Category: Inter 2nd Year

Students get through AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays”? [A.P. Mar. 17]
Answer:
Cathode rays are streams of fast-moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantized. That means the charge on anybody (oil drop) is always an integral multiple of the charge of an electron, i.e., Q = ne.

Question 3.
What is “work function” ? [T.S. Mar. 17, 15]
Answer:
The minimum energy required to liberate an electron from photometal surface is called work function, Φ₀.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficierit energy is incident on the photometal surface, electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. [A.P. Mar. 15]
Answer:
Einstein’s photoelectric equation is given by K_max = \(\frac{1}{2}\) mv²_max = hυ – Φ₀

Question 7.
Write down de-Broglie’s relation and explain the terms therein. [T.S. & A.P. Mar. 16]
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}=\frac{h}{m v}\), where h is planck’s constant.

Question 8.
State Heisenberg’s Uncertainity Principle. [Mar. 14]
Answer:
Uncertainity principle states that “it is impossible to measure both position (∆x) and momentum of an electron (∆p) [or any other particle] at the same time exactly”, i.e., ∆x . ∆p ≈ h where ∆x is uncertainty in the specification of position and ∆p is uncertainty in the specification of momentum.

Question 9.
The photoelectric cut off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? [Mar. 11]
Answer:
Cut off voltage, V₀ = 1.5 V; Maximum kinetic energy, (KE)_max = eV₀ = e × 1.5 = 1.5eV

Question 10.
An electron, an α-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Answer:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h/\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.A proton is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton.
Hence, α-particle has the shortest de Broglie wavelength.

Qeustion 11.
What is-the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts ? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelngth λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Question 12.
If the wave length of electro magnetic radiation is doubled, what happens to energy of photon ? [IPE 2015 (TS)]
Answer:
E = \(\frac{\mathrm{hc}}{\lambda}\) ⇒ E ∝ \(\frac{1}{\lambda}\), since wave length of photon is doubled, its energy becomes halved.

Question 13.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10^-15 V – s;

For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}=\frac{\mathrm{h}}{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10^-15; h = 4.12 × 10^-15 × 1.6 × 10^-19 = 6.592 × 10^-34 J – s.

Question 14.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 10³ W/m²; λ = 550 nm = 550 × 10^-9 m ,
h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s
Energy of each photon E = \(\frac{\text { hc }}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10^-19 J
No. of photons incident on the earth’s surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 10²¹ photons/m² – s.

Question 15.
Show that the wavelength of electromagnetic radiation is equal to the de-Brogile wavelength of its quantum (photon). [Mar. 14]
Answer:
Wave length of electromagnetic wave of frequency v and velocity C is given by,
λ = \(\frac{\mathrm{C}}{\mathrm{v}} \Rightarrow \lambda=\frac{\mathrm{C}}{\mathrm{v}} \times \frac{\mathrm{h}}{\mathrm{h}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{hv}}{\mathrm{C}}\right)}=\frac{\mathrm{h}}{\mathrm{p}}\)    (∵ \(\frac{\mathrm{hv}}{\mathrm{C}}\) = p)
Hence, we can say wavelength of electromagnetic radiation is equal to the de-Brogile wavelength.

Sample Problem :

Question 1.
Calculate the (a) momentum and (b) dE-Brogile wavelength of the electrons accelerated through a potential difference of 56 V. [Mar. 14]
Answer:
a) Mass of the electron, m = 9 × 10^-31 kg;
Potential difference, V = 56V
Momentum of electron, mv = \(\sqrt{2 \mathrm{eVm}}=\sqrt{2 \times\left(1.6 \times 10^{-19}\right) \times 56 \times 9 \times 10^{-31}}\) = 4.02 × 10^-24kg ms^-1

b) de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.62 \times 10^{-34}}{4.04 \times 10^{-24}}\) = 1.64 × 10^-10m

Short Answer Questions

Question 1.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

1. The experimental set up is shown in fig.
   
2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
4. The polarity of the plates C and A can be reversed by a commutator.
5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photo current stops or becomes zero is called stopping potential.
6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
7. From graph, we note that
   
   • The values of stopping potentials are different for different frequencies.
   • The value of stopping potential is more negative for radiation of highef incident frequency.
   • The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.

