with an average velocity called drift velocity V_d‘. The direction of conventional current will be opposite to the direction of drift velocity.

Let us assume that the current flows through the conductor from left ‘B’ in the plane of the paper makes an angle ‘θ’ with the direction of current ‘i’ as shown in fig.

If F’ is the force acting on the charge ‘q’ in B
∴ F = q V_d B sin θ
If ‘n’ represents number of moving electrons per unit volume (∵ n = \(\frac{\mathrm{N}}{\mathrm{V}}\))
∴ Current i = nqV_dA
If ‘N’ is the number of electrons in the length ‘l’
N = nlA .
Total force on conductor F = F’.N (∵ N = nV = n × A × l)
= (q V_dB sin θ) (nlA)
= (nqV_dA) (IB sin θ)
∴ F = ilB sin θ
Case (i) : If θ = 0°, F_Min = 0
Case (ii) : If θ = 90°, F_Max = Bil

Expression for the force between two Parallel conductors carrying conductors:
Consider two straight parallel conductors .‘AB and ‘CD’ carrying currents ‘i₁’ and “i₂’ and which are separated by a distance ‘r’ as shown in fig.

If B₁ and B₂ are magnetic inductions produced by the current carrying conductors AB and CD. Magnetic induction B₁ at a distance ‘r’ from the conductor ‘AB’ can be written as B₁ = \(\frac{\mu_0 i_1}{2 \pi r}\)
If ‘F is force acting on CD’ due to magnetic induction B₁ then
F_CD = i₂lB₁
Where l = length of the conductor
F_CD = \(\mathrm{i}_2 l\left(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\right)\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) — (1)

The direction of the force can be determined by using Flemings left hand rule.
Similarly we can find the force acting on the conductor AB due to magnetic induction B₂.
F_AB = i₁lB₂
∴ F_AB = \(\mathrm{i}_1 l\left(\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\right)\) —– (2) [∵ B₂ = \(\frac{\mu_0 i_2}{2 \pi r}\)]
From the equations (1) and (2) F_AB = F_CD = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
∴ Force between two parallel, straight conductors carrying currents
F = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
Force per unit length \(\frac{\mathrm{F}}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)

Question 2.
Obtain an expression for the torque on a current carrying loop placed in a uniform magnetic field. Describe the construction and working of a moving coil galvanometer.
Answer:
Torque acting on a coil carrying a current kept in a uniform magnetic field : Let a rectangular current loop ABCD of length l = AB = CD and width b = AD = BC carrying a current “i” be suspended in a magnetic field of flux density B.
The normal ON drawn to the plane of the coil makes an angle ‘θ’ with the magnetic field B.

Force on arm AD = i\(\overline{\mathrm{b}}\) × \(\overrightarrow{\mathrm{B}}\) acting upwards along the axis of suspension
Force on arm BC = i\(\overline{\mathrm{b}}\) × \(\overrightarrow{\mathrm{B}}\) acting downwards along the axis of suspension
Hence these two forces cancel.

Force on arm AB = ilB acting perpendicular to the plane as shown.
Force on arm CD = ilB acting perpendicular to the plane as shown.
These two forces constitute a couple on the coil.
Moment of the couple = (Force) × (Perpendicular distance between the forces) = i/B (PQ sin θ)
Torque = ilB b sinθ
But l × b = Area of coil
∴ Torque = iAB sin θ
If the loop has n’ turns the torque on the coil
\(\tau\) = n i AB sinθ
If “ϕ’ is the deflection of the coil, that is the angle between the plane of the coil and magnetic field B
\(\tau\) = n i AB cos ϕ

Moving coil galvanometer:
Principle : When a current carrying coil is placed in the uniform magnetic field, it experiences a torque.

Construction: .

1. It consists of a coil wound on a non metallic frame.
2. A rectangular coil is suspended between two concave shaped magnetic poles with the help of phosphour Bronze wire.
3. The lower portion of the coil ‘is connected to a spring.
4. A small plane mirror M is fixed to the phosphour Bronze wire to measure the deflection of the coil.
5. A small soft iron cylinder is placed with in the coil without touching the coil. ‘The soft iron cylinder increases the induction field strength.
6. The concave shaped magnetic poles render the field radial. So maximum torque acting on it.
7. The whole of the apparatus is kept inside a brass case provided with a glass window.
   

Theory:
Consider a rectangular coil of length l and breadth b and carrying current i suspended in the induction field strength B.
Deflecting torque \((\tau)\) = B i A N —–> (5)

where A = Area of the coil
N = Total number of turns.
The restoring torque developed in the suspension = C θ —-> (2)
Where C is the couple per unit twist and θ is the deflection made by the coil. When the coil is in equilibrium position
Deflecting torque = Restoring torque
B i A N = C θ
i = \(\left(\frac{C}{\mathrm{BAN}}\right) \theta\)
Where K = \(\frac{\mathrm{C}}{\mathrm{BAN}}\) = Galvanometer constant
i = K θ —-> (3)
i ∝ θ
Thus deflection of the coil is directly proportional to the current flowing through it.
The deflection in the coil is measured using lamp and scale arrangement.

Question 3.
How can a galvanometer be converted fo an ammeter ? Why is the parallel resistance smaller that the galvanometer resistance ?
Answer:
Conversion of Galvanometer into Ammeter :
Galvanometer is converted into an ammeter by connecting a suitable resistance is parallel to it.
This arrangement decreases the effective resistance.

Ammeter is used for measuring the current in an electric circuit and it is connected in series in circuit.
The inclusion of the ammeter in the circuit should not alter the current or total resistance of the circuit so it has very low resistance.
The resistance of An ideal Ammeter is zero.
Let G and S be the Galvanometer and shunt resistances respectively.
Let ‘i’ be the total current, divided at A into i_g and i_s as shown in fig.
From Kirchhoff’s I^st law i = i_g + i_s
As ‘G’ and ‘S’ are parallel
PD. across Galvanometer = P.D. across shunt
i_g G = i_s S
S = \(\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{i}_{\mathrm{s}}} \mathrm{G}\)
= \(\frac{\mathrm{Gi}_{\mathrm{g}}}{\mathrm{i}-\mathrm{i}_{\mathrm{g}}}\) [∵ i_s = i – i_g]
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)
If \(\frac{\mathrm{i}}{\mathrm{i}_g}\) = n ⇒ i_g = \(\frac{i}{n}\)
∴ The current flowing through the galvanometer be \(\left(\frac{1}{n}\right)^{\text {th }}\) of total current.
∴ S = \(\frac{\mathrm{G}}{\mathrm{n}-1}\)
If ‘R is the effective resistance between points ‘A’ and ‘B’ then

Hence current through galvanometer is proportional to the total current. Since ‘S’ is small major portion of the current flows through it and a small portion of current flows through G. So shunt protects the galvanometer from high currents. Parallel resistance is smaller than Galvanometer resistance because to protect the Galvanometer from high (large) current (or) to pass. Large currents through shunt and small current passes through the galvanometer.

Question 4.
How can a galvanometer be converted to a voltmeter ? Why is the series resistance greater that the galvanometer resistance ?
Answer:
Conversion of Galvanometer into, Voltmeter: A galvanometer-is converted into voltmeter by connecting a high resistance (R) in series with it. Voltmeter is used to measure the RD. between any two points in circuit and it is connected in parallel to the component of the circuit.

Let V be the potential difference to be measured between the points ‘A’ and ‘B’.
∴ V = (R + G) i_g [∴ V = iR]

Where G = Galvanometer Resistance
i_g = Current passing through the galvanometer
\(\frac{\mathrm{V}}{\mathrm{i}_g}\) = R + G
∴ R = \(\frac{\mathrm{V}}{\mathrm{i}_g}\) – G —- (1)
The value of ‘R’ can be calculated by using the above formula. If V_g is the maximum P.D. across the galvanometer then V_g = i_g G
∴ i_g = \(\frac{V_g}{G}\) —— (2)
Substitute ‘i_g‘ in Equ (1)
R = \(\frac{V G}{V_g}\) – G = G\(\left(\frac{V}{V_g}-1\right)\)
If \(\frac{V}{V_g}\) = n ⇒ R = G(n – 1)
Note: n = \(\frac{V}{V_g}\) is the ratio of maximum voltage to be measured to the maximum voltage across the galvanometer
Series resistance is greater than galvanometer resistance because the current in external resistance and potential difference will be decreased and to increase the resistance of the galvanometer.

Question 5.
Derive an expression for the force acting between two very long parallel current-carrying conductors and hence define the Ampere.
Answer:
Force between two parallel conductors carrying current:

Let two long parallel conductors A and B seperated by a distance r carry currents i₁ and i₂ in the same directions.
The current i₁ produces magnetic induction around the conductor A and the current i₂ produces magnetic induction B₂ around the conductor B.
If l is the length of each conductor,
The magnetic induction B₁ at distance r from conductor A is
B₁ = \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\) — (1)
The force on conductor B is given by
F₂ = i₂ l B₁ —- (2)
The direction of force is given by Fleming left hand rule
F₂ = i₂l × \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) —- (3)
The direction of force F₂ is towards conductor A. Similary the magnetic induction B₂ at distance r from conductor B is
B₂ = \(\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\) — (4)
The force on conductor A is given by
F₁ = i₁lB₂ —- (5)
Substituting B₂ value in equation (5)
F₁ = \(\frac{\mathrm{i}_1 l \times \mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
F₁ = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) — (6)
It can be seen that | F₁| = | F₂| = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
The force per unit length of the conductor is given by
\(\frac{F}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\) —–>(7)

Definition of Ampere:
If i₁ = i₂ = 1A, r = 1m
F/l = \(\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1}\) = 2 × 10^-7 Nm^-1
“When two infinitely long parallel conductors, carrying the same current are seperated by 2 × 10^-7 Nm^-1, then the current flowing through each conductor is said to be one Ampere”.

Problems

Question 1.
A current of 10A passes through two very long wires held parallel to each other and separated by a distance of 1m. What is the force per unit length between them ? (T.S. Mar. ’19 & A.P. & T.S. Mar. ’15)
Solution:
i₁ = i₂ = 10A
r = 1m
\(\frac{\mathrm{F}}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 10 \times 10}{2 \pi \times 1}\)
\(\frac{\mathrm{F}}{\mathrm{l}}\) = 2 × 10^-5 Nm^-1.

Question 2.
A moving coil galvanometer can measure a current of 10^-6 A. What is the resistance of the shunt required if it is to measure 1A ?
Solution:
Galvanometer current (i_g) = 10^-6A
i = 1A
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)
= \(\frac{\mathrm{G}}{\frac{1}{10^{-6}}-1}\)
S = \(\frac{G}{10^6-1}\)
S = \(\frac{\mathrm{G}}{99,999} \Omega\)
Where G = Galvanometer resistance.

Question 3.
A circular wire loop of radius 30 cm carries a current of 3.5 A. Find the magnetic field at a point on its axis 40 cm away from the centre.
Solution:
Radius (r) = 30 cm = 30 × 10^-2m
Current (i) = 3.5 A
x = 40 cm = 40 × 10^-2m

Textual Exercises

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil ?
Solution:
Here, n = 100, r = 8cm = 8 × 10^-2 m and I = 0.40 A
The magnetic field B at the centre
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \mathrm{In}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 3.14 \times 0.4 \times 100}{8 \times 10^{-2}}\) = 3.1 × 10^-4T
The direction of magnetic field depends on the direction of current if the direction of current is anticlockwise. According to Maxwell’s right hand rule, the direction of magnetic field at the centre of coil will be perpendicular outwards to the plane of paper.

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire ?
Solution:
Here, I = 35 A and r = 20 cm = 0.2 m
The wire is along and it is considered as an infinite length wire. The magnetic field
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 35}{0.2}\) = 3.5 × 10^-5T

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Given I = 50A and r = 2.5 m
The magnitude of magnetic field

B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) = 10^-7 × \(\frac{2 \times 50}{2.5}\)
= 4 × 10^-6T
The direction of magnetic field at point P is given by Maxwell’s right hand rule.

Question 4.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line ?
Solution:

Given I = 90 A and r = 1.5m
The magnitude of magnetic field
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 90}{1.5}\)
= 1.2 × 10^-5 T
The direction of magnetic flux is given by Maxwell’s right hand rule. So, the direction of magnetic field at point P due to the following current is perpendicularly outwards to the plane of paper.

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 300 with the direction of a uniform magnetic field of 0.15.T?
Solution:
According to the question
I = 8 A, θ = 30°, B = 0.15 T, l = 1 m
The magnitude of magnetic force

f = I(l × B) = I l B sin θ
= 8 × 1 × 0.15 × sin 30°
= \(\frac{8 \times 0.15}{2}\) = 4 × 0.15 = 0.6 N/m

Question 6.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?
Solution:
Here, the angle between the magnetic field and the direction of flow of current is 90°. Because the magneitc field due to a solenoid is along the axis of solenoid and the wire is placed perpendicular to the axis.

Given, l = 3 cm = 3 × 10^-2 m, I = 10A, B = 0.27 T
f = I l B sin 90°
= 10 × 3 × 10^-2 × 0.27 = 8.1 × 10^-2 N
According to right hand palm rule, the direction of magnetic force is perpendicular to plane of paper inwards.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Given I₁ = 8A, I₂ = 5A and r = 4 cm = 0.4m
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 8 \times 5}{0.04}\) = 2 × 10^-4N
The force on A of length 10 cm is F¹ = F × 0.1 (∵ 1m = 100 cm)
F¹ = 2 × 10^-4 × 0.1
F¹ = 2 × 10^-5N.

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. It the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Solution:
The length of solenoid, l = 80 cm = 0.8 m
Number of layers = 5
Number of turns per layer = 400

Diameter of solenoid = 1.8 cm
Current in solenoid = I = 8 A
∴ The total number of turns N = 400 × 5 = 2000
and no. of turns / length, n = \(\frac{2000}{0.8}\) = 2500
The magnitude of magnetic field inside the solenoid
B = μ₀nl = 4 × 3 .14 10^-7 × 2500 × 8
= 2. 5 × 10^-2 T
The direction of magnetic field is along the axes of solenoid.

Question 9.
A square coil of side 10 cm consists of 20 turns and carries current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil ?
Solution:
Given, side of square coil = 10 cm = 0.1 m
Number of turns (n) = 20
current in square coil I = 12 A
Angle made by coil θ = 30°
Magnetic filed B = 0.80 T

The magnitude of torque experienced by the coil
\(\tau\) = NIAB sinθ
= 20 × 12 × (10 × 10^-2)² × 0.80 × sin 30°
= 0.96 N – m

Question 10.
Two moving coil meters, M₁ and M₂ have the following particulars :
R₁ = 10 Ω, n₁ = 30,
A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T
R₂ = 14Ω, n₂ = 42,
A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T, K₁ = K₂
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Solution:
Given, R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T
R₂ = 140Ω, n₂ = 42, A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T, K₁ = K₂

Question 11.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.5 × 10^-19 C, m_e = 9.1 × 10^-31 kg)
Solution:
Given, magnetic field B = 6.5 G = 6.5 × 10^-4 T
Charge e = -1.6 × 10^-19 C
Speed of electron V = 4.8 × 10⁶ m/s
Mass of electron m_e = 9.1 × 10^-31 kg
Angle between magnetic field and electron (θ) = 90°
The force on charge particle entering in the magnetic field
F = q (V × B) = e (V × b)
The electron attains a circular path and necessarily centripetal force is provided by magnetic force.
e (V × B) = \(\frac{m V^2}{r}\)
e V B sin 90° = \(\frac{\mathrm{m} \mathrm{V}^2}{\mathrm{r}}\)
r = \(\frac{\mathrm{mV}}{\mathrm{e} \mathrm{B} \times 1}\) = \(\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}\) = 4.2 × 10^-2m = 4.2 cm

Question 12.
From Exercise 11 data, obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron ? Explain.
Solution:
Given
B = 6.5 G = 6.5 × 10^-4T, V = 4.8 × 10⁶ m/s, e = 1.6 × 10^-19C
m_e = 9.1 × 10^-3 kg
\(\frac{\mathrm{mV}^2}{\mathrm{r}}\) = q V B ⇒ \(\frac{\mathrm{mV}}{\mathrm{r}}\) = qB
If angular velocity of electron is ω, then
V = r ω
ω = \(\frac{\mathrm{qB}}{\mathrm{m}}\)
2πn = \(\frac{\mathrm{qB}}{\mathrm{m}}\) ⇒ n = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
Frequency of revolution of electron in orbit
υ = \(\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}\) = \(\frac{\mathrm{Be}}{2 \pi \mathrm{m}_{\mathrm{e}}}\) = \(\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\) = 18.18 × 10⁶ Hz.

Question 13.
(a) Circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniiorm horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil form turning.
(b) Would your answer change, If the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area ? (All other particular are also unaltered.)
Answer:
a) Given, no. of turns (n) = 30, radius (r) = 8 cm = 0.08 m
Current in coil (I) = 6A, Magnetic field (B) = 1.0 T,
Angle made by field with the normal of the coil, θ = 60°
\(\tau\) = n I A Bsinθ
= 30 × 6 × π (0.08)² × 1 × sin 60°
= 30 × 6 × 3.14 × 0.08 × 0.08 × \(\frac{\sqrt{3}}{2}\)
\(\tau\) = 3.133 N – m

b) From the formula, it is clear that the torque on the loop does not depend on the shape if area remains constant. So, the torque remains constant.

Additional Exercises

Question 1.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X Is anticlockwise, and clockwise In Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Solution:
For coil X
Radius of coil, r_x = 16 cm = 0.16 m
No. of turns n_x = 20
Current in the coil I_x = 16A (Anti clockwise) .
For coil Y
Radius of coil, r_y = 10 cm = 0.1 m
No. of turns n_y = 25
Current in the coil I_y = 18 A (clockwise)
The magnitude of the magnetic field at the centre of coil X
B_x = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_x \cdot \pi n_x}{r_x}\) = \(\frac{10^{-7} \times 2 \times 16 \times 3.14 \times 20}{0.16} \mathrm{~T}\)
B_x = 4π × 10^-4T
The magnitude field at the centre B =B_y – B_x
= (9π – 4π) × 10^-4 = 5π × 10^-4
= 1.6 × 10^-3T (towards west)

Question 2.
A magnetic field of 1oo G (1 G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^-3 m². The maximum current carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a côre is at most 1000 turns m^-1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Solution:
Magnetic field B = 100 G = 100 × 10^-4T = 10^-2 T
Maximum current I = 15A, n = 1000/m.
The magnitude of magnetic field B = μ₀ nI
nI = \(\frac{\mathrm{B}}{\mu_0}\) = \(\frac{10^{-2}}{4 \times 3.14 \times 10^{-7}}\)
⇒ nI = 7961 ≈ 8000
Here; the product of n I is 8000. So,
Current I = 8A and no. of turns = 1000
The other design is I = 10A and n = 800/m. This is the most appropriate design as the requirement.

Question 3.
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from Its centre is given by,
B = \(\frac{\mu_0 \mathbf{I R}^2 \mathbf{N}}{2\left(\mathbf{x}^2+\mathbf{R}^2\right)^{3 / 2}}\)
a) Show that this reduces to the familiar result for field at the centre of the coil.
Solution:
Given, magnetic field at distance x :
B = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\) (∵ x = 0)
∴ The magnetic field at the centre B = \(\frac{\mu_0 \mathrm{~N} \mathrm{I} \mathrm{R}^2}{2 \mathrm{R}^3}\)
B = \(\frac{\mu_0 \mathrm{~N} \mathrm{I}}{2 \mathrm{R}}\)
This result is same as the magnetic field due to current loop at its centre.

b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated bya distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = 072\(\frac{\mu_0 \mathrm{NI}}{\mathrm{R}}\) approximately.
ISuch an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] :
Solution:
Radius of two parallel co-axial coils = R, number of turns = N Current = I
Let the mid points between the coils is at point O and P be the pomt around the mid point O.
Suppose, the distance between OP = d which is very less than R (d < < R)
For the Coil A,
O_AP = \(\frac{R}{2}\) + d
the magnetic field at point P due to coil A.


as according to the question d < <R, so neglect turn d².

The direction of B_A is along PO_B. According to Maxwell’s right hand rule
for the coil B, O_BP = (Rl₂ – d)
The magnetic field at point P due to coil B

The direction of magnetic field B_B is towards PO_B. So, the resultant magnetic field at P due to coil A and coil B is

Now, use binomial theorem and neglect higher power as d < < R

Question 4.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around whIch 3500 turns of a wire are wound. If the current in the wire Is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Solution:
a) For outside the toroid, the magnetic field is zero, because the magnetic field due to toroid is only inside it and along the length of toroid.

b) Inner radius of toroid, r₁ = 25 cm = 0.25 m
Outer radius of toroid, r₂ = 26 cm = 0.26 m
Number of turns N = 3500
current in the wire, I = 11A
The mean radius of toroid r = \(\left(\frac{\mathrm{r}_1+\mathrm{r}_2}{2}\right)\) = \(\frac{2}{2}\)(0.25 + 0.26) = 0.51
∴ length of toroid = 2πr = 2π × 0.51
the magnetic field strength due to toroid is B = µ₀ n I
where n is number of turns per unit length
n = \(\frac{N}{I}\)
B = 4π × 10^-7 × \(\frac{3500}{\pi \times 0.51}\) × 11 = 3.02 × 10^-2 T

c) The magnetic field in the empty space surrounded by toroid is also zero, because the
magnetic field due to toroid is only along its length.