Question 2.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms^-1? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10^-34 J-s;
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{0.12 \times 20}=\frac{6.63 \times 10^{-34}}{2.4}\)
∴ λ = 2.762 × 10^-34 m = 2762 × 10^-21 Å.
The wave length of ball is very very small. Hence, its motion can be observed.

Question 3.
What is the effect of (i) intensity of light (ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:
1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted increases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.

ii) The effect of potential on photoelectric current:

1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
   
2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Sample Problem :

Question 1.
The work function of caesium metal is 2.14 eV When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential and (c) maximum speed of the emitted photoelectrons ? [A.P. Mar. 16]
Solution:
Given, Φ₀ = 2.14 eV; v = 6 × 10¹⁴ Hz
a) KE_max = hv – Φ₀ = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14
∴ KE_max = 0.35 eV

b) KE_max = eV₀ ⇒ 0.35 eV = eV₀
∴ V₀ = 0.35 V

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of ‘ incident light on stopping potential ?
Answer:

1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
2. The energy of photon is E = hv. Where ‘h1 is Planck’s constant; v is frequency of incident light (or radiation).
3. If the absorbed energy of photon is greater than the work function (Φ₀ = hυ₀), the electron is emitted with maximum kinetic energy i.e., k_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = eV₀ = hv – Φ₀. This equation is known as Einstein’s photoelectric equation.
4. Effect of intensity of light on photoelectric current:
   When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted decreases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.
   
5. The effect of potential on photoelectric current:
   • On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
     
   • On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v₀).
   • Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
6. The effect of frequency of incident radiation on stopping potential:
   On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means k_max increases eV₀ also increases.
   
7. From the graph, we note that
   • For a given photosensitive metal, the cut off potential (v₀) varies linearly with the frequency of the incident radiation.
   • For a given photosensitive metal, there is a certain minimum cut off frequency υ₀ (called threshold frequency) for which the stopping potential is zero.
     
8. From the graph we note that
   • The value of cut-off potential is different for radiation of different frequency.
   • The value of stopping potential is more negative for radiation of higher incident frequency.
9.  From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is. possible.

Textual Examples

Question 1.
Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. (a) What is the energy of a photon in the light beam ? (b) How many photons per second, on an average, are emitted by the source ?
Solution:
a) Each photon has an energy
E = hv = (6.63 × 10^-34 J s) (6.0 × 10¹⁴ Hz)
= 3.98 × 10^-19 J

b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = NE.
Then N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{2.0 \times 10^{-3} \mathrm{~W}}{3.98 \times 10^{-19} \mathrm{~J}}\)
= 5.0 × 10¹⁵ photons per second.

Question 2.
The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. [A.P. Mar. 16]
Solution:
a) For the cut-off or threshold frequency, the energy hv₀ of the incident radiation must be equal to work function Φ₀, so that
v₀ = \(\frac{\phi_0}{\mathrm{~h}}=\frac{2.14 \mathrm{eV}}{6.63 \times 10^{-34} \mathrm{Js}}\)
= \(\frac{2.14 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\) = 5.16 × 10¹⁴ Hz
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.

b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V₀ by the retarding potential V₀. Einstein’s Photoelectric equation is
eV₀ = hv – Φ₀ = \(\frac{\mathrm{hc}}{\lambda}\) – Φ₀
or λ = hc/(eV₀ + Φ₀)
= \(\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{(0.60 \mathrm{eV}+2.14 \mathrm{eV})}\)
= \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{(2.74 \mathrm{eV})}\)
λ = \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{2.74 \times 1.6 \times 10^{-19} \mathrm{~J}}\) = 454 nm

Question 3.
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum ? (Take h = 6.63 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J.)
(b) From which of the photosensitive materials with work functions listed in table and using the results of (i), (ii) and (iii) of (a) can you build a photoelectric device that operates with visible light ?

Solution:
a) Energy of the incident photon,
E = hv = hc/λ
E = (6.63 × 10^-34 J s) (3 × 10⁸ m/s)/λ.
= \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{\lambda}\)
i) For violet light,
λ₁ = 390 nm (lower wavelength end)
Incident photon energy,
E₁ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{390 \times 10^{-9} \mathrm{~m}}\)
5.10 × 10^-19 J = \(\frac{5.10 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
= 3.19 eV

ii) For yellow-green light,
λ₂ = 550 nm (average wavelength)
Incident photon energy,
E₂ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{550 \times 10^{-9} \mathrm{~m}}\)
= 3.62 × 10^-19 J = 2.26 eV.

iii) For red light,
λ₃ = 760 nm (higher wavelength end)
Incident photon energy,
E₃ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{760 \times 10^{-9} \mathrm{~m}}\)
= 2.62 × 10^-19 J = 1.64 eV

b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function Φ₀ of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with Φ₀ = 2.75 eV), K (with Φ₀ = 2.30 eV) and Cs (with Φ₀ = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with Φ₀ = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.

Question 4.
What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4 × 10⁶ m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s ?
Solution:
a) For the electron :
Mass m= 9.11 × 10^-31 kg, speed υ = 5.4 × 10⁶ m/s. Then, momentum
P = mυ = 9.11 × 10^-31 (kg) × 5.4 × 10⁶ (m/s)
P = 4.92 × 10^-24 kg m/s
de Broglie wavelength, λ = h/p
= \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
λ = 0.135 nm

b) For the ball:
Mass m’ = 0.150 kg, speed υ’ = 30.0 m/s.
Then momentum
p’ = m’υ’ = 0.150 (kg) × 30.0 (m/s)
p’ = 4.50 kg m/s
de Broglie wavelength λ’ = h/p’.
= \(\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
λ’ = 1.47 × 10^-34 m
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10^-19 times the size of the proton, quite beyond experimental measurement.

Question 5.
An electron, an a-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Solution:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h /\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left({ }_1^1 \mathrm{He}\right)\) is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton
Hence, α-particle has the shortest de Broglie wavelength.

Question 6.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10^-4. Calculate the particle’s mass and identify the particle.
Solution:
de Broglie wavelength of a moving particle, having mass m and velocity υ :
λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
Mass, m = h/A.
For an electron, mass m_e = h/λ_e υ_e
Now, we have υ/υ_e = 3 and
λ/λ_e = 1.813 × 10^-4
Then, mass of the particle,
m = m_e \(\left(\frac{\lambda_{\mathrm{e}}}{\lambda}\right)\left(\frac{v_{\mathrm{e}}}{v}\right)\)
m = (9.11 × 10^-31 kg) × (1/3) × (1/1.813 × 10^-4)
m = 1.675 × 10^-27 kg.
Thus, the particle, with this mass could be a proton or a neutron.

Question 7.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelength λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Students get through AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves

Very Short Answer Questions

Question 1.
Give anyone use of infrared rays. [T.S. Mar. 17; A.P. Mar. 16]
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 on Earth Satellites.
4. These are used in taking photographs during conditions of fog, smoke, etc.

Question 2.
How are infrared rays produced? How they can be detected?
Answer:
Infrared rays can be produced by vibrations of atoms and molecules. These waves can be detected by Thermopile, Bolometer, IR photographic film.

Question 3.
How are radio waves produced ? How can they detected ?
Answer:
Radio waves can be produced by rapid acceleration and deceleration of electrons in aerials (conductors). These can be detected by receivers of aerials.

Question 4.
If the wave length of E.M radiation is doubled, what happens to the energy of photon ? [IPE 2016 (TS)]
Answer:
If the wave length of electromagnetic radiation is doubled, then energy will be halved because energy is inversely proportional to.wavelength of electromagnetic waves.
E = hυ = hc/λ ⇒ E ∝ 1/λ (∵ hc is a constant)

Question 5.
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 6.
What is the ratio of speed of infrared rays and ultraviolet rays in vaccum ?
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 vaccum.

Question 7.
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 8.
What are the applications of microwaves ? [A.P. Mar. 17; 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 9.
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 10.
Give two uses of infrared rays.
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 11.
How are microwaves produced ? How can they detected ? [A.P. Mar. 16; IPE 15]
Answer:
Microwaves can be produced using Klystron valve or Magnetrons.
Microwaves can be detected using point contact diodes.