Question 5.
Answer the following questions.
a) A magnetic field that varies in magnitude form point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
Answer:
The magnetic field is in constant direction from east to west. According to the question, a charged particle travels undeflected along a striaght path with constant speed. It is only possible, if the magnetic force experienced by the charged particle is zero. The magnitude of magnetic force on a moving charged particle in a magnetic field is given by F = qvB sin θ. Here F = 0, if and only if sinθ. This indicates the angle between the velocity and magnetic fled is 0° or 180°. Thus, the charged particle moves parallel or antiparallel to the magnetic field B.

b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory Would its final speed equal the initial speed lilt suffered no collisions with the environment?
Answer:
Yes, the final speed be equal to its initial speed as the magnetic force acting on the charged particle only changes the direction of velocity of charged particle but cannot change the magnitude of velocity of charged particle.

c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Solution:
B should be in a vertically downward direction.

Question 6.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Solution:
a) Since magnetic field is perpendicular to initial velocity of electron therefore, the electron will move in circular path.
B = 0.15 T,PD = 200 V
K. E. of electron = eV = \(\frac{1}{2}\)mv²
V² = \(\frac{2 \mathrm{ev}}{\mathrm{m}}\)
V = \(\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.1 \times 10^{-31}}}\)
V = \(\frac{8}{3}\) (e = charge of electron)
V = 2.66 × 10⁷ m/s
Also
BqV = \(\frac{\mathrm{m} \mathrm{V}^2}{\mathrm{Br}}\)
∴ r = \(\frac{\mathrm{mv}}{\mathrm{Bq}}\)
or r = \(\frac{9.1 \times 10^{-31} \times 2.66 \times 10^7}{0.15 \times 1.6 \times 10^{-19}}\)
= \(\frac{91 \times 83.86 \times 10^{-5}}{15 \times 1.6}\) = 10^-3 m = 1 mm

b) When magnetic field makes an angle 30° with the initial velocity i.e. θ = 30°
Then V¹ = V sin θ = \(\sqrt{\frac{2 \mathrm{ev}}{\mathrm{m}}}\) = sin 30°
V¹ = \(\frac{8}{3}\) × 10⁷ × \(\frac{1}{2}\) = \(\frac{4}{3}\) × 10⁷ m/s
The radius of the helical path is
r = \(\frac{\mathrm{m} \mathrm{V}^{\prime}}{\mathrm{Be}}\)
= \(\frac{9 \times 10^{-31} \times\left(\frac{4}{3} \times 10^7\right)}{0.15 \times 1.6 \times 10^{-19}}\) = 0.5 mm

Question 7.
A magnetic field set up using Helmholtz coils (described in Exercise 16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^-5 V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique ?
Solution:
B = 0.75 T, E = 9 × 10⁵ Vm^-1
V = \(\frac{E}{B}\) = \(\frac{9 \times 10^5}{0.75}\) = 12 × 10⁶ m/s
K. E of charged particle
= \(\frac{1}{2} \mathrm{mv}^2\) = eV
or \(\frac{\mathrm{e}}{\mathrm{m}}\) = \(\frac{v^2}{2 v}\) = \(\frac{144 \times 10^{12}}{2 \times 15000}\) = 4.8 × 10⁷C kg^-1
Particle is deuteron; the answer is not unique because only the ratio of charge to mass is determined. Other possibe answers are He⁺⁺, Li⁺⁺⁺ etc.
∴ He⁺⁺ and Li⁺⁺⁺ also have the same value of
e/m (∵ e/m = \(\frac{2 \mathrm{e}}{2 \mathrm{~m}}\) = \(\frac{3 \mathrm{e}}{3 \mathrm{~m}}\))

Question 8.
A straight horizontal conducting rod of length 0.45 m and mass 6 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
Solution:
Rod carrying current placed in uniform magnetic field expëriences a force Bu which is balanced by weigth of rod
F = B I l
mg = B I l
B = \(\frac{\mathrm{m} \mathrm{g}}{\mathrm{I} l}\) = \(\frac{60 \times 9.8}{1000 \times 0.45 \times 5}\) = 0.26 T

b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s^-2.
Solution:
Force due to magnetic field = B I l
= 0.26 × 5 × 0.45 = 5.85 N
Weight of rod = \(\frac{60}{1000}\) × 9.8 = 0.588 N
Total force = 0.588 + 0.585 = 1.173 N

Question 9.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Solution:
F = \(\frac{\mu_0}{4 \pi} \times \frac{2 I_1 I_2}{r}\)
= 10^-7 × \(\frac{2 \times 300 \times 300 \times 100}{1.5 \times 10^{-2}}\) = 1.2 Nm^-1 (Repulsive force)

Question 10.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

a) the wire intersects the axis,
Solution:
F = BI 1 sin 900 = 1.5 × 7 × \(\frac{20}{100}\) = 2.1 N acting vertically downwards.

b) The wire is turned form N-S to north-east-northwest direction,
Solution:
Now θ = 45°, and length of the wire in the cylindrical region of the magnetic field is l, and is given by
l = l₁ sin 45°
l₁ = \(\frac{l}{\sin 45^{\circ}}\) = \(\sqrt{2 l}\)
So force F₁ = BI l sin 45°
= 1.5 × 7.0 × \(\sqrt{2} l\) × \(\frac{l}{\sqrt{2}}\)
= 10.5 × 0.2 = 2.1 N

c) The wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Solution:
When the wire is lowered through a distance 6.0 cm and reaches at CD then length of wire in magnetic field
l₂ = 2x
x.x = 4 × (10 + 6) = 64
x = 8 cm
∴ l₂ = 8 × 2 = 16 cm = \(\frac{16}{100}\)m
∴ Force on the wire
F₂ = BI l₂
= \(\frac{1.5 \times 7 \times 16}{100}\) = 1.68 N (Vertically downwards)

Question 11.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. ? What is the force on each case ? Which case corresponds to stable equilibrium?

Solution:
a) Torque = 300, G = 3000 × 10^-4 T
Area A = 10 × 5 = 50 cm² = \(\frac{50}{100 \times 100}\)m²
\(\tau\) = BIA = \(\frac{3}{10}\) × 12 × \(\frac{50}{100 \times 100}\) = 18 × 10^-2 Nm along Y-axis
b) Same as in (a)
c) 1.8 × 10^-2 Nm along x – direction
d) 1.8 × 10^-2 Nm at an angle 240° with the +x direction :
e) Zero
f) Zero
Force is zero in each case. Case (e) corresponds to stable and case (f) to unstable equilibrium.

Question 12.
A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. 1f the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^-5 m², and the free electron density in copper is given to be about 10²⁹ m^-3.)
Solution:
n = 20, r = 10 cm, B = 0.10 T, I = 50 A
a) Torque acting on the coil = nBIA sinθ = 0
b) Force acting on the coil = BIl sinθ = 0
e) Force on each electron
n = No. of electrons per unit volume
A = Area of cross – section of wire
F = Be V = \(\frac{\mathrm{BI}}{\mathrm{nA}}\) = \(\frac{0.1 \times 5}{10^{29} \times 10^{-5}}\) = 5 × 10^-25N

Question 13.
A solenoid 60 cm long and of radius 7.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two lead parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g = 9.8 m s^-2.
Solution:
B = μ₀n I
Force acts on the wire normal to its length
∴ F = BI¹ l
= μ₀n I I¹ l and n = \(\frac{\frac{900}{60}}{100}\) = 1500
= μ₀n I I¹ l = mg
I = \(\frac{2.5 \times 9.8}{1000 \times 4 \pi \times 10^{-7} \times 1500 \times 6 \times \frac{1}{15}}\) = 108 A.

Question 14.
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA How will you convert the metre into a voltameter of range 0 to 18V ?
Solution:
Here G = 12Ω, Ig = 3mA = 3 × 10^-3A, V = 18V
R = \(\frac{v}{I_g}\) – G = \(\frac{18}{3 \times 10^{-3}}\) – 12 = (6000 – 12)Ω = 5988 Ω.

Question 15.
A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A?
Solution:
Here G = 15Ω, I_g = 4 mA = 0.004 A, I = 6A
shunt S = \(\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\) = \(\frac{0.004 \times 15}{6-0.004}\) = \(\frac{0.06}{5.996}\) = ≈ 10 × 10^-3Ω = 10 m Ω
So shunt resistance = 10 m \(\simeq\) 10 mΩ

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity

Very Short Answer Questions

Question 1.
Define mean free path of electron in a conductor.
Answer:
The average distance transversed by an electron during successive collisions in a conductor is called mean free path of electron in a conductor.

Question 2.
State Ohm’s law and write its mathematical form.
Answer:
At constant temperature, the strength of the current (I) in a conductor is directly proportional to the potential difference (V) between its ends.
∴ I ∝ V ⇒ I = \(\frac{\mathrm{V}}{\mathrm{R}}\) ⇒ V = IR (Mathematical form)
where R is constant, it is called the resistance of the conductor.

Question 3.
Define resistivity or specific resistance.
Answer:
Resistivity or specific resistance (ρ) : The resistance of a conductor of unit length and unit area of cross-section is called resistivity.
If l = 1, A = 1 ⇒ ρ = \(\frac{\mathrm{R} \times 1}{1}\) = ρ ⇒ R

Question 4.
Define temperature coefficient of resistance.
Answer:
Temperature coefficient of resistance (α) : The ratio of the change in resistance per 1°C rise in temperature to the resistance at 0°C is called the temperature coefficient of resistance.
α = \(\frac{R_t-R_0}{R_0 t}\)

Question 5.
Under what conditions is the current through the mixed grouping of cells maximum ?
Answer:
The current through the mixed grouping of cells maximum, when

1. Effective emf of all the cells is high.
2. The value of external resistance is equal to the total internal resistance of all the cells.

Question 6.
If a wire is stretched to double its original length without loss of mass, how will the resistivity of the wire be influenced ?
Answer:
Resistivity of the wire remains unchanged as it does not change with change in dimensions of a material without change in its temperature.

Question 7.
Why is manganin used for making standard resistors ?
Answer:
Due to high resistivity and low temperature coefficient of resistance, manganin wire (Cu – 84% + Mn – 12% + Ni – 4%) is used in the preparation of standard resistances.

Question 8.
The sequence of bands marked on a carbon resistor are: Red, Red, Red, Silver. What is its resistance and tolerance ?
Answer:
The resistance of a carbon resistor marked with Red, Red, Red = 22 × 10²Ω = 2.2kΩ = 2200Ω

[∵ Sequence number for Red = 2 and multiplication factor = 10²]
The tolerance of carbon resistor = ± 10%

Question 9.
Write the color code of a carbon resistor of resistance 23 kilo ohms.
Answer:
Color code of a carbon resistor of 23 Kilo Ohms (= 23 × 10³Ω) are Red, Orange, Orange
[∵ Sequence number 2 for Red, 3 for orange, multiplication factor 10³ for orange]

Question 10.
If the voltage V applied across a conductor is increased to 2V, how will the drift velocity of the electrons change ?
Answer:
The drift velocity, V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) \(\tau^{\prime}\) = \(\frac{\mathrm{eV}}{\mathrm{mL}} \tau\)
\(\frac{\mathrm{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{v_1}{v_2}\)
Here V₁ = V, V₂ = 2V
\(\frac{\mathbf{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{V}{2 V}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift velocity is increased by twice.

Question 11.
Two wires of equal length, of copper and, manganin, have the same resistance. Which wire is thicker ?
Answer:
R = \(\frac{\rho \mathrm{A}}{l}\) ⇒ A = \(\frac{\mathrm{R} l}{\rho}\)
Since ρ_cu < p_manganin, copper wire is thicker than manganin wire.

Question 12.
Why are household appliances connected in parallel ?
Answer:
In parallel, the voltage (V) across each appliance is same. The current (I) through them depends upon the power (P) of the appliance. The higher power appliance draws more current and lower power appliance draws less current.
(∵ P = VI or I ∝ P)

Question 13.
The electron drift speed in metals is small (~ms^-1) and the charge of the electron is also very small (~10^-19C), but we can still obtain a large amount of current in a metal. Why ?
Answer:
Current through a metal, I = n A eV_d.
A is the area of cross-section of the metal. The electron drift speed, V_d (~10^-5, ms^-1) is small. The charge of electron, e (~1.6 × 10^-19C) is also very small. But we can still obtain a large amount of current in a metal due to presence of large number of free electrons (n) is a conductor (~ 10²⁹ m^-3).

Short Answer Questions

Question 1.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor R.

1. If the current in the circuit is 0.5 A. Calculate the value of R.
2. What is the terminal voltage of the battery when the circuit is closed.

Answer:
Given, E = 10V, r = 3Ω, I = 0.5A, R = ?, V = ?

1. E = I(R + r) or R + r = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) =20Ω ⇒ R = 20 – 3 = 17Ω
2. Terminal voltage, V = IR = 0.5 × 17 = 8.5Ω

Question 2.
Draw a circuit diagram showing how a potentiometer may be used to find Internal resistance of a cell and establish a formula for It.
Answer:
Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
   
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (RB) through a key K₂.
3. With key K₂ open, balance is obtained at length
   l₁ (AN₁). Then ε = ϕl₁ —– (1)
4. When key K₂ is closed, the cell sends a current (I) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂). Then V = ϕl₂ —– (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —- (3)
7. But ε = I(r + R) and V = IR. This gives
   \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{I_1}{I_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = R\(\left(\frac{l_1}{l_2}-1\right)\)

Question 3.
Derive an expression for the effective resistance when three resistors are connected in
(i) series
(ii) parallel. (T.S. Mar. ’19)
Answer:
Effective resistance when three resistors are connected:
(i) In series:

1. Three resistors R₁, R₂ and R₃ are connected in series as shown in fig. V₁, V₂, V₃ are the potential differences across R₁, R₂ and R₃. I is the current flowing through them.
2. Applying Ohm’s law to R₁, R₂ and R₃, Then V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
3. In series, V = V₁ + V₂ + V₃
   IR_S = IR₁ + IR₂ + IR₃ [∵ V = IR_S]
   ∴ R_S = R₁ + R₂ + R₃

ii) In parallel:

1. Three resistors, R₁, R₂ and R₃ are connected in parallel as shown in fig. Potential differences across each resistor is V. I₁, I₂, I₃ are the currents flowing through them.
2. Applying Ohmes law to R₁, R₂ and R₃, then
   V = I₁R₁ = I₂R₂ = I₃R₃
   ⇒ I₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\), I₂ = \(\frac{\mathrm{V}}{\mathrm{R}_2}\), I₃ = \(\frac{\mathrm{V}}{\mathrm{R}_3}\)
3. In parallel, I = I₁ + I₂ + I₃
   ⇒ \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) + \(\frac{\mathrm{V}}{\mathrm{R}_2}\) + \(\frac{\mathrm{V}}{\mathrm{R}_3}\) [∵ I = \(\frac{V}{R_p}\)]
   ∴ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

Question 4.
‘m’ cells each of emf E and internal resistance ‘r’ are connected in parallel. What is the total emf and internal resistance ? Under what conditions is the current drawn from mixed grouping of cells a maximum ?
Answer:
Cells in parallel:

1. When ‘m’ identical cells each of emf ‘ε’ and internal resistance r are connected to the external resistor of resistance R as shown in fig., then the cells are connected in parallel.
2. As the cells are connected in parallel, their equivalent internal resistance r_p is given by
   \(\frac{1}{\mathrm{r}_{\mathrm{p}}}\) = \(\frac{1}{\mathrm{r}}\) + \(\frac{1}{\mathrm{r}}\) + ….. upto m terms = \(\frac{\mathrm{m}}{\mathrm{r}}\)
   ∴ r_p = \(\frac{\mathrm{r}}{\mathrm{m}}\)
   
3. As R and r_p are in series, so total resistance in the circuit = R + \(\frac{\mathrm{r}}{\mathrm{m}}\).
4. In parallel combination of identical cells, the effective emf in the circuit is equal to the emf due to a single cell, because in this combination, only the size of the electrodes increases but not emf.
5. Therefore, current in the resistance R is given by I = \(\frac{\varepsilon}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{m}}}\) = \(\frac{\mathrm{m} \varepsilon}{\mathrm{m} R+\mathrm{r}}\)
6. When the external resistance is negligible is comparison to the internal resistance (R<<r), the current drawn from mixed grouping of cells a maximum.

Question 5.
Define electric resistance and write it’s SI unit. How does the resistance of a conductor vary if
(a) Conductor is stretched to 4 times of it’s length
(b) Temperature of a conductor is increased.
Answer:
Electric resistance (R) : The resistance offered by a flow of electrons in a conductor is called electric resistance.
S.l unit of resistance is ohm (Ω).
The resistance of a conductor
R = \(\frac{\rho l}{\mathrm{~A}}\) = \(\frac{\rho l^2}{\mathrm{~V}}\) ⇒ R ∝ l²
a) in first case, R₁ = R, l₁ = l
b) In second case, l₂ = 4l, R₂ =?
\(\frac{R_2}{R_1}\) = \(\frac{l_2^2}{l_1^2}\) ⇒ \(\frac{R_2}{R}\) = \(\left(\frac{4l}{l_4}\right)^2\) ∴ R₂ = 16R

b) Variation of Resistance with temperature is given by R_t = R₀ (1 + α t)
If temperature increases, resistance also increases.

Question 6.
When the resistance connected In series with a cell is halved, the current is equal to or slightly less or slightly greater than double. Why?
Answer:
‘When he resistance R is connected to cell of emf, ε in series, the current is given by
I = \(\frac{\varepsilon}{\mathrm{R}+\mathrm{r}}\)

where r is internal resistance of cell.
When the resistance is halved \(\left(\frac{\mathrm{R}}{2}\right)\), the current flows through the circuit is I’ = \(\frac{\varepsilon}{\frac{R}{2}+r}\)

1. If r is negligible comparison with \(\frac{\mathrm{R}}{2}\), I¹ = \(\frac{2 \varepsilon}{R}\)
   ∴ I¹ = 2 I [∵ \(\frac{\varepsilon}{R}\) also equal to 1]
2. If r < < \(\frac{\mathrm{R}}{2}\), the current I¹ is slightly greater than 2.
3. If r is just slightly greater than R, the current (I¹) is slightly less than 2.

Question 7.
Two cells of emfs 4.5V and 6.0V and infernal resistance 6Ω and 3Ω respectively have their negative terminals joined by a wire of 18Ω and positive terminals by a wire of 12Ω resistance. A third resistance wire of 24Ω connects middle points of these wires. Using Kirchhoffs laws, find the potential difference at the ends of this third wire.
Answer:

1. Let the currents through the various arms, of the network be as shown in fig:
2. Applying KVL to closed mesh ABCDA, we have
   4.5 – 6I₁ – 18I₁ – 24 (I₁ + I₂) = 0
   ⇒ 48I₁ + 24I₂ = 4.5 —–> (i)
3. For a closed mesh CDEFC, we have
   – 24(I₁ + I₂) – 12I₂ + 6 – 3I₂ = 0
   24I₁ + 39I₂ = 6 —–> (ii)
   
4. (ii) × 2 – (i) ⇒
   
   ∴ I₂ = \(\frac{7.5}{54}\) = 0.139 A —–> (iii)
   Substituting (iii) in (i), we get
   48I₁ + 78 × 0.139 = 12
   48I₁ = 12 – 10.84 = 1.158
   I₁ = \(\frac{1.158}{48}\) = 0.024 A
5. Potential difference across third wire = (I₁ + I₂) × 24 = 0.163 × 24 = 3.912 Volt.

Question 8.
Three resistors each of resistance 10 ohm are connected, in turn, to obtain
(i) minimum resistance
(ii) maximum resistance. Compute
(a) The effective resistance in each case
(b) The ratio of minimum to maximum resistance so obtained.
Answer:
Given, Resistance of each resistor R = 10Ω, no. of resistors, n = 3
i) If three resistors are connected in parallel, we get minimum resistance.
∴ Minimum resistance R_min = R_p = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3} \Omega\) = 3.33Ω

ii) If three resistors are connected in series, we get maximum resistance.
∴ Maximum resistance R_max = R_s = n R = 3 × 10 = 30Ω

a) The effective resistance to get minimum resistance,
R_eff = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3}\) = 3.33Ω (In parallel)
The effective resistance to get maximum resistance
R_eff = n R = 3 × 10 = 30Ω (In series)

b) \(\frac{R_{\min }}{R_{\max }}\) = \(\frac{\left(\frac{10}{3}\right)}{(3 \times 10)}\) = \(\frac{10}{90}\) ∴ \(\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\max }}\) = \(\frac{1}{9}\)

Question 9.
State Kirchhoffs law for an elêctrical network. Using these laws deduce the condition for balance in a Wheatstone bridge. (Ã.P. Mar. 19, ‘16 & Mar. 14)
Answer:
1) Kirchhoff s first law (Junction rule or KCL) : The algebraic sum of the currents at any junction is zero. ∴ ΣI = 0
(or)
The sum of the currents flowing towards a junction is equal to the sum of currents away from the junction.

2) Kirchhoffs second law (Loop rule or KVL): The algebraic sum of potential around any closed loop is zero.
∴ Σ(IR) + ΣE = 0
Wheatstone bridge : Wheatstone’s bridge circuit consists of four resistances R₁, R₂, R₃ and R₄ are connected to form a closed path. A cell of emf e is connected between the point A and C and a galvanometer is connected between the points B and D as shown in fig. The current through the various branches are indicated in the figure. The current through the galvanometer is I_g and the resistance of the galvanometer is G.
Applying Kirchhoffs first law at the junction D, I₁ – I₃ – I_g = 0 —— (1)
at the junction B, I₂ + I_g – I₄ = 0 —— (2)
Applying Kirchhoffs second law to the closed path ADBA,
-I₁R₁ + I₂R₂ = 0
or
⇒ I₁R₁ + I_gG = I₂R₂ —– (3)
applying kirchhoffs second law to the closed path DCBD,
-I₃R₃ + I₄R₄ + I_gG = 0
⇒ I₃R₃ – I_gG = I₄R₄ —– (4)
When the galvanometer shows zero deflection the points D and B are at the same potential. So I_g = 0.

Substituting this value in (1), (2), (3) and (4).
I₁ = I₃ ——- (5)
I₂ = I₄ —— (6)
I₁R₁ = I₂R₂ —– (7)
I₃R₃ = I₄R₄ —– (8)
Dividing (7) by (8)
\(\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_3 \mathrm{R}_3}\) = \(\frac{I_2 R_2}{I_4 R_4}\) ⇒ \(\frac{R_1}{R_3}\) = \(\frac{R_2}{R_1}\) [∵ I₁ = I₃ & I₂ = I₄]
∴ Wheatstone’s Bridge principle : R₄ = R₃ × \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 10.
State the working principle of potentiometer. Explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer. (T.S. Mar. 19 & A.P. Mar. 16)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant,
i.e. ε ∝ l ⇒ ε = ϕl where ϕ is potential gradient.
Comparing the emf of two cells ε₁ and ε₂ :

1. To compare the emf of two cells of emf E₁ and E₂ with potentiometer is shown in diagram.
   
2. The points marked 1, 2, 3 form a two way key.
3. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to ε₁.
4. The Jockey is moved along the wire till at a point N₁ at a distance l₁ from A, there is no deflection in the galvanometer. Then ε₁ ∝ l₁ ⇒ ε₁ = ϕl₁ —– (1)
5. Similarly, if another emf ε₂ is balanced against
   l₂ (AN₂) then ε₂ ∝ l₂ ⇒ ε₂ = ϕl₂ —— (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon_1}{\varepsilon_2}\) = \(\frac{l_1}{l_2}\)

Question 11.
State the working principle of potentiometer explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell. (A.P. & T.S. Mar. ’15)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant.
i.e. E ∝ l ⇒ e = ϕl
where ϕ is potential gradient.

Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (R.B) through a key K₂.
   
3. With key K₂ open, balance is obtained at length l₁(AN₁). Then ε = ϕl₁ ——-> (1)
4. When key K₂ is closed, the cell sends a current (T) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
   Then V = ϕl₂ ——> (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —– (3)
7. But ε = I (r + R) and V = IR. This gives
   \(\frac{\varepsilon}{V}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{l_1}{l_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = \(\mathrm{R}\left(\frac{l_1}{l_2}-1\right)\)

Question 12.
Show the variation of current versus voltage graph for GaAs and mark the
(i) Non-linear region
(ii) Negative resistance region.
Answer:
The relation between V and I is not unique. That is, there is more than one value of V for the same current I. material exhibiting such behaviour is GaAs (i.e., a light emitting diode).

Question 13.
A student has two wires of iron and copper of equal length and diameter. He first joins two wires in series and passes an electric current through the combination which increases gradually. After that he joins two wires in parallel and repeats the process of passing current. Which wire will glow first in each case ?
Answer:
1) In series combination, there will be same current through Iron and as well as copper wire. Since the rate of heat production, P = I² R or P ∝ R (for the given value of I). The resistance of Iron wire is more than that of copper for the given length and diameter. Hence in Iron wire, the rate of heat production increases gradually. In series combination Iron will glow first.

2) In parallel combination of Iron and copper wire, there will be same P.D (V) across them.
Since the rate of heat production, P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) or P ∝ \(\frac{1}{R}\) (for the given value of V). The resistance of Iron is more than that of copper for the given length and diameter. Hence in copper wire, the rate of heat production is more.
In parallel combination copper will glow first.

Question 14.
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance.
Answer:
Given three resistances are identical.
Hence R₁ = R₂ = R₃ = x (say)
Total resistance in parallel, R_p = \(\frac{\mathrm{R}}{3}\)
If three identical resistances are connected in parallel, then
\(\frac{1}{R_p}\) = \(\frac{1}{\mathrm{R}_1}\) + \(\frac{1}{\mathrm{R}_2}\) + \(\frac{1}{\mathrm{R}_3}\)
\(\frac{1}{\left(\frac{\mathrm{R}}{3}\right)}\) = \(\frac{1}{x}\) + \(\frac{1}{x}\) + \(\frac{1}{x}\) ⇒ \(\frac{3}{\mathrm{R}}\) = \(\frac{1+1+1}{\mathbf{x}}\)
∴ x = R.

Long Answer Questions

Question 1.
Under what condition is the heat produced in an electric circuit
a) directly proportional
b) inversely proportional to the resistance of the circuit ?
Compute the ratio of the total quantity of heat produced in the two cases.
Answer:
Expression of heat produced by electric current:
Consider a conductor AB of resistance R.

Let V = P.D applied across the ends of AB.
I = current flowing through AB.
t = time for which the current is flowing.
∴ Total charge flowing from A to B in time t is q = It. By definition of P.D, work done is carrying unit charge from A to B = V
Total work done in carrying a charge q from A to B is
W = V × q = V It = I² Rt
(∵ V = IR)
This work done is called electric work done. If this electric work done appears as heat, then amount of heat produced (H) is given by H = W = I² Rt Joule.
This is a statement of Joule’s law of heating.
a) If same current flows through an electric circuit, heat is developed.
i.e., H ∝ R.

b) If same P.D applied across the the electric circuit heat is developed.
i.e., H₂ ∝ \(\frac{1}{\mathrm{R}}\).

c) The ratio of H₁ and H₂ is given by
\(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = \(\frac{\mathrm{R}}{\frac{1}{\mathrm{R}}}\)
∴ \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = R²

Question 2.
Two metallic wires A and B are connected in parallel. Wire A has length L and radius r, wire B has a length 2L and radius 2r. Compute the ratio of the total resistance of the parallel combination and resistance of wire A.
Answer:
1) For metal Wire ‘A’
Length = L
Radius = r
Area = πr²
Resistance, R_A = \(\frac{\rho_{\mathrm{A}} \mathrm{L}}{\pi \mathrm{r}^2}\) —– (i)
Where ρ_A is specific resistance.
For metal Wire ‘B’
Length = 2L
Radius = 2r
Area = π(2r)² = 4πr²
Resistance, R_B = \(\frac{\rho_{\mathrm{B}}(2 \mathrm{~L})}{4 \pi \mathrm{r}^2}\) = \(\frac{\rho_{\mathrm{B}} \mathrm{L}}{2 \pi \mathrm{r}^2}\) —– (ii)
Where ρ_B is specific resistance.

2) Total resistance of wire A and wire B in parallel combination is given by

3)

4) The ratio of the total resistance parallel combination to resistance of wire A, is given by

5)
∴ \(\frac{R_p}{R_A}\) = \(\frac{\rho_{\mathrm{B}} \pi \mathrm{r}^2}{\mathrm{~L}\left(2 \rho_{\mathrm{A}}+\rho_{\mathrm{B}}\right)}\)

Question 3.
In a house three bulbs of 100W each are lighted for 4 hours daily and six tube lights of 20W each are lighted for 5 hours daily and a refrigerator of 400W is worked for 10 hours daily for a month of 30 days. Calculate the electricity bill if the cost of one unit is Rs. 4.00.
Answer:
No. of bulbs in a house, N = 3
Rated power on each bulb, P = 100 W
Time of lighted t = 4H
Energy consumption of 3 bulbs per day = \(\frac{\mathrm{Npt}}{1000}\) KWH
Energy consumption of 3 bulbs for 30 days = \(\frac{30 \mathrm{~N} \mathrm{Pt}}{1000}\) KWH
E_B = \(\frac{30 \times 3 \times 100 \times 4}{1000}\) = 36KWH
Similarly, Energy consumption of 6 tube lights for 30 days, ET = \(\frac{30 \times 6 \times 20 \times 5}{1000}\) KWH = 18 KWH.
And similarly, Energy consumption of one Refrigerator for 30 days,
E_R = \(\frac{30 \times 400 \times 10}{1000}\)KWH = 120 KWH
∴ The total energy consumption, E = E_B + E_T + E_R = (36 + 18 + 120) KWH
∴ E = 174 KWH = 174 units [∵ 1KWH = 1 unit]
Cost of 1 unit = Rs. 4.00/-
Cost of 174 units = No. of units × cost of 1 unit
= 174 × 4 = Rs. 696/-
∴ Electricity bill for one month of that house = Rs. 696/-

Question 4.
Three resistors of 4 ohms, 6 ohms and 12 ohms are connected in parallel. The combination of above resistors is connected in series to a resistance of 2 ohms and then to a battery of 6 volts. Draw a circuit diagram and calculate.
a) Current in main circuit.
b) Current flowing through each of the resistors in parallel.
c) p.d and the power used by the 2 ohm resistor.
Answer:
Circuit diagram for the given data is

a) Effective resistance when R₁, R₂ and R₃ are connected in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) ⇒ \(\frac{1}{R_p}\) = \(\frac{1}{4}\) + \(\frac{1}{6}\) + \(\frac{1}{12}\)
⇒ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{3+2+1}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{12}\)
∴ R_p = 2Ω
Total resistance in the circuit R = R_p + R₄ = 2 + 2 = 4Ω
∴ Current in mam circuit I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{6}{4}\) = 1.5A

b) Current flowing through R₁, I₁ = \(\frac{I R_P}{R_1}\) = \(\frac{1.5 \times 2}{4}\) = 0.75A
Current flowing through R₂, I₂ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_2}\) = \(\frac{1.5 \times 2}{6}\) = 0.5A
Current flowing through R₃, I₃ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_3}\) = \(\frac{1.5 \times 2}{12}\) = 0.25A

c) RD across 2Ω resistor (i.e., R₄), V₄ = IR₄ = 1.5 × 2 = 3 Volt.
Power used by 2Ω resistor, P = V₄I = 3 × 1.5 = 4.5 W.

Question 5.
Two lamps, one rated 100 Ω at 220 V and the other 60W at 220 V are connected in parallel to a 220 volt supply. What current is drawn from the supply line?
Answer:
Data for

Since R₁ and R₂ are connected in parallel effective resistance

Question 6.
Estimate the average drift speed of conduction electrons in a copper wire of cross – sectional area 3.0 × 10^-7 m² carrying a current of 5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³ and its atomic mass is 63.5 u.
Answer:
Given, Cross-sectional area of copper wire, A = 3 × 10^-7m²
carrying current of copper, I = 5A
Charge of electron, e = 1.6 × 10^-19C
Density of conduction electrons = No. of atoms per cubic meter,

∴ Average drift speed of conduction electrons.
V_d = \(\frac{\mathrm{I}}{\mathrm{neA}}\) = \(\frac{5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3 \times 10^{-7}}\)
⇒ V_d = \(\frac{5}{8.5 \times 1.6 \times 3 \times 10^2}\) = 0.1225 × 10^-2m/s
∴ V_d = 1.225 mm/s

Question 7.
Compare the drift speed obtained above with
i) Thermal speed of copper atoms at ordinary temperatures.
ii) Speed of propagation of electric field along the conductor which causes the drift motion.
Answer:
i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from

∴ drift speed of electron (V_d) = 1.047 × 10^-8
= 10^-8 times of thermal speed at ordinary temperature.

ii) The electric field travels along conductór with speed of EMW
C = 3 × 10⁸ m/s
V_d = 1.225 × 10^-3m/s
\(\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{C}}\) = \(\frac{1.225 \times 10^{-3}}{3 \times 10^8}\)
V_d = 0.408 × 10^-11 C
∴ Drift speed is, in compansion of C, extremely smaller by a factor of 10^-11.

Problems

Question 1.
A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there Is no change in its density on stretching, calculate the resistance of the stretched wire.
Solution:
Given R₁ = 10Ω,
l₁ = 1
l₂ = 3l, R₂ ?

R₁ = \(\frac{\rho}{\mathrm{V}} l_1^2\)
R₂ = \(\frac{\rho}{\mathrm{V}} l_2^2\)
R₃ = \(\left(\frac{l_2}{l_1}\right)^2\) ⇒ \(\frac{\mathrm{R}_2}{10}\) = \(\left(\frac{3 l}{l}\right)^2\)
∴ R₂ = 10 × 9 = 90Ω.

Question 2.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter? (A.P. Mar. ’19 & T.S. Mar. ’16, Mar. ’14)
Solution:
Resistance of long wire = 4R
Hence the resistance of half wire =
\(\frac{4 R}{2}\) = 2R

Now these two wire are connected in parallel. Hence the effective resistance between the ends of the diameter
R_P = \(\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}\) ⇒ R_p = \(\frac{2 \mathrm{R} \times 2 \mathrm{R}}{2 \mathrm{R}+2 \mathrm{R}}\)
∴ R_p = R.

Question 3.
Find the resistivity of a conductor which carries a current of density of 2.5 × 10⁶A m^-2 when an electric field of 15 Vm^-1 is applied across it.
Solution:
Given current density
J = \(\frac{1}{A}\) = 2.5 × 1o⁶ Am^-2
Applied electric field E = 15Vm^-1
Resistivity of conductor,
ρ = \(\frac{E}{J}\) = \(\frac{15}{2.5 \times 10^6}\)
∴ ρ = 6 × 10^-6Ωm.

Question 4.
What is the color code for a resistor of resistance 350mΩ with 5% tolerance ?
Solution:
Resistance of resistor = 350mΩ with 5% tolerance
= 350 × 10^-3Ω
= 35 × 10^-2Ω
First significant figure (3) indicates 1^st band
Second significant figure (5) indicates 2^nd band
Third significant figure (10^-2) indicates 3^rd band
We know that
0 1 2 3 4 5 6 7 8
B B R O Y of Great Britian has Very Good Wife wearing Gold silver Necklace
9 10^-1 ← 10^-2
3 indicates orange
5 indicates green
10^-2 indicates Silver
5% tolerance substance is Gold.
∴ Color code of given resistor is orange, green, silver, gold.

Question 5.
You are given 8Ω resistor. What length of wire of resistivity 120 Ωm should be joined in parallel with it to get a value of 6Ω ?
Solution:
Given, Resistance of resistor R = 8Ω

Resisty of wire ρ = 120
Let l length of the resistance x is to be connected to get effective resistance,
R_p = 6Ω
Then \(\frac{1}{\mathrm{R}}\) + \(\frac{1}{x}\) = \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\)
\(\frac{1}{8}\) + \(\frac{1}{x}\) = \(\frac{1}{6}\) ⇒ \(\frac{1}{x}\) = \(\frac{1}{6}\) – \(\frac{1}{8}\) = \(\frac{2}{48}\)
∴ x = 24Ω
And x l = ρ
24 l = 120
∴ l = 5m

Question 6.
Three resistors 3Ω, 6Ω and 9Ω are connected a battery. In which of them will the power dissipation be maximum if:
a) They all are connected in parallel
b) They all are connected in series ? Give reasons.
Solution:
Given R₁ = 3Ω, R₂ = 6Ω, R₃ = 9Ω
a) Effective resistance in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{9}\)
\(\frac{1}{R_p}\) = \(\frac{6+3+2}{18}\)
∴ R_P = \(\frac{18}{11} \Omega\)

∴ Dissipated power in parallel,
P_P ∝ \(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) ⇒ P_P ∝ \(\frac{1}{\left(\frac{18}{11}\right)}\) ∴ P_P ∝ \(\frac{11}{18}\) —– (1)

b) Effective resistance in series is given by R_s = R₁ + R₂ + R₃ = 3 + 6 + 9 = 18Ω
∴ Dissipated power in series,
P_S ∝ R_S ⇒ P_S ∝ 18 —- (2)
From equations (1) and (2) power dissipation is maximum in series and minimum in parallel.

Reasons:

1. In series connection, P ∝ R and V ∝ R. Hence dissipated power (P) and potential difference (V) is more because current is same across each resistor.
2. In parallel connection, P ∝ \(\frac{1}{R}\) and I ∝ \(\frac{1}{R}\). Hence dissipated power (P) and potential difference (V) is less because voltage is same across each resistor.

Question 7.
A silver wire has a resistance of 2.1Ω at 27.5°C and a resistance of 2.7Ω at 100°C. Determine the temperature coeff. of resistivity of silver.
Solution:
For silver wire, R₁ = 2.1Ω, t₁ = 27.5°C
R₂ = 2.7Ω, t₂ = 100°C, α = ?
α = \(\frac{R_2-R_1}{R_1 t_2-R_2 t_1}\) = \(\frac{2.7-2.1}{2.1 \times 100-2.7 \times 27.5}\)
= \(\frac{0.6}{210-74.25}\) = \(\frac{0.6}{135.75}\)
∴ Temperature coefficient of resistivity
∝ = 0.443 × 10^-2/°C

Question 8.
If the length of a wire conductor is doubled by stretching it while keeping the potential difference constant, by what factor will the drift speed of the electrons change ?
Solution:
Taking l₁ = l,
l₂ = 2l
Since V_d ∝ l, \(\frac{\mathrm{v}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{l_2}{l_1}\)
\(\frac{\mathrm{V}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{2 l}{l}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift speed of electrons changes by a factor 2.

Question 9.
Two 120V light bulbs, one of 25W and’ another of 200W are connected in series. One bulb burnt out almost instantaneously. Which one was burnt and why ?
Solution:
Given, For first bulb,
P₁ = 25W,
V₁ = 120V

Resistance of first bulb R₁ = \(\frac{\mathrm{v}_1^2}{\mathrm{P}_1}\)
R₁ = \(\frac{(120)^2}{25}\) —– (1)
For second bulb, P₂ = 200W, V₂ = 120V
Resistance of second bulb,
R₂ = \(\frac{(120)^2}{200}\) —— (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = 8 ⇒ R₁ = 8R₂
As R₁ > R₂, 25 W bulb burnt out almost instantaneously, since two bulbs have rated at same voltage.

Question 10.
A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance.
Solution:
Given, % change in length, \(\frac{\mathrm{d} l}{l}\) = 5%
Resistance of wire R = \(\frac{\rho l^2}{\mathrm{~V}}\)
% Change in Resistance of wire,
\(\frac{d R}{R}\) = 2\(\frac{\mathrm{d} l}{l}\) = 2 × 5% = 10%

Question 11.
Two wires A and B of same length and same material, have their cross sectional areas in the ratio 1:4. What would be the ratio of heat produced in these wires when the Voltage across each is constant ?
Solution:
Given l_A = l_B, ρ_A = ρ_B, V_A = V_B,
A_A : A_B = 1 : 4
Rate of heat produced in a wire,
H = i²R = \(\frac{\mathrm{V}^2}{\mathrm{R}^2}\) = \(\frac{V^2 A}{\rho l}\)
Since V, ρ, l are same for both wires A and B, H ∝ A (area of crossection)
For two wires A and B,
\(\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}\) = \(\frac{A_A}{A_B}\) = \(\frac{1}{4}\)
∴ H_A : H_B = 1 : 4.

Question 12.
Two bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation in these ?
Solution:
Given, R₁ : R₂ = 1 : 2, In parallel series
Dissipated power P = \(\frac{V^2}{R}\)
⇒ P ∝ \(\frac{\mathrm{I}}{\mathrm{R}}\) [ ∵ V = constant]
The ratio of dissipated powers in two bulbs is given by,
\(\frac{P_1}{P_2}\) = \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\) = \(\frac{2}{1}\)
∴ P₁ : P₂ = 2 : 1

Question 13.
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180cm of the potentiometer wire. (A.P. Mar. ’16)
Solution:
Length of potentiometer wire L = 5m
Potential difference V = 6 Volt
Potential gradient ϕ = \(\frac{\mathrm{V}}{\mathrm{L}}\) = \(\frac{6}{5}\) = 1.2 V / m
Balancing length l = 180cm
= 1.80m
Emf of the cell E = ϕl
= 1.2 × 1.8 = 2.16V

Question 14.
A battery of emf 2.5 V and internal resistance r is connected in series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r.
Solution:
Circuit diagram for the given data is shown below.

Given, E = 2.5 V; R = 4 5Ω;
r_A = 1A; I = 50mA;
r = ?
E = I (R + r_A + r)
2.5 = 50 × 10^-3 (45 + 1 + r)
46 + r = \(\frac{2.5}{50 \times 10^{-3}}\) = \(\frac{2.5 \times 10^3}{50}\) = 50
∴ r = 50 – 46 = 4Ω.