Question 12.
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 = i_d = ε₀ = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
∴ i = i_d = 0.6 A
∴ Displacement current (i_d) = 0.6 A.

Question 13.
What physical quantity is the same for X-rays of wavelength 10^-10m, red light of wavelength 6800 Å and radiowaves of wavelength 500in ?
Answer:
The speed in vaccum is same for all the given wavelengths, which is 3 × 10⁸ m/s.

Question 14.
A radio can tune into any station in the 7.5 MHz to 12MHz band. What is the corresponding wavelength band ?
Answer:
λ₁ = \(\frac{3 \times 10^8}{7.5 \times 10^6}\) = 40 m
λ₂ = \(\frac{3 \times 10^8}{12 \times 10^6}\) = 25 m
Thus wavelength band is 40m to 25m.

Question 15.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vaccum 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 16.
Define displacement current
Answer:
Displacement current (I_d) is equal to ε₀ times to the rate of change of electric flux. Displacement current is not the current produced due to charge carried. But it is due to varying electric flux. It is the current in the sense that it produces a magnetic field.
I_d = ε₀ \(\frac{\mathrm{d} \phi_{\varepsilon}}{\mathrm{dt}}\)

Short Answer Questions

Question 1.
State six characteristics of electromagnetic waves.
Answer:
Characteristics of electromagnetic waves :

1. Electromagnetic waves are produced by accelerated charges.
2. Electromagnetic waves are transverse in nature.
3. Electromagnetic waves donot require material medium for their propagation.
4. Electromagnetic waves obey principle of superposition of waves.
5. Velocity of E.M waves in vaccum depends on permittivity and permeability of free space.
6. Electromagnetic waves carry energy and momentum.
7. Electromagnetic waves exert pressure when they strike a surface.

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.

Problems

Question 1.
A plane electromagnetic wave travels in vaccum along 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:
Electric and magnetic fields \(\overline{\mathrm{E}}\) and \(\overline{\mathrm{B}}\) of an electromagnetic wave must be perpendicular to the propagation of electromagnetic wave. Hence they lie in X – Y plane mutually perpendicular to each other.
Frequency of wave, v = 30MHz = 30 × 10⁶Hz.; Velocity of light, C = 3 × 10⁸m/s
Wavelength of the wave, λ = \(\frac{C}{V}=\frac{3 \times 10^8}{30 \times 10^6}\) = 10m

Question 2.
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:
According to Maxwell, a charged particle oscillating with a frequency produces electro-magnetic waves of same frequency. Hence frequency of EM waves produced is, 10⁹Hz.

Textual Examples

Question 1.
A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, It is connected for charging in series with a resistor R = 1 M Q across a 2V battery (fig)- Calculate the magnetic field at a point P halfway between the centre and the periphery of the plates, after t = 10^-3 s. (The charge on the capacitor at time τ is q (t) = CV [1 – exp (-t/τ)], where the time constant τ is equal to CR.).
Solution:
The time constant of the CR circuit is τ = CR = 10^-3s. Then, we have
q(t) = CV [1 – exp (-t/τ) ]
= 2 × 10^-9 [1 – exp (-t /10^-3)
The electric field in between the plates at time t is
E = \(\frac{q(t)}{\varepsilon_0 A}=\frac{q}{\pi \varepsilon_0}\); A = π (1)² m² = area of the plates.

Consider now a circular loop of radius (1/2)m parallel to the plates passing through R The magnetic field B at all points on the loop is along the loop arid of the same value.
The flux Φ_E through this loop is
The flux Φ_E = E × area of the loop
= E × π × (\(\frac{1}{2}\))² = \(\frac{\pi \mathrm{E}}{4}=\frac{\mathrm{q}}{4 \varepsilon_0}\)
The displacement current
i_d = e₀ \(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\) = \(\frac{1}{4} \frac{\mathrm{dq}}{\mathrm{dt}}\) = 0.5 × 10^-6 exp (-1)
at t = 10^-3s. Now, applying Ampere-Maxwell law to the loop, we get
B × 2π × (\(\frac{1}{2}\)) = m₀.(i_c + i_d) = m₀(0 + i_d) = 0.5 × 10^-6 m₀ exp(-1)
or, B = 0.74 × 10^-13 T.