Question 15.
Amount of charge passing through the cross section of a wire is q(t) = at² + bt + c. Write the dimensional formula for a, b and c. If the values of a, b and c in SI unit are 6, 4, 2 respectively, find the value of current at t = 6 seconds.
Solution:
Charge passing through wire is given by q(t) = at² + bt + c
According to principle of homogenity, Dimensional formula of q(t) = dimensional formula of at²
IT = aT²
∴ Dimensional formula a = IT^-1
Dimensional formula of q(t) = Dimensional formula of bt
IT = bT
∴ Dimensional formula of b = I
Dimensional formula of q(t) = Dimensional formula of C.
IT = C
∴ Dimensonal formula of C = IT
Current, I = \(\frac{\mathrm{dq}(\mathrm{t})}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(at² + bt + c)
= 2at + b
Here a = 6 and b = 4
⇒ I = 12t + 4
∴ Current at t = 6 sec,
I = 12 × 6 + 4 = 76 A.

Textual Exercises

Question 1.
The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Solution:
Here E = 12 V, r = 0.4Ω
Maximum Current, I_max = \(\frac{\mathrm{E}}{r}\) = \(\frac{12}{0.4}\) = 30A

Question 2.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Solution:
Here E = 10 V, r = 3Ω, I = 0.5 A, R = ?, V = ?
I = \(\frac{E}{(R+r)}\) or (R + r) = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) = 20 or
R = 20 – r = 20 – 3 = 17Ω
Terminal voltage V = IR = 0.5 × 17 = 8.5 Ω.

Question 3.
a) Three resistors 1Ω, 2Ω, and 3Ω are combined in series. What is the total resistance of the combination ?
Solution:
Here R₁ = 1Ω, R = 2 Ω, R₃ = 3Ω, V = 12V
In series, total resistance R_S = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6Ω.

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
Current through the circuit I = V/Rs = 12/6 = 2A
∴ Potential drop across R₁ = IR₁ = 2 × 1 = 2V
Potential drop across R₂ = IR₂ = 2 × 2 = 4V
Potential drop across R₃ = IR₃ = 2 × 3 = 6V

Question 4.
a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination ?
Solution:
Here R₁ = 2Ω, R₂ = 4Ω, R₃ = 5Ω, V = 20V
In parallel combination total resistance R_P is given by
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) = \(\frac{10+5+4}{20}\) = \(\frac{19}{20}\) or R_P = \(\frac{20}{19} \Omega\)

b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Solution:
Current through R₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) = \(\frac{20}{2}\) = 10A
Current through R₂ = \(\frac{20}{4}\) = 5 A
Current through R₃ = \(\frac{20}{5}\) = 4A
Total current = \(\frac{20}{\left(\frac{20}{9}\right)}\) = 19A

Question 5.
At room temperature (27.0°C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^-4°C^-1.
Solution:
Here R₂₇ = 100Ω, R₁ = 117Ω, t = ? α = 1.70 × 10^-4°C^-1
We know that
α = \(\frac{R_t-R_{27}}{R_{27}(t-27)}\) or t – 27 = R_t – R₂₇
t = \(\frac{\mathrm{R}_1-\mathrm{R}_2}{\mathrm{R}_{27} \times \alpha}\) + 27 = \(\frac{117-100}{100 \times 1.7 \times 10^{-4}}\) + 27
= 1000 + 27 = 1027°C

Question 6.
A negligibly small current is passed through a wire of length 15m and uniform cross-section 6.0 × 10^-7 m², and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment ?
Solution:
Here L = 15 m, A = 6.0 × 10^-7m², R = 5.0 Ω, ρ = ?

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5°C, and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:
Here R_27.5 = 2.1Ω, R₁₀₀ = 2.7Ω; α = ?
α = \(\frac{\mathrm{R}_{100}-\mathrm{R}_{27.5}}{\mathrm{R}_{27.5} \times(100-27.5)}\) = \(\frac{2.7-2.1}{2.1 \times(100-27.5)}\) = 0.0039°C^-1

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4°C^-1.
Solution:

Question 9.
Determine the current in each branch of the network shown in Fig.

Solution:
The current’s through the various arms of the circuit have been shown in figure.
According to Kirchhoff’s second law;
-10 + 10 (i₁ + i₂) + 10i₁ + 5(i₁ – i₃) = 0
(or) 10 = 25i + 10i – 5i₃
(or) 2 = 5i₁ + 2i₂ – i₃
(or) 2 = 5i₁ + 2i₂ – i₃ ………. → (i)
In a closed circuit ABDA ~
10i₁ + 5i₃ – 5i₂ = 0
(or) 2i₁ + i₃ – i₂ = 0
(or) i₂ = 2i₁ + i₃ …….. → (ii)
In a closed circuit BCDB
5(i₁ – i₃) – 10 (i₂ + i₃) – 5i₃ = 0
(or) 5i₁ – 10i₂ = 20i₃ =0
i₁ = 2i₁ + 4i₃…….. → (iii)
From (ii) and (iii)
i₁ = 2 (2i₁ – i₃) + 4i₃ = 4i₁ + 6i₃
(or) 3i₁ = -6i₃
(or) i₁ = -2i₃
Putting this value in (ii) : i₂ = 2(-2i₃) i₃ = -3i₃
Putting values in (i)
2 = 5(-2i₃) + 2(-3i₃) – i₃ (or) 2 = -17i₃
(or) i₃ = -2/17A
From (iv) i₁ = -(-2/17) = -4/17A
from (v) i₂ = 3(-2/17) = 6/17A
i₁ + i₂ = (4/17) + (6/17) = (10/17)A
i₁ + i₃ = (4/17) + (-2/17) = (6/17) A
i₂ + i₃ = (6/17) + (-2/17) = 4/17A.

Question 10.
a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips ?

A meter bridge. Wire Ac is 1m long R is a resistance to be measured and S is a standard resistant
Solution:
Here l = 39.5cm, R = X = ?, S = Y = 12.5 Ω
As S = \(\frac{100-l}{l} \times \mathrm{R}\)
∴ 12.5 = \(\frac{100-39.5}{39.5} \times \mathrm{X}\)
or X = \(\frac{12.5 \times 39.5}{60.5}\) = 8.16Ω
Thick copper strips are used to minimise resistance of the connections which are not accounted in the formula.

b) Determine the balance point of the bridge above if X and Y are interchanged.
Solution:
As X and Y are interchanged therefore, l₁ and l₂ (i.e.) lengths are also interchanged.
Hence L = 100 – 39.5 = 60.5 cm.

c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ?
Solution:
The galvanometer will show no current.

Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit ?
Solution:
Here emf of the the battery = 8.0V; voltage of d.c. supply = 120V
Internal resistance of battery r = 0.5Ω; external resistance R = 15.5Ω
Since a storage battery of emf 8V is charged with a.d.c supply of 120 V the effective emf in the circuit is given by ε = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0Ω
∴ Current in the circuit during charging is given by
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{112}{16}\) = 7.0A
∴ Voltage across R = IR = 7.0 × 15.5 = 108.5 V
During charging the voltage of the d.c supply in a circuit must be equal to the sum of the voltage drop across R and terminal voltage of the battery
∴ 120 = 108.5 V or V= 120 – 108.5 = 11.5V
The series resistor limits the current drawn from the external source of d.c supply. In its absence the current will be dangerously high.

Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?
Solution:
Here ε₁ = 1.25 V, l₁ = 35.0 cm, ε₂ = ?.l₂ = 63.0 cm.
As \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{l_2}{l_1}\) or
ε₂ = \(\frac{\varepsilon_1 \times l_2}{l_1}\) = \(\frac{1.25 \times 63}{35}\) = 2.25V

Question 13.
The number density of free electrons in a copper conductor estimated in textual example 6.1 is 8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0m long to its other end ? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Solution:
Here n = 8.5 × 10²⁸ m^-3; L = 3.0 m; A = 2.0 × 10^-6 m²; I = 3.0A, t = ?
As I = n A eVd
∴ V_d = \(\frac{1}{\mathrm{nAe}}\)
Now, t = \(\frac{1}{\mathrm{v}_{\mathrm{d}}}\)
=
= \(\frac{3.08 \times 8.5 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
= 2.72 × 10⁴ S
= 7hour 33 minutes.

Additional Exercises

Question 1.
The earth’s surface has a negative surface charge density of 10^-9 C m^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶m).
Solution:
Here r = 6.37 × 10⁶ m; Q = 10^-9 cm²; I = 1800 A
Area of the globe A = 4πr² = 4 × 3.14 × (6.37 × 10⁶)²
= 509.64 × 10³C
t = \(\frac{Q}{I}\) = \(\frac{509.64 \times 10^3}{1800}\) = 283.1 S

Question 2.
a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω are joined in series to provide a supply to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
Solution:
Here ε = 2.0V; n = 6; r = 0.015Ω; R = 8.5 Ω
Current I = \(\frac{n E}{R+n r}\) = \(\frac{6 \times 2.0}{8.5+6 \times 0.015}\) = 1.4A
Terminal voltage,V = IR = 1.4 × 8.5 = 11.9V.

b) A secondary cell after long use has an emf 011.9 V and a large internal resistance of 380Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car?
Solution:
Here E = 1.9 V; r = 380Ω
I_max = \(\frac{\varepsilon}{\mathrm{r}}\) = \(\frac{1.9}{380}\) = 0.005A
This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds.

Question 3.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ_Al = 2.63 × 10^-8 Ωm, ρ_Cu = 1.72 × 10^-8 Ωm, Relative density of Al = 2.7, of Cu = 8.9.)
Solution:
Given, for aluminium wire; R₁ = R; l₁ = l
Relative density d₁ = 2.7.
For copper wire R₂ = R, t₂ = 1, d₂ = 8.9
Let A₁, A₂ be the area of cross section for aluminium wire and copper wire.
We know, R₁ = \(\rho_1 \frac{l_1}{\mathrm{~A}_1}\) = \(\frac{2.63 \times 10^{-8} \times l}{A_1}\)
and mass of the aluminium wire m₁ = A₁l₁ × d₁ = A₁l₁ × 2.7
R₂ = ρ₂ = \(\frac{l_2}{\mathrm{~A}_2}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\)
Mass of copper wire m₂ = A₂l₂ × d₂ = A₂l × 8.9
Since two wires are of equal resistance R₁ = R₂
\(\frac{2.63 \times 10^{-8} \times 1}{\mathrm{~A}_1}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\) or \(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\) = \(\frac{1.72}{2.63}\)
from (ii) and (iv) we have
\(\frac{\mathrm{m}_2}{\mathrm{~m}_1}\) = \(\frac{\mathrm{A}_2 l \times 8.9}{\mathrm{~A}_1 l \times 2.7}\) = \(\frac{8.9}{2.7} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}\)
= \(\frac{8.9}{2.7} \times \frac{1.72}{2.63}\) = 2.16

It shows that copper wire is 2.16 times heavier than aluminium wire since for the same value of length and resistance aluminium wire has lesser mass than copper wire, therefore aluminium wire is preferred for overhead power cables. A heavy cable may sag down owing to its own weight.

Question 4.
What conclusion can you draw from the following observations on a resistor made of alloy manganin ?

Answer:
Since the ratio of voltage and current for different readings is same so ohm’s law is valid to high accuracy. The resistivity of the alloy manganin is nearly independent of temperature.

Question 5.
Answer the following Questions.
a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed ?
Solution:
Only current through the conductor of non-uniform area of cross section is constant as the remaining quantities vary inversly with the area of cross-section of the conductor.

b) Is Ohm’s law.universally applicable for all conducting elements ? If not, give examples of elements which do not obey Ohm’s law.
Solution:
Ohm’s law is not applicable for non-ohmic elements. For example, vaccum tubes, semi-conducting diode, liquid electrolyte etc.

c) A low voltage supply from which one needs high currents must have very low internal resistance. Why ?
Solution:
As, I_max = emf internal resistance so for maximum current internal resistance should be least.

d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why ?
Solution:
A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.

Question 6.
Choose the correct alternative :
a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).
Solution:
a) Greater
b) Lower
c) Nearly independent
d) 10²²

Question 7.
a) Given n resistors each of resistance R, how will you combine them to get the
i) maximum,
ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance ?
Solution:
For maximum effective resistance the n resistors must be connected in series.
Maximum effective resistance R_S = nR
For minimum effective resistance the n resistors must be connected in parallel.
Maximum effective resistance R_P = R/n
∴ \(\frac{R_S}{n p}\) = \(\frac{\mathrm{nR}}{\mathrm{R} / \mathrm{n}}\) = n²

b) Given the resistances of 1Ω, 2Ω, 3Ω, how will be combine them to get an equivalent resistance of
(i) (11/3) Ω
(ii) (11/5) Ω,
(iii) 6Ω,
(iv) (6/11)Ω?
Solution:
It is to be noted that
a) the effective resistance of parallel combination of resistors is less than the individual resistance and
b) the effective resistance of series combination of resistors is more than individual resistance.

case (i) Parallel combination of 1Ω and 2Ω is connected in series with 3Ω.
Effective resistance of 1Ω and 2Ω in parallel will be given by
RP = \(\frac{1 \times 2}{1+2}\) = \(\frac{2}{3} \Omega\)
∴ Equivalent resistance of \(\frac{2}{3} \Omega\) and 3] and 3Ω in series
= \(\frac{2}{3}\) + 3 = \(\frac{11}{3}\)Ω

Case(ii) : Parallel combination of 2Ω and 3Ω is connected in series with 1Ω.
Equivalent resistance of 2Ω and 3Ω in parallel
= \(\frac{2 \times 3}{2+3}\) = \(\frac{6}{5} \Omega\)
Equivalent resistance of \(\frac{6}{5} \Omega\) and 1Ω in series = \(\frac{6}{5}\) + 1 = \(\frac{11}{5}\)Ω

Case (iii) : All the resistances are to be connected in series now
∴ Equivalent resistance = 1 + 2 + 3 = 6Ω

Case (iv) : All the resistances are to be connected in parallel
∴ Equivalent resistance (R) is given by
\(\frac{1}{\mathrm{R}}\) = \(\frac{1}{1}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\)
= \(\frac{6+3+2}{6}\) = \(\frac{11}{6}\) (Or) r = \(\frac{6}{11} \Omega\)

c) Determine the equivalent resistance of networks shown in Fig.

Answer:
a) The given network is a series combination of 4 equal units. Each unit has 4 resistances in which 2 resistances (1Ω each in series) are in parallel with 2 other resistances (2Ω each in series).
∴ Effective resistances of two resistances (each of 1Ω) in series = 1 + 1 = 2Ω.
Effective Resistance of two resistances (each of 2Ω) in series = 2 + 2 = 4Ω
If R is the resistance of one unit of resistances then
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\) or R_P = \(\frac{4}{3} \Omega\)
∴ Equivalent resistance in network = 4 R_P = 4 × \(\frac{4 \Omega}{3}\) = \(\frac{16 \Omega}{3}\)

b) Total resistances each of value R are connected in series. Their effective resistance = 5R.

Question 8.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the Infinite network shown in Fig. Each resistor has 1Ω resistance.

Solution:
Let x be the equivalent résistance of infinite network. Since the net work is infinite, therefore, the addition of one more unit of three resistances each of value of 1Ω across the terminals will not alter the total resistance of network i.e. it should remain x.

Therefore, the network would appear as shown in the figure and its total resistance should remain x.
There the parallel combination of x and 1Ω is in series with two resistors of 1Ω each.
The resistance of parallel combination is

\(\frac{1}{\mathbf{R}_{\mathbf{P}}}\) = \(\frac{1}{x}\) + \(\frac{1}{1}\) = \(\frac{1+x}{x}\)
R_p = \(\frac{x}{(1+x)}\)
∴ Total resistance of network will be given by
x = 1 + 1 + \(\frac{x}{x+1}\) = 2 + \(\frac{x}{x+1}\)
x(x + 1) = 2(x + 1) + x
or x² + x = 2x + 2 + x or x² – 2x – 2 = 0
or x = \(\frac{2 \pm \sqrt{4+8}}{2}\) = \(\frac{2 \pm \sqrt{12}}{2}\)
Total resistance of the circuit shows a full scale deflection for a current of 2.5 mA. How will you convert the meter into
= \(\frac{2 \pm 2 \sqrt{3}}{2}\) = 1 ± \(\sqrt{3}\)
The value of resistance cannot be negative, therefore the resistance of network
= 1 + \(\sqrt{3}\) = 1 + 1.73 Ω = 2.73 Ω
Total resistance of the cfrcuit = 2.73 + 0.5
= 3.23Ω
∴ Current draw I = \(\frac{12}{3.23}\) = 3.72 amp

Question 9.
Figure shows a potentiometer with a cell of 2.0’V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

a) What is the value ε ?
Solution:
Here ε₁ = 1.02 V, L₁ = 67.3cm, ε₂ = e = ?, L₂ = 82.3 cm
Since \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\)
∴ ε = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1} \times \varepsilon_1\) = \(\frac{82.3}{67.3} \times 1.02\) = 1.247V

b) What purpose does the high resistance of 600 kΩ have ?
Answer:
The purpose of using high resistance of 600k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.

c) Is the balance point affected by this high resistance ?
Answer:
No, the balance point is not affected by the presence of this resistance.

d) Is the balance point affected by the internal resistance of the driver cell ?
Answer:
No, the balance point is not affected by the internal resistance of the driver cell.

e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V ?
Answer:
No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f of the driver cell is less than the emf of the other cell.

f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple) ? If not, how will you modify the circuit ?
Answer:
The circuit will not work for measuring extremely small emf because in that case the balance point will be just close to the end A. To modify the circuit we have to use a suitable high resistance in series with the cell of 2.0V This would decrease the current in the potentiometer wire. Therefore potential difference 1cm df wire will decrease. Hence extremely small emf can be measured.

Question 10.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?

Answer:
Here L₁ = 58.3cm; L₂ = 68.5 cm; R = 10Ω; I X = ?. Let be the current in the potentiometer wire and ε₁ and ε₂ be the potential drops across R and X respectively when connected in circuit by closing respective keky. Then

If there is no balance point with given cell of emf it means potential drop across R or X greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced which is possible by reducing the current in R and X for that either suitable resistance should be put in series with R and X or a cell of smaller emf E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.

Question 11.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Solution:
Here l₁ = 76.3 cm, l₂ = 64.8 cm.
r = ?, R = 9.5 Ω
Now, r = \(\left(\frac{l_1-l_2}{l_2}\right) R\) = \(\left(\frac{76.3-64.8}{64.8}\right)\) 9.5 = 1.68 Ω

Textual Exercises

Question 1.
a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10^-7 m² carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³, and its atomic mass is 63.5 u.
b) Compare the drift speed obtained above with,
i) thermal speeds of copper atoms at ordinary temperatures,
ii) speed of propagation of electric field along the conductor which causes the drift motion.
Solution:
a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed ud is given by Eq.
IΔt = + neA /v_d/Δt
V_d = (I/neA)
Now, e = 1.6 × 10^-19 C, A = 1.0 × 10^-7 m², I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic meter (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of ‘copper has a mass of 9.0 × 10³ kg. Since 6.0 × 10²³ copper atoms have a mass of 63.5 g,
n = \(\frac{6.0 \times 10^{23}}{63.5}\) × 9.0 × 10⁶
= 8.5 × 10²⁸ m^-3 Which gives,
\(v_{\mathrm{d}}\) = \(\frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10 \times 10^{-7}}\)
= 1.1 × 10^-3 m s^-1 = 1.1 mm s^-1

b) i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from
[<(1/2) Mυ² > = (3/2) K_BT] and is thus typically of the order of \(\sqrt{\mathrm{k}_{\mathrm{B}} \mathrm{T} / \mathrm{M}}\), where K_B is the Boltzmann constant. For copper at 300 K, this is about 2 × 10² m/s. This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about 10^-5 times the typical thermal speed at ordinary temperatures,

ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 × 10⁸ m s^-1. The drift speed is, in comparison, extremely small, smaller by a factor of 10^-11.

Question 2.
a) In Textual Example 1, the electron drift speed is estimated to be only a few mm s^-1 for currents in the range of a few amperes ? How then is current established almost the instant a circuit is closed ?
b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed ?
c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor ?
d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction ?
e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the i) absence of electric field,
ii) presence of electric field ?
Solution:
a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However it does take a little while for the current to reach its steady value.
b) Each ‘free’ electron does accelerate, increasing its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
c) Simple, because the electron number density is enormous, ~10²⁹ m^-3.
d) By no means. The drift velocity is superposed over the large random velocities of electrons.
e) In the absence of electric field, the paths are straight lines, in the presence of electric field, the paths are, in general curved.

Question 3.
An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element ? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10^-4 °C^-1.
Solution:
When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
R₂ = \(\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}\) = 85.8Ω
Using the relation
R₂ = R₁ [1 + α(T₂ – T₁)]
with α = 1.70 × 10^-4°C^-1, we get
T₂ – T₁ = \(\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}\) = 820°C
that is, T₂ = (820 + 27.0)°C = 847 °C

Question 4.
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.39 Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath.
Solution:
R₀ = 5Ω, R₁₀₀ = 5.23 Ω and R_t = 5.795Ω
Now, t = \(\frac{R_t-R_0}{R_{100}-R_0} \times 100\), R_t = R₀ (1 + αt)
= \(\frac{5.795-5}{5.23-5} \times 100\)
= \(\frac{0.795}{0.23} \times 100\) = 345.65°C

Question 5.
A network of resistors is connected to a 16 V battery with internal resistance of 1Ω, a shown in Fig.
a) Compute the equivalent resistance of the network.
b) Obtain the current in each resistor.
c) Obtain the voltage drops V_AB, V_BC and V_CD.

Solution:
a) The network is a simple series and parallel combination of resistors. First the two 4Ω resistors in parallel are equivalent to a resistor
= [(4 × 4)/(4 + 4)]Ω = 2Ω
In the same way, the 12Ω and 6Ω resistors in parallel are equivalent to a resistor of
[(12 × 6)/(12 + 6)]Ω = 4Ω
The equivalent resistance R of the network is obtained by combining these resistors (2Ω and 4Ω) With 1Ω in series, that is,
R = 2Ω + 4Ω + 1Ω = 7Ω

b) The total current I in the circuit is
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{16 \mathrm{~V}}{(7+1) \Omega}\) = 2A
Consider the resistors between A and B. If I₁ is the current in one of the 4 Ω resistors and I₂ the current in the other.
I₁ × 4 = I₂ × 4
That is, I₁ = I₂, which is otherwise obvious from the symmetry of the two arms. But I₁ + I₂ = I = 2A. Thus,
That is, current in each 4Ω resistor is 1 A. Current in lfi resistor between B and C would be 2 A.
Now, consider the resistances between C and D. If I₃ is the current in the 12Ω resistor, and I₄ in the 6Ω resistor,
I₃ × 12 = I₄ × 6 i.e., I₄ = 2I₃
But, I₃ + I₄ = I = 2A
Thus, I₃ = \(\left(\frac{2}{3}\right)\)A, I₄ = \(\left(\frac{4}{3}\right) \mathrm{A}\)
That is, the current in the 12Ω resistor is (2/3)A, while the current in the 6Ω resistor is (4/3) A.

c) The voltage drop across AB is
V_AB = I₁ × 4 = 1 A × 4Ω = 4V
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is,
V_A = 2A × 2Ω = 4V
The voltage drop across BC is
V_BC = 2A × 1Ω = 2V
Finally, the voltage drop across CD is,
V_CD = 12Ω × I₃ = 12Ω × \(\left(\frac{2}{3}\right) \mathrm{A}\) = 8V.
This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D, that is.
V_CD = 2A × 4Ω = 8V
Note that the total voltage drop across AD is 4V + 2V + 8V = 14V. Thus, the terminal voltage of the battery is 14V, while its emf is 16V The loss of the voltage (= 2V) is accounted for by the internal resistance ID of the battery [2A × 1Ω = 2V].

Question 6.
A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω Fig. Determine the equivalent resistance of the network and the current along each edge of the cube.
Solution:

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths AA’. AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A’, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC’EA, and apply Kirchhoff’s second rule :
-IR – (1/2)IR – IR + ε = 0
where R is the resistance of each edge and ε the emf of battery. Thus, ε = \(\frac{5}{2}\)IR
The equivalent resistance R_eq of the network is R_eq = \(\frac{\varepsilon}{3 I}\) = \(\frac{5}{6}\)R
For R = IΩ, R,sub>eq = (5/6) Ω and for ε = 10V, the total current (=3I) in the network is 3I = 10V/(5/6) p = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the Fig.

Question 7.
Determine the current in each branch of the network shown in Fig.

Solution:
Each branch of the network is assigned an unknown current to be determined by the application of Krichhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknown I₁, I₂ and I₃ which can be found by applying the second rule of Krichhoff to three different closed loops. Kirchhoff s second rule for the closed loop ADCA gives,
10 – 4(I₁ – I₂) + 2 (I₂ + I₃ – I₁) – I₁ = 0
that is, 7I₁ – 6I₂ – 2I₃ = 10 ——-> (1)
For the closed loop ABCA, we get
10 – 4I₂ – 2 (I₂ + I₃) – I₁ = 0
that is, I₁ + 6I₂ + 2I₃ = 10 —-—> (2)
For the closed loop BCDEB, we get
5 – 2 (I₂ + I₃) – 2 (I₂ + I₃ – I₁) = 0
that is, 2I₁ – 4I₂ – 4I₃ = -5 ——–> (3)
Equations (1, 2, 3) are three simultaneous equations in three unknowns. These can be solved by the usual method to give.
I₁ = 2.5A, I₂ = \(\frac{5}{8}\)A, I₃ = 1\(\frac{7}{8}\)A
The currents in the various branches of the network are
AB : \(\frac{5}{8}\) A, CA : 2\(\frac{1}{2}\) A, DEB : 1\(\frac{7}{8}\) A
AD = 1\(\frac{7}{8}\) A, CD : 0 A, BC : 2\(\frac{1}{2}\) A

It is easily verified that Krichhoffs second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
5V + \(\left(\frac{5}{8} \times 4\right)\) – \(\left(\frac{15}{8} \times 4\right) \mathrm{V}\)
equal to zero, as required by Krichhoffs second rule.

Question 8.
The four arms of a Wheatstone bridge (Fig.) have the following resistances :
AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution:
Considering the mesh BADB, we have
100I₁ + 15I_g – 60I₂ = 0
or 20I₁ + 3I_g – 12I₂ = 0 —–> (1)
Considering the mesh BCDB, we have
10(I₁ – I_g) – 15I_g – 5(I₂ + I_g) = 0
10I₁ – 30I_g – 5I₂ = 0
2I₁ – 6I_g – I₂ = 0 ——> (2)
Considering the mesh ADCEA,
60I₂ + 5(I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2 —–> (3)
Multiplying equation (2) by 10 ‘
20I₁ + 60I_g – 10I₂ = 0
From Equations. (4) and (1) we have  ——> (4)
63I_g + 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into Equation (3) we get.
13(31.5I_g) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA.

Question 9.
In a metre bridge (Fig.), the null point is found at a distance of 36.7 cm from A. If now a resistance of 12Ω is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Solution:
From the first balance point, we get
\(\frac{\mathrm{R}}{\mathrm{S}}\) = \(\frac{33.7}{66.3}\) —–> (1)
After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S_eq, where
S_eq = \(\frac{12 \mathrm{~S}}{\mathrm{~S}+12}\) —–> (2)
and hence the new balance condition now gives
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}\) = \(\frac{R(S+12)}{12 S}\)
Substituting the value of R/S from Equation (1), we get
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{S}+12}{12}\) . \(\frac{33.7}{66.3}\)
Which gives S = 13.5 Ω. Using the value of R/S above, we get R = 6.86 Ω.

Question 10.
A resistance of R Ω draws current from a potentiometer. The potentiometer has a total resistance R₀ Ω. (Figure) A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Solution:
While the slide is in the middle of, the potentiometer only half of its resistance (R₀/2) will be between the points A and B. Hence, the total resistance between A and B, say, R₁, will be given by the following expression.
\(\frac{1}{\mathrm{R}_1}\) = \(\frac{1}{R}\) + \(\frac{1}{\left(\mathrm{R}_0 / 2\right)}\)
R₁ = \(\frac{\mathrm{R}_0 \mathrm{R}}{\mathrm{R}_0+2 \mathrm{R}}\)
The total resistance between A and C will be süm of resistance between A and B and B and C, i.e., R₁ + R₀/2
∴ The current flowing through the potentiometer will be
I = \(\frac{\mathrm{V}}{\mathrm{R}_1+\mathrm{R}_0 / 2}\) = \(\frac{2 \mathrm{~V}}{2 \mathrm{R}_1+\mathrm{R}_0}\)
The voltage V₁ taken from the potentiometer will be the product of current I and resistance R₁,

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance

Very Short Answer Questions

Question 1.
Can there be electric potential at a point with zero electric intensity? Give an example.
Answer:
Yes, There can exist potential at a point where the electric intensity is zero.

Ex :

1. Between two similar charges intensity of electric field is zero. But potential is not zero.
2. Inside the charged spherical conductor electric field intensity is zero but potential is not zero.

Question 2.
Can there be electric intensity at a point with zero electric potential ? Give an example.
Answer:
Yes, electic intensity need not be zero at a point where the potential is zero.

Ex :
1) At mid point between two equal opposite charges potential is zero. But intensity is not zero.

Question 3.
What are meant by equipotential surfaces ?
Answer:
Surface at every point of which the value of potential is the same is defined as equipotential surface
For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces.

Question 4.
Why is the electric field always at right angles to the equipotential surface ? Explain.
Answer:
No work is done in moving a charge from one point on equipotential surface to the other. Therefore, component of electric field intensity along the equipotential surface is zero. Hence, the surface is perpendicular to the field lines.

Question 5.
Three capacitors of capacitances 1μF, 2μF and 3μF, are connected in parallel
(a) What is the ratio of charges ?
(b) What is the ratio of potential difference ?
Answer:
When capacitors are connnected in parallel
(a) q₁ : q₂ : q₃ = V: C₂ V: C₃ V = 1μF : 2μF : 3μF

∴ q₁ : q₂ ; q₃ = 1 : 2 : 3
(b) V₁ : V₂ : V₃ = V : V : V = 1 : 1 : 1

Question 6.
Three capacitors of capacitances 1μE, 2μF and 3μF are connected in series
(a) What is the ratio of charges ?
(b) What is the ratio of potential differences ?
Answer:
When capacitors are connected in series

(a) q₁ : q₂ : q₃ = q : q : q = 1 : 1 ; 1
(b) V₁ : V₂ : V₃ = \(\frac{\mathrm{q}}{\mathrm{C}_1}: \frac{\mathrm{q}}{\mathrm{C}_2}: \frac{\mathrm{q}}{\mathrm{C}_3}\) = \(\frac{1}{1}: \frac{1}{2}: \frac{1}{3}\)
∴ V₁ : V₂ : V₃ = 6 : 3 : 2

Question 7.
What happens to the capacitance of a parallel plate capacitor if the area of its plates is doubled ?
Answer:
\(\frac{C_2}{C_1}\) = \(\frac{A_2}{A_1}\) [∵ C₂ = 2C₁]
Given A₂ = 2A₁ ; \(\frac{C_2}{C_1}\) = \(\frac{2 \mathrm{~A}_1}{\mathrm{~A}_1}\) [∴ C₂ = 2C₁]
Therefore capacity increases by twice.

Question 8.
The dielectric strength of air is 3 × 10⁶.Vm^-1 at certain pressure, A parallel plate capacitor with air in between the plates has a plate separation of 1 cm. Can you charge the capacitor
to 3 × 10⁶V?
Answer:
Dielectric strength of air E₀ = 3 × 10⁶ Vm^-1
Electric field intensity between the plates, E = \(\frac{E_0}{K}\) = 3 × 10⁶ Vm^-1 [∵ for air K = 1]
Distance between two plates, d = 1 cm = 102m
Electric potential difference between plates, V = Ed = 3 × 10⁶ × 10^-2
∴ V = 3 × 10⁴ Volt.
Hence we cant charge the capacitor upto 3 × 10⁶ Volt.

Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge. (T.S. Mar. ’16)
Answer:
Expression for the electric potential due to a point charge:

1. Electric potential at a point is defined as the amount of workdone in moving a unit +ve charge from infinity to that point.
   
2. Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
3. Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, Whereas the potential difference is +ve in, the direction A to B.
4. Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P
   

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Expression for the electrostatic potential energy of a system of two point charges:

1. Let two point charges q₁ and q₂ are separated by distance ‘r’ in space.
2. An electric field will develop around the charge q₁
   
3. To bring a charge q2 from infinity to the point B some work must be done.
   workdone = q₂ V_B
   But V_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1}{\mathrm{r}}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
4. This amount workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.
   ∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
5. If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
6. Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Expression for potential energy of an electric dipole placed in a uniform electric field:

1. Consider a electric dipole of length 2a having charges + q and -q.
2. The electric dipole is placed in uniform electric field E and its axis makes an angle θ with E.
   
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   Torque \(\tau\) = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}}\) ⇒ BC = 2a sinθ
   ∴ Torque \(\tau\) = qE × 2a sinθ = PE sin θ [∴ p = 2aq]
4. Suppose the dipole is rotated through an angle dθ, the workdone dw is given by
   dw = tdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂,
   workdone W = \(\int_{\theta_1}^{\theta_2} \mathrm{PE} \sin \theta d \theta\) = PE(cosθ₁ – cosθ₂)
6. This workdone (W) is then stored as potential energy(U) in the dipole.
   ∴ U = PE(cosθ₁ – cosθ₂)
7. If θ₁ = 90°and θ₂ = 0°, U = -PE cosθ.
   In vector form U = \(-\vec{P} \cdot \vec{E}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. (Mar.’16 (AP) Mar ’14)
Answer:
Expression for the capacitance of a parallel plate capacitor:

1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is + q and surface charge density of charge = σ
   ∴ q = Aσ
   
4. The electric intensity at poiñt x, E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   
6. Capacitance of the capacitor Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathbf{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field. (A.P. Mar. ’19)
Answer:

1) When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts

opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.

2) Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.

3) If E₀ the external field strength and E₁ is the electric field strength induced, then the net field \(\overrightarrow{\mathrm{E}}_{\text {net }}\) = \(\overrightarrow{\mathrm{E}}_0\) + \(\overrightarrow{\mathrm{E}}_1\)
(E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Long Answer Questions

Question 1.
Define electric potential. Derive and expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.
Answer:
Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.
Expression for the potential at a point due to a dipole:

1. Consider A and B having -q and + q charges separated by a distance 2a.
2. The electric dipole moment P = q × 2a along AB
3. The electric potential at the point ‘P’ is to be calculated.
4. P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
5. BN and AM are perpendicular to OP.
6. Potential at P due to charge + q at B,
   
7. Potential at P due to charge -q at A, V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{AP}}\right]\)
   ∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{MP}}\right]\) [∵ BP = NP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{N P}-\frac{q}{M P}\right]\) …… (1)
9. In Δle ONB, ON = OB cosθ = a cosθ; ∴ NP = OP – ON = r – a cosθ ….. (2)
10. In Δle AMO, OM = AO cosθ = a cosθ; ∴ MP = MO + OP = r + a cos θ ….. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathbf{r}^2}\)
13. Electric potential on the axial line of dipole :

(i) When θ = 0°, point p lies on the side of + q
∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]

ii) When θ = 180°, point p lies on the side of -q.
∴ V = \(\frac{-\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

b) Electric potential on the equitorial line of the diopole:
when θ = 90°, point P lies on the equitorial line.
∴ V = o [∵ cos 90° = 0]

Question 2.
Explain series and parallel combination of capacitors. Derive the formula for equivalent capacitance in each combination. (AP. & T.S. Mar. ‘15)
Answer:
Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.

In this combination

1. Charge on each capacitor is equal.
2. PD’s across the capacitors is not equal.

Consider three capacitors of capacitanceš C₁, C₂ and C₃ are connected in series across a battery of P.D ‘V’ as shown in figure.

Let ‘Q’ be the charge on each capacitor.
Let V₁, V₂ and V₃ be the P.D’s of three
V = V₁ + V₂ + V₃ —— (1)
P.D. across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
P.D. across II^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across III^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1), V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_2}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_3}\) = Q\(\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)

For ‘n’ number of capacitors, the effective capacitance can be written as

Parallel Combination : The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,

1. The PD’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal. Consider three capacitors of capacitancé C₁, C₂ and C₃ are connected in parallel across a RD ‘V’ as shown in fig.

The charge on Ist capacitor Q₁ = C₁ V
The charge on IInd capacitor Q₂ = C₂ V
The charge on IIIrd capacitor Q₂ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{Q}{V}\) = C₁ + C₂ + C₃
C = C₁ + C₂ + C₃ [ ∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]
for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as C = C₁ + C₂ + C₃ + …. + C_n

Question 3.
Derive an expression for the energy stored in a capacitor. What is the energy stored when the space between the plates is filled with a dielectric
(a) with charging battery disconnected?
(b) with charging battery connected in the circuit?
Answer:
Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’
∴ Average potential difference V_A = \(\frac{\mathrm{O}+\mathrm{V}}{2}\) = \(\frac{\mathrm{V}}{2}\)
Hence workdone to move the charge Q = W = V_A × Q = \(\frac{\mathrm{VQ}}{2}\)
This is stored as electrostatic potential energy U’
∴ U = \(\frac{\mathrm{QV}}{2}\)
We know Q = CV then ‘U’ can be written as given below.
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Effect of Dielectric on energy stored :

Case (a) : When the charging battery is disconnected from the circuit:
Let the capacitor is charged by a battery and the disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{K}\) times and charge remains constant.
Capacity increases by ‘K times

∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b): When the charging battery is connected in the circuit:
Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V ,
New charge on the plates Q’ = KQ
Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC
Energy stored in the capacitor U’ = \(\frac{1}{2}\)Q’V = \(\frac{1}{2}\)(KQ) V = KU
U’ = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
An elementary particle of mass ‘m’ and charge +e initially at a very large distance is projected with velocity ‘v’ at a much more massive particle of charge + Ze at rest. The closest possible distance of approach of the incident particle is
Solution:
For an elementary particle, mass = m; charge = +e; velocity = v.
For much more massive particle, charge = + Ze
From law of conservation of energy, we have
K.E of elementary particles = Electrostatic potential energy of elementary particle at a closest distance (d)

∴ The closest possible distance of approach of the incident particle,
d = \(\frac{Z e^2}{2 \pi \varepsilon_0 m v^2}\)

Question 2.
In a hydrogen atom the electron and proton are at a distance of 0.5 A. The dipole moment of the system is
Answer:
In a hydrogen atom the charge of an electron = -1.6 × 10^-19C

In a hydrogen atom the charge of proton,
q_p = +1.6 × 10^-19C
The distance between the proton and an electron
2a = 0.5A = 0.5 × 10^-10 m
The dipole moment of the system,
P = 2a × q_p = 0.5 × 10^-10 × 1.6 × 10
∴ P = 8 × 10^-30cm

Question 3.
There is a uniform electric field in the XOY plane represented by (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is
Answer:
Given, uniform Electric field intensity,
\(\overrightarrow{\mathrm{E}}\) = (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1
Electric potential at the origin = 200V
Position vector d\(\begin{aligned}
&\rightarrow \\
&\mathrm{r}
\end{aligned}\) = (2\(\hat{i}\) + 1\(\hat{j}\)) m
We know that,

V_p – V_o = -(80 + 30) = -110Volt
V_p = V_o – 110 = (200 – 110) Volt = 90 Volt
∴ potential at point P, V_p = 90Volt.

Question 4.
An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is
Answer:
Charge at the centroid of an equilateral triangle = +q
The charge + q divides the line segment in ratió 2 : 1.
That means r_max = 2 and r_min = 1
V_min = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_{\max }}\) and V_max = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_{\min }}\)

Question 5.
ABC is an equilateral triangle of side 2m. There is a uniform electric field of intensity 100V/m in the plane of the triangle and parallel to BC as shown. If the electric potential at A is 200 V, then the electric potentials at B and C.
Answer:
Given length of side of an equilateral triangle a = 2m
E = 100V/m; V_A = 200V
Let D be the mid point between B and C Potential at D = V_D = 200V

From fig V_B – V_D = Ed
⇒ V_B – 200 = 100 × 1
∴ Potential at B, V_B = 300 V And V_D – V_C = Ed
200 – V_C = 100 × 1
∴ Potential at C, V_C = 100V.

Question 6.
An electric dipole of moment p is placed in a uniform electric field E, with p parallel to E. It is then rotated by an angle q. The work done is
Solution:
Let AB be a electric dipole having charges -q and + q
Electric dipole moment of AB = p
Electric field = E The workdone by a dipole, when it is rotated through an angle q from E,

W = \(\int_0^q p E \sin \theta d \theta\)
⇒ W = pE \([\cos \theta]_0^q\) = pE (cos0° – cosq)
∴ W = pE(1 – cosq)

Question 7.
Three identical metal plates each of area ‘A’ are arranged parallel to each other, ’d’ is the distance between the plates as shown. A battery of V volts is connected; as shown. The charge stored in the system of plates is

Answer:
Area of each plate = A
distance between two plates = d capacity of each, parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Two capacitors are connected in parallel as shown in figure
Equivalent capacity of two capacitors connected in parallel, C_p = 2C = \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Charge stored in the system of plates,
q = C_pV = \(\frac{2 \varepsilon_0 A}{d} V\)
∴ q = \(\frac{2 A \varepsilon_0 \mathrm{~V}}{\mathrm{~d}}\)

Question 8.
Four identical metal plates each of area A are separated mutually by a distance d and are connected as shown. Find the capacity of the system between the terminals A and B.

Solution:
Area of each plate of a capacitor = A
Distance between two parallel plate capacitors = d
Capacity of each parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Given fig is

The equivalent circuit of the above fig is

Question 9.
In the circuit shown the battery of ‘V’ volts has no internal resistance. All three condensers are equal in capacity. Find the condenser that carries more charge ?

Answer:
The equivalent circuit to the given circuit is as shown

In series combination of capacitors, charge q flows through each capacitor. Then q₁ = q = C₁V₁; q₂ – q = C₂V₂; q₃ = q = C₃V₃
∴ q₁ = q₂ = q₃
Hence three capacitors C₁, C₂ and C₃ carry the same charge.