Question 2.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x – direction. At a particular point in space and time, E = \(6.3 \hat{\mathbf{j}}\) V/m. What is B at the point ?
Solution:
Using Eq. B₀ = [E₀/c] the magnitude of B is
B = \(\frac{\mathrm{E}}{\mathrm{c}}\)
= \(\frac{6.3 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 2.1 × 10^-8 T
To find the direction, we note that E is along y-direction and the wave propagates along x- axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, \((+\hat{\mathrm{j}})\) × \((+\hat{\mathrm{k}})\) = i, B is along the z-direction. Thus. B = 2.1 × 10^-8 \(\hat{\mathrm{k}}\) T

Question 3.
The magnetic field in a plane electromagnetic wave is given by
B_y = 2 × 10^-7 sin (0.5 × 10³ × + 1.5 × 10¹¹ t)T.
a) What is the wavelength and frequency of the wave ?
b) Write an expression for the electric field.
Solution:
a) Comparing the given equation with
B_y = B₀ sin \(\left[2 \pi\left(\frac{\mathrm{x}}{\lambda}+\frac{\mathrm{t}}{\mathrm{T}}\right)\right]\)
We get, λ = \(\frac{2 \pi}{0.5 \times 10^3}\) m = 1.26 cm,
and \(\frac{1}{\mathrm{~T}}\) = v = (1.5 × 10¹¹)/2π = 23.9 GHz

b) E₀ = B₀C = 2 × 10^-7 T × 3 × 10⁸ m/s = 6 × 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the di-rection of magnetic field. Therefore, the electric field component along the z-axis is obtained as E_z = 60 sin (0.5 × 10³x + 1.5 × 10¹¹t) V/m.

Question 4.
Light with an energy flux of 18 W/cm² falls on a non reflecting surface at normal incidence. If the surface has an area of 20 cm² find the average force exerted on the surface during a 30 minute time span.
Solution:
The total energy falling on the surface is
U = (18 W/cm²) × (20 cm²) × (30 × 60)
= 6.48 × 10⁵ J
Therefore, the total momentum delivered (for complete absorption) is
P = \(\frac{\mathrm{U}}{\mathrm{C}}=\frac{6.48 \times 10^5 \mathrm{I}}{3 \times 10^8 \mathrm{m} / \mathrm{s}}\) = 2.16 × 10^-3 kg m/s
The average force exerted on the surface is
F = \(\frac{\mathrm{p}}{\mathrm{t}}=\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}\) = 1.2 × 10^-6 N

Question 5.
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Solution:
The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3m, the surface area of the surrounding sphere is A = 4 πr² = 4π (3)² = 113m²
The intensity at this distance is
I = \(\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^2}\) = 0.022 W.m²
Half of this intensity is provided by the electric field and half by the magnetic field.
\(\frac{1}{2} I=\frac{1}{2}\left(\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2 \mathrm{C}\right)\)
= \(\frac{1}{2}\) (0.022 W/m²)
E_rms = \(\sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^8\right)}}\) V/m = 2.9 V/m
The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E₀ is
E₀ = \(\sqrt{2}\)E_rms = \(\sqrt{2}\) × 2.9 V/m
= 4.07 V/m
Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.
Now, let us calculate the strength of the magnetic field. It is
B_rms = \(\frac{E_{\mathrm{rms}}}{\mathrm{C}}=\frac{2.9 \mathrm{Vm}^{-1}}{3 \times 10^8 \mathrm{~ms}^{-1}}\) = 9.6 × 10^-9 T.
Again, since tbs field in the light beam is sinusoidal, the peak magnetic field is B₀ = \(\sqrt{2}\) B_rms = 1.4 × 10^-8 T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

Students get through AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 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. [T.S. Mar. 16]
Answer:
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
V_p = 200V, V_s = 2000V, N_p = 10
N_s = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) × N_p = \(\frac{2000}{200}\) × 10
N_s = 100.

Question 2.
What type of transformer is used in a 6V bed lamp ? [A.P. Mar. 17]
Answer:
Step down transformer is used in 6V bed lamp.