Question 10.
Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over, the battery is removed. Now a dielectric slab of K = 2 is inserted between the plates of A so as to fill the completely. The energy lost by the system during the sharing of charges is
Solution:
i) With battery of parallel combination:
C₁ = C;C₂ = 2C; V = V
C_p = C₁ + C₂ = 3C; q = 3Cv

ii) Without battery of parallel combination : .

Question 11.
A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged?
Answer:
For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V\(V_1^2\) = \(\frac{1}{2}\) CV² …….. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}\) = \(\frac{1}{4} \mathrm{CV}^2\); Let potential difference across the capacitor = V₂

Textual Exercises

Question 1.
Two charges 5 × 10^-8 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Here q1 = 5 × 10^-8C, q₂ = -3 × 10^-8C
Let the potential be zero at a distance x cm from the charge q₁ = 5 × 10^-8C.
∴ r₁ = x × 10^-2m
r₂ = (16 – x) × 10^-2m
Now V = \(\frac{\mathrm{q}_1}{4 \pi \varepsilon_0 \mathrm{r}_1}\) + \(\frac{\mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_2}\)
= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]\)
∴ \(\frac{\mathrm{q}_1}{\mathrm{r}_1}\) = \(\frac{-\mathrm{q}_2}{\mathrm{r}_2}\)
= \(\frac{5 \times 10^{-8}}{x \times 10^{-2}}\) = \(\frac{-\left(-3 \times 10^{-8}\right)}{(16-x) 10^{-2}}\) or \(\frac{5}{x}\) = \(\frac{3}{16-x}\)
3x = 80 – 5x
8x = 80, x = 10cm

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer:
In fig. O is centre of hexagon ABCDEFA of each side 10 cm, As it clear from the figure, OAB, OBC etc. are equilateral triangles.
Therefore
OA = OB = OC = OD = OE = OF = r = 10 cm = 10^-1m
As potential is scalar, there for C potential at O is

Question 3.
Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.
a) Identify an equipotential surface of the system.
b) What is the direction of the electric field at every point on this surface ?
Solution:
a) The plane normal to AB and passing through its middle point has zero potential everywhere.
b) Normal to the plane in the direction AB.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field.
a) inside the sphere
b) just outside the sphere
c) At a point 18 cm from the centre of the sphere ?
Solution:
a) Here r = 12 cm = 12 × 10^-2m,
q = 1.6 × 10^-7C. Inside the sphere, E = 0

b) Just coincide the sphere (say on the surface of the sphere)
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
= 9 × 10⁹ × \(\frac{1.6 \times 10^{-7}}{\left(12 \times 10^{-2}\right)^2}\) = 10⁵ N/c

c) At r = irm = 18 × 10^-2
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) = \(\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2}\)
= 4.4 × 10⁴ N/C

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
Solution:
C₁ = \(\frac{\varepsilon_0 A}{d}\) = 8pF
C₂ = k\(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}\) = \(\frac{6 \times 2 \varepsilon_0 A}{d}\) = 12 × 8 = 96pF.

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
a) What is the total capacitance of the combination ?
b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Solution:
a) \(\frac{1}{C_S}\) = \(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ; C_s = 3pF
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

b) P.d across each capacitor =
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
a) What is the total capacitance of the combination ?
b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
a) C_p = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V₁ = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₃ = C₃V = 4 × 100 = 400pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Solution:
Here A = 6 × 10^-3m², d = 3mm = 3 × 10^-3m, C = ?
V = 100V, q = ?
C₀ = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{3 \times 10^{-3}}\)
= 1.77 × 10^-11F
q = C₀V= 1.77 × 10^-11 × 100
= 1.77 × 10^-9C

Question 9.
Explain what would happen if in the capacitor given in Exercise 8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.
a) While the voltage supply remained connected.
b) after the supply was disconnected.
Solution:
a) Capacity increases to C = KC₀
= 6 × 1.77 × 10^-11F
charge increases to
q¹ = C¹V = 6 × 1.77 × 10^-11 × 10²C

b) After the supply was disconnected new capacity C = KC₀ = 6 × 1.77 × 10^-11F
New voltage V¹ = \(\frac{q}{C^1}\) = \(\frac{1.77 \times 10^{-9}}{6 \times 1.77 \times 10^{-11}}\)
= 16.67V

Question 10.
A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor ?
Solution:
Here C = 12pF = 12 × 10^-12F, V = 50Volt, E = ?
E = \(\frac{1}{2} \mathrm{CV}^2\) = \(\frac{1}{2}\left(12 \times 10^{-12}\right)(50)^2\)
= 1.5 × 10^-8J

Question 11.
A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Solution:
Here C₁ = C₂ = 600pF = 600 × 10^-12
F = 6 × 10^-10F,
V₁ = 200 V, V₂ = o

Additional Exercises

Question 1.
A charge of 8mC is located at the origin. Calculate the workdone in taking a small charge of’ -2 × 10^-9 C from a point P(0, 0, 3 cm) to a point Q(0,4 cm, 0), Via a point R(0,6 cm, 9 cm).
Solution:
From fig. a charge q = 8mc = 8 × 10^-3C is located at the origin O. Charge to be carried is
q₀ = -2 × 10^-9C from P to Q
Where OP = r_p = 3 cm
= 3 × 10^-2m and OQ = r_Q = 4cm = 4 × 10^-2m
As electrostatic forces are conservative forces, workdone is independent of the path. Therefore there is no relevance of point R.

Question 2.
A cube of side b has a charge q at each of its vertices, Determine the potential and electric field due to this charge array at the centre of the cube.
Answer:
We know that the length of diagonal of thè cube of each side b is \(\sqrt{3 b^2}\) = \(\mathrm{b} \sqrt{3}\)
Distance between centre of the cube and each vertex r = \(\frac{b \sqrt{3}}{2}\)
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}\)
and 8 charges each of valué q are present at the eight vertices, of the cube therefore

∴ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{8 q}{b \sqrt{3} / 2}\) or V = \(\frac{4 \mathrm{q}}{\sqrt{3} \pi \varepsilon_0 \mathrm{~b}}\)
Further electric field intensity at the centre due to all the eight charges is zero because the fields due to individual charges cancel in pairs.

Question 3.
Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field
a) at the mid-point of the line joining the two charges and
b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Solution:
Here q₁ = 1.5,C = 1.5 × 10^-6 C,
q₁ = 2.5μC = 2.5 × 10^-6C
Distance between the two spheres = 30cm
from figure

b) Let P be the point in a plane normal to the line passing through the mid point,
where OP 10cm = 0. 1m
From figure,
Now PA = PB

Resultant field intensity at P is
E = \(\sqrt{\mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos \theta}\)
E = 6.58 × 10⁵ Vm^-1
Let θ be the angle which resultant intensity \(\overrightarrow{\mathrm{E}}\) makes with \(\overrightarrow{\mathrm{E}_1}\).

Question 4.
A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
a) A charge q is placed at the centre of the shell. What ¡s the surface charge density on the inner 4nd outer surfaces of the shell?
Answer:
a) The charge of + Q resides on the Outer surface of the shell. The charge q placed at the centre of the shell induces charge -q on the inner surface and charge + q on the outer surface of the shell, from figure.
∴ Total charge on inner surface of the shell is -q and total charge on the outer surface of the shell is (Q + q)
σ₁ = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}\) and σ₁ = \(\frac{\mathrm{Q}_1+\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

b) Is the electric field inside a cavity (with no charge) zero, even If the shell is not spherical, but has any irregular shape? Explain.
Solution:
Electric field intensity inside a cavity with no charge is zero, eveñ when the shell has

any irregular shape. If we were to take a closed loop part of which is inside the cavity along a field line and the rest outside it then network done by the field in carrying a rest charge over the closed loop will not be zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity with no charge is always zero.

Question 5.
a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to
another given by (E₂ – E₁). \(\hat{\mathbf{n}}\) = \(\frac{\sigma}{\varepsilon_0}\)
Where \(\hat{\mathbf{n}}\) is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of \(\hat{\mathbf{n}}\) is from side 1 to side 2.)
Hence show that just out side a conductor, the electric field is σ\(\hat{\mathrm{n}} / \varepsilon_0\)
b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
(Hint: for (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.)
Answer:
a) Normal component of electric field intensity due to a thin infinite plane sheet of charge on left side.
\(\overrightarrow{\mathrm{E}}_1\) = –\(\frac{\sigma}{2 \varepsilon_0} \hat{\mathbf{n}}\)
and on right side 2 = \(\overrightarrow{\mathrm{E}_2}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{\mathrm{n}}\)
Discontinuity in the normal component from one side to the other is

b) To show that the tangential component of electostatic field is continous from one side of a charged surface to another, we use the fact that workdone by electrostatic field on a closed loop is zero.

Question 6.
A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Solution:
From figure A is a long charged cylinder of linear charge density λ, lengh l and radius a. A hollow co-axial conducting cylinder B of length L and radius b surrounds A.

The charge q = λL spreads uniformly on the outer surface of λ. It induces – q charge on the cylinder B. which spreads on the inner surface of B. An electric field \(\overrightarrow{\mathrm{E}}\) is produced in the space between the two cylinders which is directed radically outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The Electric flux through the end faces of the cylindrical Gaussian surface is zero as \(\overrightarrow{\mathrm{E}}\) is parallel to them. According to Gauss’s theorem

Question 7.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53Å:
a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation ?
Solution:
a) Here q₁ = -1.6 × 10^-19C; q₂ = + 1.6 × 10^-19C.
r = 0.53λ = 0.53 × 10^-10m
Potential energy = P.E at ∞ -P. E at r
= 0 – \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r}\) = \(\frac{-9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{0.53 \times 10^{-10}}\)
= -43.47 × 10^-19 Joule
= \(\frac{-43.47 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\) = -27.16 eV

b) K.E in the Orbit = \(\frac{1}{2}\) (27.16) eV
Total energy = K.E + P.E
= 13.58 – 27.16 = – 13.58 eV
Work required to free the electron = 13.58eV

c) Potential energy at a seperation of r₁ (= 1.06A) is
= \(\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_1}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10}}\)
= 21.73 × 10^-19J = 13.58eV
Potential energy of the system, when zero of P.E is taken at r₁ = 1.06A is
= RE at r₁ -P.E at r = 13.58 – 27.16 = -13.58eV.
By shifting the zero of potential energy work required to free the electron is not affected. It continues to be the same, being equal to + 13.58 eV

Question 8.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion \(\mathbf{H}_2^{+}\). In the ground state of an \(\mathbf{H}_2^{+}\), the two protons are separated by roughly 1.5A. and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
Here q₁ = charge on electron (= -1.6 × 10^-19C)
q₂, q₃ = charge on two protons, each = 1.6 × 10^-19C
r₁₂ = distance between
q₁ and q₂ = 1A = 10^-10m
r₂₃ = distance between
q₂ and q₃ = 1.5A = 1.5 × 10^-10m
r₃₁ = distance between
q₃ and q₁ = 1A = 10^-10m.
Taking zero of potential energy at infinity, we have

Question 9.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends’ of a conductor is higher than on its flatter portions.
Solution:
The charge flows from the sphere at higher potential to the other at lower potential till their potentials become equal. After sharing the charges on two spheres would be.
\(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{C_1 V}{C_2 V}\) where C₁. C₂ are the capacities of two spheres.
But \(\frac{\mathrm{C}_1}{\mathrm{C}_2}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\) ∴ \(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{a}{b}\)
Ratio of surface density of charge on the two spheres

Hence ratio of electric fields at the surface of two spheres
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{\sigma_1}{\sigma_2}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)
A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions.

Question 10.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> I.
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Here -q is at (0, 0, -a) and +q is at (0, 0, a)
a) Potential at (0, 0, z) would be

Potential at (x, y, 0) i.e at a point 1 to z-axis where charges are located is zero

b) we have proved that
V = \(\frac{P \cos \theta}{4 \pi \varepsilon_0\left(r^2+a^2 \cos ^2 \theta\right)}\)
If \(\frac{r}{a}\) >> 1 then a << r ∴ V = \(\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}\)
∴ V = \(\frac{1}{\mathrm{r}^2}\)
i.e potential is inversely proportional to square of the distance

c) Potential at (5, 0, 0) is

As work done = charge (V₂ – V₁)
W = zero

As work done by electrostatic field is independent of the path connecting the two points therefore work done will
continue to be zero along every path.

Question 11.
Figure shows a charge array known as an electric quadrupole, For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1. and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer:
As is clear from figure an electric quadrupole may be regarded as a system of three charges +q, -2q and + q at A, B and C respectively.
Let AC = 2a we have to calculate electric potential at any point P where BP = r, using superposition principle. Potential at p is given by

∴ \(\frac{\mathrm{a}^2}{\mathrm{r}^2}\) is negligibly small V = \(\frac{\mathrm{q} \cdot 2 \mathrm{a}^2}{4 \pi \varepsilon_0 \mathrm{r}^3}\)
clearly V ∝ \(\frac{1}{\mathrm{r}}\)
In case of an electric dipole V ∝ \(\frac{1}{\mathrm{r}^2}\) and in case of an electric monopole
(i.e a single charge), V ∝ \(\frac{1}{\mathrm{r}}\)

Question 12.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Here total capacitance, C = 2μF
Potential difference v = 1KV = 1000 volt
Capacity of each capacitor C₁ = 1μF
Maximum potential difference across each V = 400 volt

Let n capacitors of 1μF each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row
= 1000 Volt
∴ Potential difference across each capacitor = \(\frac{1000}{\mathrm{n}}\) = 400
∴ n = \(\frac{1000}{400}\) = 2.5
As n has to be a whole number (not less than 2.5) therefore n = 3
capacitance of each row of 3 condensors of 1μF
Each is series = 1/3
Total capacitance of m such rows in parallel = \(\frac{\mathrm{m}}{3}\)
∴ \(\frac{\mathrm{m}}{3}\) = 2(μf) or m = 6
∴ Total number of capacitors =
n × m = 3 × 6 = 18.
Hence 1μF capacitors should be connected in six parallel rows, each row containing three capacitors in series.

Question 13.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Answer:
Here A = ? C = 2F,
d = 0.5 cm = 5 × 10^-3m
As C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
∴ A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
= 1.13 × 10⁹m²
Which is too large.
That is why ordinary capacitors are in the range of μF or less. However in electrolytic capacitors d is too small. Therefore their capacitance is much larger (=0.1F)

Question 14.
Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer:
Here C₂ and C₃ are in series
A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
∴ \(\frac{1}{C_1}\) = \(\frac{1}{200}\) + \(\frac{1}{200}\) = \(\frac{2}{200}\) = \(\frac{1}{100}\)
C_S = 100pF
Now C_S and C₁ are in parallel
∴ C_p = C_s + C₁ = 100 + 100 = 200pF
Again C_p and C₄ are in series
∴ \(\frac{1}{\mathrm{C}_{\mathrm{s}}}\) = \(\frac{1}{\mathrm{C}_{\mathrm{p}}}\) + \(\frac{1}{\mathrm{C}_4}\) = \(\frac{1}{200}\) + \(\frac{1}{100}\) = \(\frac{3}{200}\)
∴ C = \(\frac{200}{3}\)pF = 66.7 × 10^-12F
As C_p and C₄ are in series
∴ V_p + V₄ = 300
Charge on C₄ is q₄
= CV = \(\frac{200}{3}\) × 10^-12 × 300 = 2 × 10^-8C
Potential difference across
C₄ is V₄ = \(\frac{\mathrm{q}_4}{\mathrm{C}_4}\) = \(\frac{2 \times 10^{-8}}{100 \times 10^{-12}}\) = 200V
from (i) V_p = 300 – V₄ = 300 – 200 = 100
Potential difference across
C₁ is V₁ = V_p = 100V
Charge on C₁ = q₁ = C₁V₁
= 100 × 10^-12 × 100 = 10^-8C.
Potential diff across
C₂ and C₃ in series = 100V
charge on C₂;
q₂ = C₂ V₂ = 200 × 10^-12 × 50 = 10^-8C
Charge on C₃;
q₃ = C₃V₃ = 200 × 10^-12 × 50 = 10^-8C

Question 15.
The plates of parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u, Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:
a) Here A = 90 cm² = 90 × 1o^-4m²
= 9 × 10^-3m²
d = 2.5mm = 2.5 × 10^-3m

Question 16.
A 4µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Here C₁ = 4µF = 4 × 10^-6F, V₁ = 200volt. Initial elctrostatic energy stored in C₁ is E₁
= \(\frac{1}{2} C_1 V_1^2\) = \(\frac{1}{2}\) × 4 × 10^-6 × 200 × 200
E₁ = 8 × 10^-2 Joule
When 4µF capacitor is connected to uncharged capacitor of 2µF charge flows and both acquire a common potential.

∴ final electrostatic energy of both capacitors
E₂ = \(\frac{1}{2}\)(C₁ + C₂)V²
= \(\frac{1}{2}\) × 6 × 10^-6 × \(\frac{800}{6}\) × \(\frac{800}{6}\)
E₂ = 5.33 × 10^-2Joule.
Energy dissipated in the form of heat and electro magnetic radiation.
E₁ – E₂ = 8 × 10^-2 – 5.33 × 10^-2
= 2.67 × 10^-2 Joule.

Question 17.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Answer:
If F is the force on each plates of parallel plate capacitor, then work done in increasing the seperation between the plates by Δx = fΔx

This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is = A Δx
If U = energy density = energy stored/ volume then the increase in potential energy = U.AΔx
∴ fΔx = U. AΔx

The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is. E. Therefore the average value of the field (i.e E/2) contributes to the force.

Question 18.
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.) Show that the capacitance of a spherical capacitor is given by C = \(\frac{4 \pi \varepsilon_0 r_1 r_2}{r_1-r_2}\)

where r₁ and r₂ are the radii of outer and inner spheres, respectively.
Solution:
As is clear from the figure +Q charge spreads uniformly on inner surface of outer sphere of radius r₁. The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r₂. The outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r₂ and E = 0 for r < r₂ and E = 0 for r > r₁

In the space between the two spheres electric intensity E exists as shown. Potential difference between the two spheres.

Question 19.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere ?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Solution:
Here r_a = 12cm = 12 × 10^-2m
r_b = 13cm = 13 × 10^-2m
q = 2.5μC = 2.5 × 10^-6C E_r = 32
(a) C = ?

(b)
V = ?
V = \(\frac{q}{c}\) = \(\frac{2.5 \times 10^{-6}}{5.5 \times 10^{-9}}\) = 4.5 × 10²Volt

(c) Capacity of an isolated sphere of radius R
R = 12 × 10^-2m is
C¹ = \(4 \pi \varepsilon_0 R\) = \(\frac{1}{9 \times 10^9}\) × 12 × 10^-12
= 1.33 × 10^-11 Farad.

The capacity of an isolated sphere is much smaller because in a capacitor outer sphere is earthed potential difference decreases and capacitance increases.

Question 20.
Answer carefully:
(a) Two large conducting sphers carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁ Q₂/\(4 \pi \varepsilon_0 \mathbf{r}^2\). where r is the distance between their centres ?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point ?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacitance of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer:
a) When the charged spheres are brought close together the charge distributions on them become non-uniform. Therefore, coloumb’s law is not valid hence the magnitude of force is not given exactly by this formula.
b) No Gauss’s law will not be true if coloumb’s law involved 1/r³ dependence instead of 1/r² dependence.
c) The line of force gives the direction of accelaration of charge. If the electric line of force is linear the test charge will move along the line if the line of force is not linear the charge will not go along the line.
d) As force due to the field is discreted towards the nucleus and the electron does not move in the direction of this force, therefore work done is zero when the orbit is circular. This is true even when orbit is elliptical as electric forces are conservative forces.
e) No electric potential is continuous.
f) The capacity of a single conductor implies that the second conductor is infinity.
g) This is because a molecule of water in its normal state has an unsymmetrical shape and therefore it has a permanent dipole moment.

Question 21.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effets (i.e., bending of field lines at the ends).
Answer:
Here L = 15cms = 15 × 10^-2m
r_a = 1.4cm = 1.4 × 10^-2m,
r_b = 1.5cm = 1.5 × 10^-2m
q = 3.5 μC = 3.5 × 10^-6 coloumb, C = ? V = ?

Question 22.
A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e, without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?
Answer:
Here V = 1kV= 1000Volt; K = ε_r = 3
Dielectric strength = 10⁷V/m
As electric field at the most should be 10% of dielectric strength due to reasons of safety.
E = 10% of 10⁷V/m = 10⁶V/m A = ?
C = 50pF = 50 × 10^-12F

Question 23.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z – direction.
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
By definition an equipotential surfaces is that every point of which potential is the same. In the four cases given above:
a) Equipotential surfaces are planes parallel to x – y plane. These are equidistant.
b) Equipotential surfaces are planes parallel to x – y plane. As the field increases uniformly distance between the planes decreases.
c) Equipotential surfaces concentric spheres with origin at the centre.
d) Equipotential surfaces have the shape which changes periodically at far off distances from the grid.

Question 24.
In a Van de Graff type generator a spherical metal shell is to be a 15 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm^-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build and electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer:
Here V = 15 × 10⁶ volt .
Dielectric strength = 5 × 10^-7 Vm^-1
minimum rodius, r = ?
max. Electric field E = 10% (dielectric strength)
E = \(\frac{10}{100}\) × 5 × 10⁷ = 5 × 10⁶VM^-1
As E = \(\frac{\mathrm{V}}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{V}}{\mathrm{E}}\) = \(\frac{15 \times 10^6}{5 \times 10^6}\) = 3m
obviously we cannot build an electrostatic generator, using a very small shell.

Question 25.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radium r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.
Answer:
As the charge resides always on the outer surface of the shell therefore, when the sphere and shell are connected by a wire, charge will flow essentially from the sphere to the shell, whatever be the magnitude and sign of charge q₂.

Question 26.
Answer the following :
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^-1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house on evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge it self completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lighting?
(Hint : The earth has and electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9Cm^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Answer:
a) since our body and the surface of earth both are conducting therefore our body and the ground form an equipotential surface. As we step out into the open from our house the original equipotential surfaces of open air change, keeping out body and the ground at the same potential that is why we do not get an electric shock.

b) Yes, the man will get a shock This is because the steady discharging current of the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the condenser formed by the aluminium sheet and the ground and the insulating slab.

c) The atmosphere is being discharged continuously by understorms and lightning all over the globe. It is also discharging due to the small conductivity of air. The two opposing processes, on an overage, are in equilibrium. Therefore the atmosphere. Keeps charged.

d) During lightning the electric energy of the atmosphere is dissipated in the form of light, heat and sound.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields

Very Short Answer Questions

Question 1.
What is meant by the statement ‘charge is quantized’ ?
Answer:
The minimum charge that can be transferred from one body to the other is equal to the charge of the electron (e = 1.602 × 10^-19C). A charge always exists an integral multiple of charge of electron (q = ne). Therefore charge is said to be quantized.

Question 2.
Repulsion is.the sure test of charging than attraction. Why ?
Answer:
A charged body may attract a neutral body and also an opposite charged body. But it always repels a like charged body. Hence repulsion is the sure test of electrification.

Question 3.
How many electrons constitute 1 C of charge ?
Answer:
n = \(\frac{q}{e}\) = \(\frac{1}{1.6 \times 10^{-19}}\) = 6.25 × 10¹⁸ electrons

Question 4.
What happens to the weight of a body when it is charged positively ?
Answer:
When a body positively charged it must loose some electrons. Hence weight of the body will decrease.

Question 5.
What happens to the force between two charges if the distance between them is
a) halved
b) doubled ?
Answer:
From Coulombs law, F ∝ \(\frac{1}{\mathrm{~d}^2}\), so
a) When distance is reduced to half, force increases by four times.

b) When distance is doubled, then force is reduced by four times.

Question 6.
The electric lines of force do not intersect. Why ?
Answer:
They do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible.

Question 7.
Consider two charges + q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
Answer:
Charges are scalars, but the .electrical intensities are vectors and add vectorially.

Question 8.
Electrostatic field lines of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force ?
Answer:
It is conservative force.

Question 9.
State Gauss’s law in electrostatics.
Answer:
Gauss’s law : It states that “the total electric flux through any closed surface is equal to – \(\frac{1}{\varepsilon_0}\) times net charge enclosed by the surface”.
\(\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

Question 10.
When is the electric flux negative and when is it positive ?
Answer:
Electric flux ϕ = \(\vec{E} \cdot \vec{A}\). If angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{A}}\) is 180°, then flux will have a ‘-ve’ sign. We consider the flux flowing out of the surface as positive and flux entering into the surface as negative.

Question 11.
Write the expression for electric intensity due to an infinite long charged wire at a distance radial distance r from the wire.
Answer:
The electric intensity due to an infinitely long charged wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) the conductor.
Where λ = Uniform linear charge density
r = Distance of the point from the conductor.

Question 12.
Write the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
The electric intensity due to an infinite plane sheet of charge is E = \(\frac{\sigma}{2 \varepsilon_0}\).

Question 13.
Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.
Answer:
a) Intensity of electric field at any point inside a spherical shell is zero.
b) Intensity of electric field at any point outside a uniformly charged spherical shell is
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity.
Answer:
Coulomb’s law – Statement: Force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the straight line joining the two charges.
Explanation : Let us consider two charges q₁ and q₂ be separated by a distance r.
Then F ∝ q₁q₂ and F ∝ \(\frac{1}{\mathrm{r}^2}\) or F ∝ \(\frac{q_1 q_2}{r^2}\)

∴ F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}\) where \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C^-2
In vector form, in free space \(\overrightarrow{\mathrm{F}}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2} \hat{\mathrm{r}}\). Here \(\hat{\mathrm{r}}\) is a unit vector.
ε₀ is called permittivity of free space.
ε₀ = 8.85 × 10^-12 C²/N-m² or Farad/meter.
Where ε is called permittivity of the medium.

Question 2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
Answer:
Intensity of electric field (E) : Intensity of electric field at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
Expression :
1) Intensity of electric field is a vector. It’s direction is along the direction’ of motion of positive charge.
2) Consider point charge q. Electric field will exist around that charge. Consider any point P in that electric field at a distance r from the given charge. A test charge q₀ is placed at R
3) Force acting on q₀ due to q is F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2}\)
4) Intensity of electric field at that point is equal to the force experienced by a test charge q₀.
Intensity of electric field, E = \(\frac{\mathrm{F}}{\mathrm{q}_0}\)

E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\) N/C or V/m

Question 3.
Derive the equation for the couple acting on a electric dipole in a uniform electric field.
Answer:
1) A pair of opposite charges separated by a small distance is called dipole.
2) Consider the charge of dipole are -q and +q coulomb and the distance between them is 2a.
3) Then the electric dipole moment P is given by P = q × 2a = 2aq. It is a vector. It’s direction is from -q to +q along the axis of dipole.
4) It is placed in a uniform electric field E, making an angle 0 with field direction as shown in fig.
5) Due to electric field force on +q is F = +qE and force on -q is F = -qE.

6) These two equal and opposite charges constitute torque or moment of couple.
i. e., torque, \(\tau\) = ⊥^r distance × magnitude of one of force
∴ \(\tau\) = (2a sin θ)qE = 2aqE sin θ = PE sin θ
In vector form, \(\vec{\tau}\) = \(\overrightarrow{\mathrm{P}}\) × \(\overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.
Answer:
Electric field at a point on the axis of a dipole :
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre ‘O’.

2) We shall calculate electric field E at point P on the axial line of dipole, and at a distance OP = r.
3) Let E₁ and E₂ be the intensities of electric field at P due to charges +q and -q respectively.
4)

The resultant intensity at P is E = E₁ – E₂ [∵ They are opposite and E₁ > E₂]

If r > > a then a² can be neglected in comparision to r².

In vector form, \(\overrightarrow{\mathrm{E}}\) = \(\frac{2 \overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 5.
Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole. (A.P. Mar. ’19, ’15)
Answer:
Electric field intensity on equitorial line of electric dipole:
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre at ‘O’.
2) We shall calculate electric field E at P on equitorial line of dipole and at a distance OP = r.

3) Let E₁ and E₂ be the electric fields at P due to charges +q and -q respectively.
4) The ⊥^r components (E₁ sin θ and E₂ sin θ) cancel each other because they are equal and opposite. The ||^el components (E₁ cos θ and E₂ cos θ) are in the same direction and hence add up.
5) The resultant field intensity at point P is given by E = E₁ cos θ + E₂ cos θ

6) From figure, cos θ = \(\frac{a}{\left(r^2+a^2\right)^{1 / 2}}\)
∴ E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
7) If r >> a, then a² can be neglected in comparison to r². Then
E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
In vector form \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and explain its importance.
Answer:
Gauss’s law : The total-electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface.
Total electric flux,

Here q is the total charge enclosed by the surface ‘S’, \(\oint\) represents surface integral of the closed surface.

Importance :

1. Gauss’s law is very useful in. calculating the electric field in case of problems where it is possible to construct a closed surface. Such surface is called Gaussian surface.
2. Gauss’s law is true for any closed surface, no matter what its shape or size.
3. Symmetric considerations in many problems make the application of Gauss’s law much easier.

Long Answer Questions

Question 1.
Define electric flux. Applying Gauss’s law and derive the expression for electric intensity due to an infinite long straight charged wise. (Assume that the electric field is everywhere radial and depends only on the radial distance r of the point from the wire.)
Answer:
Electric flux : The number of electric lines of force passing perpendicular to the area is known as electric flux (ϕ). Electric flux ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\). So flux is a scalar.

Expression for E due to an infinite long straight charged wire :

1) Consider an infinitely long thin straight wire with uniform linear charge density ‘λ’.
2) Linear charge density λ = \(\frac{\text { change q }}{\text { length } l}\) ⇒ λl —– (1)
3) Construct a coaxial cylindrical gaussion surface of length T and radius ‘r’. Due to symmetry we will assume that electric field is radial i.e., normal to the conducting wire.
4) The flat surfaces AB and CD are ⊥^r to the wire. Select small area ds₁ and ds₂ on the surface as AB and CD.
They are ⊥^r to \(\overrightarrow{\mathrm{E}}\). So flux coming out through them is zero.
Since flux ϕ = \(\oint \vec{E} \cdot d \vec{s}\) = Eds cos 90° = 0

5) So flux coming out through the cylindrical surface ABCD is taken into account.

6) From Gauss’s law

7) From (2) and (3), E(2πrl) = \(\frac{Q}{\varepsilon_0}\) = \(\frac{\lambda /}{\varepsilon_0}\) (∵ Q = λl)
∴ E = \(\frac{\lambda l}{2 \pi \varepsilon_0 \mathrm{r} l}=\frac{1}{2 \pi \varepsilon_0} \frac{\lambda}{\mathrm{r}}\)

8) Therefore electric intensity due to an infinitely long conducting wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\).

Question 2.
State Gauss’s law in electrostatics. Applying Gauss’s law derive the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
Gauss’s law: The total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface, i.e.,

Expression for E due to an infinite plane sheet of charge :

1. Consider an infinite plane sheet of charge. Let the charge distribution is uniform on this plane.
2. Uniform charge density on this surface σ = \(\frac{\mathrm{dq}}{\mathrm{dS}}\) where dq is the charge over an infinite small area ds.
3. Construct a horizontal cylindrical Gaussian surface ABCD perpendicular to the plane with length 2r.
4. The flat surfaces BC and AD are parallel to the plane sheet and are at equal distance from the plane.
5. Let area of these surfaces are dS₁ and dS₂. They are parallel to \(\overrightarrow{\mathrm{E}}\). So flux through these two surfaces is
   ——– (1)
   Where S is area of plane surface AD or BC. Both are equal in area and intensity.
6. Consider cylindrical surface of AB and CD. Let their areas are say dS₃ and dS₄. These surfaces are ⊥^lr to electric intensity \(\overrightarrow{\mathrm{E}}\).
7. So angle between \(\overrightarrow{\mathrm{E}}\) and d\(\overrightarrow{\mathbf{s}_3}\) or dS₄ is 90°. Total flux through these, surfaces is zero. Since
   
8. From Gauss’s law total flux, ϕ = \(\oint \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = 2ES = \(\frac{\mathrm{q}}{\varepsilon_0}\)
   ∴ 2ES = \(\frac{\sigma S}{\varepsilon_0}\) [∵ \(\text { Q }\) = σ × S]
9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \(\frac{\sigma}{2 \varepsilon_0}\)

Question 3.
Applying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at
(i) a point outside the shell
(ii) a point on the surface of the shell and
(iii) a point inside the shell.
Answer:
Expression for E due to a charged conducting spherical shell:

1. Consider a uniformly charged spherical shell. Let total charge on it is ‘q’ and its radius is R.
   
2. Since the shell is uniformly charged, the intensity of electric field at any point depends on radial distance ‘r’ from centre ‘O’. The direction of E is away from the centre along the radius.

i) E at a point outside the shell:

1) Consider a point at a distance ‘r’ outside the sphere. Construct a Gaussian surface with ‘r’ as radius (where r > R).

2) Total flux coming out of this sphere is



3) Therefore at any point outside the sphere, E = \(\frac{\sigma}{\varepsilon_0} \frac{\mathrm{R}^2}{\mathrm{r}^2}\)

ii) E at a point on the surface of shell:

1) Construct a Gaussian surface with radius r = R.

2)


3) Therefore intensity at any point on surface of the sphere E = \(\frac{\sigma}{\varepsilon_0}\)

iii) E at a point inside the shell :

1) Consider a point P inside the shell. Construct a Gaussian surface with radius r (where r < R). There is no charge inside the shell. So from Gauss’s law \(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

2) Therefore, intensity of electric field at any point inside a charged shell is zero.

Textual Exercises

Question 1.
Two small identical balls, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal lengths. The balls position themselves at equilibrium such that the angle between the threads is 60°. If the distance between the balls is 0.5 m, find the charge on each ball.
Solution:
Given m = 0.20 g = 0.2 × 10^-3 kg; θ = 60° ⇒ α = \(\frac{\theta}{2}\) = 30°
r = 0.5 m, Let q₁ = q₂ = q

Question 2.
An infinite number of charges-each of magnitude q are placed on x-axis at distance of 1, 2, 4, 8, …….. meter from the origin respectively. Find intensity of the electric field at origin.
Solution:
Let q₁ = q₂ = q₃ = q₄ = ……. = q
r₁ = 1; r₂ = 2; r₃ = 4; r₄ = 8, …….
The resultant electric field at origin ‘O’ is given by
E = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_2}{\mathrm{r}_2^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{q_3}{r_3^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_4}{\mathrm{r}_4^2}\) + ……..

Question 3.
A clock face has negative charges -q, -2q, -3q, ….. -12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial ?
Solution:
Let distance of each charge from unit charge at centre ‘O’ = r.
Resultant electric field of each charge, E = \(\frac{1}{4 \pi \varepsilon_0} \frac{6 q}{r^2}\) [∵ -6q – (-12q)]
Let OX be the reference axis. The angles of resultant fields with OX-axis are shown.

Resultant field along OX-axis = \(\left(0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\)i = (2 + \(\sqrt{3}\))i
Resultant field along OY-axis = \(\left(1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{j}}\)
= 1\(\hat{\mathrm{i}}\)
∴ Resultant electric field, E_R(OH) = (2 + \(\sqrt{3}\))\(\hat{i}\) + 1\(\hat{j}\)
The direction of resultant field (OH) is given by, tan θ = \(\frac{|\mathrm{OY}|}{|\mathrm{OX}|}\)
⇒ tan θ = \(\frac{1}{2+\sqrt{3}}\) = tan 15°
⇒ θ = 15°, with OX-axis
∴ The hour hand shows at the centre of the dial is at 9.30.

Question 4.
Consider a uniform electric field E = 3 × 10³ N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x – axis ?
Solution:
a) Given E = 3 × 10³ N/C
S = 10² cm² = 10² × (10^-2m)² = 10^-2m²
θ = 0°
ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 0°
∴ ϕ = 30 Nm²C^-1

b) If θ = 60°, ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 Nm²C^-1

Question 5.
There are four charges, each with a magnitude Q. Two are positive and two are negative. The charges are fixed to the comers of a square of side ‘L’, one to each comer, in such a way that the force on any charge is directed toward the center of the square. Find the magnitude of the net electric force experienced by any charge ?
Solution:

Question 6.
The electric field in a region is given by \(\overrightarrow{\mathbf{E}}\) = a\(\hat{\mathbf{i}}\) + b\(\hat{\mathbf{j}}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = a\(\hat{\mathrm{i}}\) + b\(\hat{\mathrm{j}}\)
\(\vec{S}\) = L²\(\hat{\mathrm{i}}\)

ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (a\(\hat{i}\) + b\(\hat{j}\)) .L²\(\hat{i}\)
∴ ϕ = aL² [∴ \(\hat{i}\). \(\hat{i}\) = 1 and \(\hat{i}\). \(\hat{j}\) = 0]

Question 7.
A hollow spherical shell of radius r has a uniform charge density σ. It is kept in a cube of edge 3r such that the centre of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
Solution:
For spherical shell, charge = q (say)
Radius = r
Charge density = σ = \(\frac{q}{A}\) = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}^2}\)
∴ Charge on spherical shell, q = 4πr²σ
Flux through one of the face of a cube,
ϕ_E = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{1}{6} \times \frac{4 \pi r^2 \sigma}{\varepsilon_0}\) = \(\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}\)

Question 8.
An electric dipole consists of two equal and opposite point charge +Q and -Q, separated by a distance 2l. P is a point collinear with the charges such that its distance from the positive charge is half of its distance from the negative charge. Calculate electric intensity at P.
Solution:
Distance of P from -Q = d (say)
Distance of P from +Q = d/2

Question 9.
Two infinitely long thin straight wires having uniform linear charge densities λ and 2λ are arranged parallel to each other at a distance r apart. Calculate intensity of the electric field at a point midway between them.
Solution:
Distance between two parallel infinite long thin straight wires = r.
Electric field due to infinite long thin straight wire, E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\)

∴ Electric intensity at mid point, E = E₂ – E₁ = 2E₁ – E₁ = E
∴ E = \(\frac{\lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 10.
Two infinitely long thin straight wires having uniform linear charge densities e and 2e are arranged parallel to each other at a distance r apart. Find the intensity of the electric field at a point midway between them.
Solution:
For first infinitely long straight wire, linear charge density λ = e.
For second infinitely long straight wire, linear charge density λ’ = 2e
Distance between two infinite parallel straight wires = r.
Distance of point P from 1^st and 2^nd wire = \(\frac{\mathrm{r}}{2}\)
Electric field intensity at P due 1^st wire, E₁ = \(\frac{\lambda}{2 \pi \varepsilon_0\left(\frac{\mathrm{r}}{2}\right)}=\frac{\mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\) —— (1)

Electric field intensity at P due 2^nd wire, E₂ = \(\frac{\lambda^{\prime}}{2 \pi \varepsilon_0\left(\frac{r}{2}\right)}=\frac{2 \mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\)
∴ E₂ = 2E₁ [∵ from(1)]
∴ Electric field intensity at middle point due to second infinitely long wire
E₂ = \(\frac{2 \lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 11.
An electron of mass m and charge e is fired perpendicular to a uniform electric field of intensity E with an initial velocity u. If the electron tranverses a distance x in the field in the direction of firing, find the transverse displacement y it suffers.
Solution:
Given m_e = m; q = e; d = x; u_x = u; u_y = 0
Electric field between the plates = E
Time taken travel in the field, t = \(\frac{d}{u_x}\) = \(\frac{\mathbf{X}}{\mathbf{u}}\)

Force on electron F = qE = eE
Acceleration of electron, a = \(\frac{F}{m}\) = \(\frac{\mathrm{eE}}{\mathrm{m}}\)
Transverse displacement of electron y = u_yt + \(\frac{1}{2} \mathrm{at}^2\)
⇒ y = 0 + \(\frac{1}{2}\left(\frac{e E}{m}\right)\left(\frac{x}{u}\right)^2\)
∴ y = \(\frac{\mathrm{eEx}^2}{2 \mathrm{mu}^2}\)

Additiona Exercises

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air ?
Solution:
Given, q₁ = 2 × 10^-7 C; q₂ = 3 × 10⁷ C; d = 30 cm = 30 × 10^-2 m = 3 × 10^-1m

As q₁, q₂ are positive charges, the force between them is repulsive.

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres ?
(b) What is the force on the second sphere due to the first ?
Solution:
a) Given q₁ = 0.4 μc ;
= 0.8 × 10^-6C
q₂ = 0.8 μc; F = 0.2 N = 0.4
= 0.4 × 10^-6m
0.2 = \(\frac{9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{\mathrm{r}^2}\)
r² = 16 × 9 × 10^-4
r = 4 × 3 × 10^-2 = 12 × 10^-2 m
∴ Distance between two charges, r = 12 cm

b) Electrostatic force between two charges obeys the Newton’s third law. i.e., force on q₁ due to q₂ = force on q₂ due to q₁
f₁₂ = f₂₁ = 0.2N

Question 3.
Check that the ratio ke²/G m_em_p is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify ?
Solution:
i) In electrostatics, F_e = \(\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}^2}\) ……. (1)
Where q₁ = q₂ = e
In gravitation, F_g = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}}{\mathrm{r}^2}\) …. (2)
Where m₁ = m_e ; m₂ = m_p

Thus the given ratio is dimensionless.

ii) We know that e = 1.6 × 10^-19 C ; G = 6.67 × 10^-11 N-m²C²

Question 4.
a) Explain the meaning of the statement ‘electric charge of a body is quantized’.
b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e, large scale charges ?
Answer:
a) The electric charge of a body is quantized means that the charge on a body can occur in some particular values only. Charge on any body is the integral multiple of charge on an electron because the charge of an electron is the elementary charge in nature. The charge on any body can be expressed by the formula q = ± ne. Where n = number of electrons transferred and e = charge on one electron. The cause of quantization is that only integral number of electrons can be transferred from one body to other

b) We can ignore the quantization of electric charge when dealing with macroscopic charges because the charge on one electron is 1.6 × 10^-19 C in magnitude, which is very small as compared to the large scale change.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
According to law of conservation of charge, “charge can neither be created nor be destroyed but it can be transferred from one body to another body”. Before rubbing the two bodies they both are neutral i.e., the total charge of the system is zero. When the glass rod is rubbed with a silk cloth, some electrons are transferred from glass rod to silk cloth. Hence glass rod attains positive charge and silk cloth attains same negative charge.

Again the total charge of the system is zero, i.e., the charge before rubbing is same as the charge after rubbing. This is consistent with the law of conservation of charge. Here we can also say that charges can be created only in equal and unlike pairs.

Question 6.
Four point charges q_A = 2 µC, q_B = -5 µC, q_C = 2 µC and q_D = -5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?
Solution:
Let the centre of the square is at O.
The charge placed on the centre is µC
AB = BC = CD = DA = 10 cm; AC = \(\sqrt{2}\) × 10 = 10\(\sqrt{2}\)cm
AO = BO = CO = DO = \(\frac{10 \sqrt{2}}{2}\) = 5\(\sqrt{2}\) cm


Here we observe that, F_A = -F_C and F_D = -F_B
∴ The net resultant force on 1 µC is
F = F_A + F_B + F_C + F_D
= -F_C + F_B + F_C – F_B
= 0.

Question 7.
a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?
b) Explain why two field lines never cross each other at any point ?
Answer:
a) An electrostatic field line represents the actual path travelled by a unit positive charge in an electric field. If the line have sudden breaks it means the unit positive test charge Jumps from one place to another which is not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So, the field line cannot have any sudden breaks.

b) If two field lines cross each other, then we can draw two tangents at the point of intersection which indicates that (as tangent drawn at any point on electric line of force gives the direction of electric field at that point) there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two field lines never cross each other at any point.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vaccum.
a) What is the electric field at the midpoint O of the line AB joining the two charges ?
b) If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge ?
Solution:
a) Given q_A = 3 μC = 3 × 10^-6 C; q_B = -3 μC = -3 × 10^-6C

From fig. AO = OB = 10 cm = 0.1 m
Electric field at midpoint ‘O’ due to q_A

The direction of E_A is A to O.
Electric field at midpoint ‘O’ due to q_B at B is

The direction of E_B is O to B.
Now we see that E_A and E_B are in same direction. So, the resultant electric field at O is E. Hence,
E = E_A + E_B = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C :
The direction of E will be from O to B or toward B.

b) Let test charge q₀ = -1.5 × 10^-9 C is placed at midpoint O’.
Electric field intensity at ‘O’ is E = 5.4 × 10⁶

Force F = E_q = 5.4 × 10⁶ × -1.5 × 10^-9 N
= -8.1 × 10³N
The direction of force is from O to A.

Question 9.
A system has two charges q_A = 2.5 × 10^-7 C, and q_B = -2.5 × 10^-7 C located at points A(0, 0, -15 cm) and B(0, 0, +15 cm). What are the total charge and electric dipole moment of the system ?
Solution:
Given A(0, 0, -15 cm) and B(0, 0, 15 cm)
q_A = 2.5 × 10^-7C
q_B = -2.5 × 10^-7 C
AB = 2a = length of the dipole
= 30 cm = 30 × 10^-2 m

The total charge q on the dipole is
q = q_A + q_B = 2.5 × 10^-7C – 2.5 × 10^-7C = 0
The electric dipolemoment
P = Any charge (q_A) × length of dipole (2a)
= 2.5 × 10^-7 × 10 × 10^-2
∴ P = 7.5 × 10^-8 C-m
The direction of P is from negative charge to positive charge that is along B to A.

Question 10.
An electric dipole with dipole moment 4 × 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Solution:
Given, P = 4 × 10^-9 C-m; E = 5 × 10⁴ N/C; θ = 30°,
Torque, \(\tau\) = PE sin θ

= 4 × 10^-9 × 5 × 10⁴ sin 30° = \(\frac{20 \times 10^{-5}}{2}\) = 10^-4N-m
The direction of torque is ⊥^r to both electric field and dipole moment.

Question 11.
A polythene piece rubbed with wool is found to have a negative charge 3 × 10^-7 C.
a) Estimate the number of electrons transferred (from which to which ?)
b) Is there a transfer of mass from wool to polythene ?
Solution:
a) Given, charge on Polythene, q = -3 × 10^-7 C
e = -1.6 × 10^-19 C
No. of electrons transferred, n = \(\frac{\mathrm{q}}{\mathrm{e}}\) = \(\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}\)
∴ n = 1.875 × 10¹² [∵ q = ± ne]
Electrons are transferred from wool to polythene.
So wool gets positive charge and polythene gets negative charge.

b) The number of electrons transferred = 1.875 × 10¹²
The mass of one electron, m_e = 9.1 × 10^-3 kg
Mass transferred from wool to polythene M = n × m_e
M = 1.875 × 10¹² × 9.1 × 10^-31 = 1.8 × 10^-18 kg

Question 12.
a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each 6.5 × 10^-7 C ? The radii of A and B are negligible compared to the distance of
separation.
b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
Solution:

a) Given, q_A = 6.5 × 10^-7C ; q_B = 6.5 × 10^-7C
r = AB = 50 cm = 50 × 10^-2m

b)

This force is also repulsive in nature because both the charges are similar (positive) in nature.

Question 13.
Suppose the spheres A and B in Exercise – 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with second and finally removed from both. What is the new force of repulsion between A and B?
Solution:
Given q_A = 6.5 × 10^-7C;
q_B = 6.5 × 10^-7 C; q_C = 0

After contact of A and C, the charges will be divided equally on both of them. Then final charge on A, then
\(\mathrm{q}_{\mathrm{A}}^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{A}}+\mathrm{q}_{\mathrm{C}}}{2}\) = \(\frac{6.5 \times 10^{-7}+0}{2}\)
= 3.25 × 10^-7C
Similarly charge on C, \(\mathrm{q}_{\mathrm{c}}^{\prime}\) = 3.25 × 10^-7 C
After contact of B and C, the charges will be divided equally on both of them.

Then final charge on B, \(q_B^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{B}}+\mathrm{q}_{\mathrm{C}}^{\prime}}{2}\) = \(\frac{6.5 \times 10^{-7}+3.25 \times 10^{-7}}{2}\) = 4.875 × 10^-7 C
Similarly final charge one, \(q_C^{\prime \prime}\) = 4.875 × 10^-7 C

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?

Answer:
We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.

Here, particle 1 and particle 2 are attracted towards positive plate that means particle 1 and particle 2 are negatively charged. Particle 3 is attracted towards negatively charged plate so it is positively charged. As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio.
y ∝ \(\frac{\mathrm{q}}{\mathrm{m}}\)
Here, the deflection in particle 3 is maximum, so the charge to mass ratio of particle 3 is maximum.

Question 15.
Consider a uniform electric field E = 3 × 10³ \(\hat{\mathbf{i}}\) N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis ?
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C

a) As the surface is in Y – Z plane, so the area vector (normal to the square) is along X – axis
Area S = 10 × 10 = 100 cm² = 10^-2 m²

Area vector \(\vec{S}\) = 10^-2 \(\hat{\mathbf{i}}\) m²
ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (3 × 10³ \(\hat{\mathbf{i}}\)). (10^-2i)
∴ ϕ = 3 × 10³ × 10^-2 = 30N-m²/c

b) \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C ; \(\vec{S}\) = \(\hat{\mathbf{i}}\) m² ; θ = 60°
ϕ = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{S}}\) = ES cos 60° = 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 N – m²/C

Question 16.
What is the net flux of the uniform electric field of Exercise -15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes ?
Answer:
As we know that the number of lines entering in the cube is the same as that the number of lines leaving the cube. So, no flux is remained on the cube and hence, the net flux over the cube is zero.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ Nm²/C.
(a) What is the net charge inside the box ?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box ? Why or Why not ?
Solution:
a) Given, ϕ = 8.0 × 10³ N – m²/C
ε₀ = 8 × 10³ × 8.854 × 10^-12
∴ q = 0.07 μc
The flux is outward hence the charge is positive in nature

b) Net outward flux = 0
Then, we can conclude that the net charge inside the box is zero. i.e., the box may have either zero charge or have equal amount of positive and negative charges. It means we cannot conclude that there is no charge inside the box.

Question 18.
A point charge +10 μC is a ‘distance 5 cm directly above the centre of a square of side 10 cm, as shown in fig. What is the magnitude of the electric flux through the square ? (Hint: Think of the square as one face of a cube with edge 10 cm).

Solution:
Let the charge q is placed at the centre of cube as shown in fig.
The total flux enclosed through the cube is ϕ = \(\frac{q}{\varepsilon_0}\)
The flux enclosed by one face ϕ = \(\frac{1}{6}\) of total flux.
[∵ Cube has 6 faces]

ϕ = \(\frac{\phi}{6}\) = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\)
Here q = 10 μC = 10 × 10^-6C ; ε₀ = 8.854 × 10^-12C² – N^-1-m^-2
∴ ϕ = \(\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}\)
= 1.88 × 10⁵ N-m²/C

Question 19.
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface ?
Solution:
Given, q = 2.0 μC = 2.0 × 10^-6C
ε₀ = 8.854 × 10^-12 C²-N^-1 – m^-2
The net flux through the surface,
ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}\) = 2.26 × 10⁵N-m²/C

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge,
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ?
(b) What is the value of the point charge ?
Solution:
a) From Gauss’s law, ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\)
Electric flux ϕ depends on charge q.
It is independent of radius of Gaussian surface. Hence the radius of Gaussian surface were doubled, flux does not change.

b) ϕ = – 1.0 × 10³ N-m²/c ; ε₀ = 8.854 × 10^-12 e²-N^-1-m^-2
q = ϕε₀ = -1.0 × 10³ × 8.854 × 10^-12 = -8.85 × 10^-9C.
∴ The value of point charge, q = -8.85 × 10^-9C

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?
Solution:
E = 1.5 × 10³ N/C; r = 20 cm = 20 × 10^-2m.
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)
1.5 × 10³ = \(\frac{9 \times 10^9 \times \mathrm{q}}{\left(20 \times 10^{-2}\right)^2}\)
q = \(\frac{1.5 \times 10^3 \times 20 \times 20 \times 10^{-4}}{9 \times 10^9}\) = 6.67 × 10^-9C.

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density, of 80.0 μC/m².
(a) Find the charge on the sphere,
(b) What is the total electric flux leaving the surface of the sphere ?
Solution:
a) Given D = 2.4 m; r = \(\frac{\mathrm{D}}{2}\) = 1.2 m
σ = 80 µc/m² = 80 × 10^-6 C/m²
σ = \(\frac{\mathrm{q}}{4 \pi r^2}\) ⇒ q = σ 4πr²
⇒ q = 80 × 10^-6 × 4 × 3.14 × 1.2 × 1.2
∴ q = 1.45 × 10^-3C

b) ϕ = \(\frac{Q}{\varepsilon_0}\) = \(\frac{1.4 \times 10^{-3}}{8.854 \times 10^{-12}}\) = 1.6 × 10⁸N-m²/C
Thus, the flux leaving the surface of sphere is 1.6 × 10⁸ N – m²/c

Question 23.
An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
Solution:
Given r = 2 cm = 2 × 10^-2m ; E = 9 × 10⁴ N/C


Thus, the linear charge density is 10^-7 C/m.

Question 24.
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^-22 C/m². What is E :
(a) in the outer region of the first plate,
(b) in the outer region of the second plate and
(c) between the plates ?
Solution:
Given σ_A = 127.0 × 10^-22 C/m²
σ_B = 17.0 × 10^-22 C/m²

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2; e = 1.60 × 10^-19C).
Solution:
Given n = 12; E = 2.55 × 10⁴ N/C
p = 1.26 g/cm³ = 1.26 × 10³ kg/m³
e = 1.6 × 10^-19C ; g = 9.81 ms^-2

As the oil drop is stationary,
Electrostatic force = Gravitational force
⇒ qE = mg
neE = \(\frac{4}{3} \pi r^3 \mathrm{\rho g}\)
r³ = \(\frac{3 \mathrm{neE}}{4 \pi \rho \mathrm{g}}\) = \(\frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1.26 \times 10^3 \times 9.8}\)
r = 0.94 × 10^-18
r = [0.94 × 10^-18]^{\(\frac{1}{3}\)} = 9.81 × 10^-7m
∴ Radius of the drop = 9.81 × 10^-7 m.

Question 26.
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines ?

Solution:
a) According to the properties of electric lines of force, the lines should be always ⊥r to the surface of a conductor as they starts or they ends. Here, some of the lines are not ⊥r to the surface, thus it cannot represent the electrostatic field lines.

b) According to the property of electrostatic field lines, they never start from negative charge, here some of the lines start from negative charge. So, it cannot represent the electrostatic field lines.

c) As the property of electric field lines that they start outwards from positive charge. Hence, it represents the electrostatic field lines.

d) By the property of electric field lines, two electric field lines never intersect each other. Here, two lines intersect. So it does not represent the electric field lines.

e) By the property of electric field lines that they are not in the form of closed loops. Here, the lines form closed loop. So, it does not represent the electric field lines.

Question 27.
In a certain region of space, electric field is along the Z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive Z-direction, at the rate of 10⁵ NC^-1 per metre. What are the force and torque experienced by a system having a total dipolemoment equal to 10^-7 Cm in the negative Z-direction ?
Solution:
The electric field increases in positive Z – direction. dE
\(\frac{\mathrm{dE}}{\mathrm{dZ}}\) = 10⁵ N/C-m
The direction of dipolemoment is in the negative Z-direction

So the negative charge q is placed at A and positive charge q is placed at B as the direction of dipole moment is from negative charge to positive charge.
P_Z = -10^-7C-m

The negative sign shows its direction in negative Z – axis. According to the basic definition of electric field, F = qdE Now, multiplying and dividing by dz,
F = q\(\frac{\mathrm{dE}}{\mathrm{dz}} \cdot \mathrm{dz}\) .dz = q.dz\(\frac{\mathrm{dE}}{\mathrm{dz}}\)
qdz = dipolement p_z, as the length of the dipole is dz.

∴ F = P_z. \(\frac{\mathrm{dE}}{\mathrm{dz}}\) = -10^-7 × 10⁵ = -10^-2N
Torque, \(\tau\) = PE sin θ (∵ θ = 180° angle between P and E)
\(\tau\) = PE sin 180° = 0
Thus the force is -10^-2 N and the torque is 0.

Question 28.
a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor, (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q (Fig. (b)). (c) A sensitive instrument is o he shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution:
a) As we know the property of conductor that the net electric field inside a charged conductor is zero, i.e., E = 0.
Now let us choose a Gaussian surface lying completely inside the conductor enclosing the cavity.
So, from Gauss’s theorem \(\oint \text { E. dS }\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)
As E = 0 ⇒ \(\frac{q}{\varepsilon_0}\) = 0 ⇒ q = 0
That means the charge inside the cavity is zero. Thus, the entire charge Q on the conductor must appear on the outer surface of the conductor.

b) As the conductor B carrying a charge +q inserted in the cavity, the charge -q is induced on the metal surface of the cavity and then charge +q induced on the outside surface of the conductor A. Initially the outer surface of A of A has a charge Q and now it has a charge +q induced. So the total charge on the outer surface of A is Q + q.

c) To protect any sensitive instrument from electrostatic field, the sensitive instrument must be put in the metallic cover. This is known as electrostatic shielding.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/εε₀)\(\hat{\mathbf{n}}\), where \(\hat{\mathbf{n}}\) is the unit vector in the outward normal direction and σ is the surface charge density near the hole.
Solution:
Surface charge density near the hole = σ
Unit vector = \(\hat{\mathbf{n}}\) (normal directed outwards)
Let P be the point on the hole.
The electric field at point P closed to the surface of conductor, according to Gauss’s theorem,

\(\oint \mathrm{E} \cdot \mathrm{dS}\) = \(\frac{q}{\varepsilon_0}\)
Where q is the charge near the hole.
E ds cos θ = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\) (∴ σ = \(\frac{\mathrm{q}}{\mathrm{dS}}\) ∴q = σ dS) where dS = area

∴ Angle between electric field and area vector is 0°.
EdS = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0} \hat{\mathrm{n}}\)

This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite. So, there is no electric field inside the conductor. Outside the conductor, the electric fields are equal and are in the same direction.
So, the electric field at P due to each part = \(\frac{1}{2} \mathrm{E}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{n}\)

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Solution:
Let us consider a long thin wire of linear charge density λ. We have to find the resultant electric field due to this wire at point P.
Now, consider a very small element of length dx at a distance x from C.

The charge on this elementary portion of length dx
q = λ dx ——- (1)

Electric field intensity at point P due to the elementary portion
dE = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{(\mathrm{OP})^2}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d x}{(\mathrm{OP})^2}\) [∵ from (1)]
Now, in ΔPCO (PO)² = (PC)² + (CO)²
(OP)² = r² + x²
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d \mathbf{x}}{\left(x^2+r^2\right)}\) ——- (2)

The components of dE are dE cos θ along PD and dE sin θ along PF.
Here, there are so many elementary portion. So all the dE sin θ components balance each other. The resultant electric field at P is due to only dE cos θ components.
The resultant electric field due to elementary component, dE’ = dE cos θ

dE’ = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda d x}{\left(x^2+r^2\right)} \cos \theta\) —— (3)
In ΔOCP tan θ = \(\frac{x}{r}\) ⇒ x = r tan θ
Differentiating with respect to θ, we get dx = r sec² θ dθ
Putting in equation (3), we get

As the wire is of infinite length, so integrate within the limits –\(\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we get

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge +(2/3)e and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Solution:
For the protons, the charge on it is +e let the number of up quarks are a, then the number of down quarks are (3 – a) as the total number of quarks are 3.
So, a_x up quark charge + (3 – a) down quark charge = +e
a × \(\frac{2}{3} \mathrm{e}\) + (3 – a)\(\left(\frac{-\mathrm{e}}{3}\right)\) = e
\(\frac{2 \mathrm{ae}}{3}\) – \(\frac{(3-\mathrm{a}) \mathrm{e}}{3}\) = e
2a – 3 + a = 3
3a = 6
a = 2
Thus, in the proton there are two up quarks and one down quark.
∴ Possible quark composition for proton = uud

For the neutron, the charge on neutron is 0.
Let the number of up quarks are b and the number of down quarks are (3 – b)
So, b_x up quark charge + (3 – b) × down quark charge = 0
b\(\left(\frac{2 \mathrm{e}}{3}\right)\) + (3 – b)\(\left(\frac{-\mathrm{e}}{3}\right)\) = 0
2b – 3 + b = 0
3b = 3
∴ b = 1
Thus, in neutron, there are one up quark and two down quarks.
∴ Possible quark composition for neutrons = udd.

Question 32.
a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Solution:
a) Let us consider that initially the test charge is in the stable equilibrium. When the test charge is displaced from the null point (where, E = 0) in any direction, it must experience a restoring force towards the null point.
This means that there is a net inward flux through a closed surface around the null point According to the Gauss’s theorem, the net electric flux through a surface net enclosing any charge must be zero. Hence, the equilibrium is not stable.

b) The middle point of the line joining two like charges is a null point. If we displace a test Charge slightly along the
line, the restoring force try to bring the test charge back to the centre. If we displace the test charge normal to the line, the net force on the test charge takes it further away from the null point. Hence the equilibrium is not stable.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed V_x (as in the fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m \(\mathbf{V}_{\mathbf{x}}^2\)).
Compare this motion with motion of a projectile in gravitational field discussed in section 4.10 of 1^st Year Textbook of Physics.
Solution:
Mass of particle = m
Charge of particle = -q
Speed of particle = V_x
Length of plates = L

Electric field between the plates = E (from positive plate to negative plate).
Let the deflection in the path of the charge – q is Y, because the force acting in +Y axis direction. The direction of force is from negative plate to positive plate because the charge is negative in nature.

Let us discuss the motion in Y axis direction. Initial velocity u = 0
Acceleration a = \(\frac{F}{m}\) = \(\frac{+\mathrm{qE}}{\mathrm{m}}\)
Deflection Y = ?
Time = \(\frac{\text { Distance }}{\text { Velocity }}\) = \(\frac{\mathrm{L}}{\mathrm{V}_{\mathrm{x}}}\)
Using second equation of motion,
S = ut + \(\frac{1}{2} \mathrm{at}^2\)at
Putting the values y = 0 + \(\frac{1}{2} \times\left(+\frac{\mathrm{qE}}{\mathrm{m}}\right) \frac{\mathrm{L}^2}{\mathrm{~V}_{\mathrm{x}}^2}\)
Y = \(\frac{\mathrm{qEL}^2}{2 \mathrm{mV}_{\mathrm{x}}^2}\)

In the case of projectile motion y = \(\frac{1}{2} \mathrm{gt}^2\). Thus, it is exactly similar to the projectile motion in the gravitational field.

Question 34.
Suppose that the particle is an electron projected with velocity V_x = 2.0 × 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate ? (|e| = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg.)
Solution:
Given V_x = 2 × 10⁶ m/s; E = 9.1 × 10² N/C
q = e = 1.6 × 10^-19 C; m_e = 9.1 × 10^-31 kg
d = 0.5 cm = 0.5 × 10^-2 m = 5 × 10^-3 m
The electron will strike the upper plate at its other end at X = L as it get deflected.