Question 3.
What is the phenomenon involved in the working of a transformer ? [Mar. 16(A.P.) 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 = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)

Question 5.
Write the expression for the reactance of i) an inductor and (ii) 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. [T.S. Mar. 15]
Answer:

1. In pure resistor A.C. e.m.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_rms I_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 sinΦ) component of current is zero. This component of current is known as wattless current. (I_rms sinΦ) is the wattless component of current.

Question 9.
When does a LCR series circuit have minimum impedance ?
Answer:
In LCR series circuit, Impendence (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 (cosΦ) = 1
Phase difference between voltage and current is zero. (Φ = 0)

Short Answer Questions

Question 1.
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 V_S be the emf s across the primary and secondary.
\(\frac{V_S}{V_p}=\frac{\text { Output emf }}{\text{Input emf}}=\frac{-N_2 \frac{d \phi}{d t}}{-N_1 \frac{d \phi}{d t}}=\frac{N_2}{N_1}\)
∴ \(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = Transformer ratio
Efficiency of transformer :
It is the ratio of output power to the input power.
η = \(\frac{\text { Outputpower }}{\text { Input power }}\) × 100

Problems

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
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 = i₀sin(ωt – Φ))
υ = 40 sin(100t)V (∵ V = V₀sin(ωt ))
i₀ = \(\sqrt{2}\) , V₀ = 40, ω = 100, Φ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0}\) cosΦ = \(\frac{40}{\sqrt{2}}\) cos\(\frac{\pi}{4}\), R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\), R = 20 Ω

Question 4.
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 = \(\sqrt{V_R^2+\left(V_L^2-V_C^2\right)}\) ; V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\) ; V = \(\sqrt{35^2}\); V = 35 Volt.

Question 5.
What is step up transformer ? How it differs from step down transformer ?
Solution:
The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio.
T = \(\frac{N_S}{N_p}=\frac{\text {No. of turns in the secondary }}{\text {No. of turns in the primary }}\)
If N_S > N_P, then the transformer is called step up transformer.
If N_S < N_P, then the transformer is called step down transformer.

Textual Examples

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V.
The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?
Solution:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 4.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?
Solution:
The capacitive reactance is
X_c = \(\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}\)
= 212 Ω
The rms current is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{220 \mathrm{~V}}{212 \Omega}\)
= 1.04 A
The peak current is
i_m = \(\sqrt{2}\)I = (1.41)(1.04A) = 1.47A
This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 5.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.
Solution:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Question 6.
A resistor of 200Ω and a capacitor of 15.0 μF are connected in series to a 220V, 50 Hz ac source,
(a) Calculate the current in the circuit;
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.
Solution:
Given
R = 200Ω. C = 15.0 μF = 15.0 × 10^-6F
V = 220V, v = 50Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
Z = \(\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{C}}^2}=\sqrt{\mathrm{R}^2+(2 \pi v C)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+\left(2 \times 3.14 \times 50 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+(212 \Omega)^2}\) = 291.5Ω
Therefore, the current in the circuit is
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{220 \mathrm{~V}}{291.5 \Omega}\) = 0.755A

(b) Since the current is the same throughout the circuit, we have
V_R = IR = (0.755 A) (200Ω) = 151V
V_C = IX_C = (0.755A) (212.3Ω) = 160.3V
The algebraic sum of the two voltages, V_R and V_C is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox ? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
V_R+C = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 7.
a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
Solution:
a) We know that P = IV cosΦ where cosΦ is the power factor. To supply a given power at a given voltage, if cosΦ is small, we have to increase current accordingly. But this will lead to large power loss (I^R) in transmission.

b) Suppose in a circuit, current I lags the voltage by an angle Φ.
Then power factor cosΦ = R/Z
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram in the figure how this can be achieved. Let us resolve I into two components, I_P

along the applied voltage V and I_q perpendicular to the applied voltage. I_q is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I’_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I’_q cancel each other and P is effectively I_P V.

Question 8.
A sinusoidal voltage of peak value 283 V , and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω. L = 25.48 mH. and C = 796μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Solution:
a) To find the impedance of the circuit, we first calculate X_L and X_C.
X_L = 2πvL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω
X_C = \(\frac{1}{2 \pi v \mathrm{C}}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4Ω
Therefore,
z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\sqrt{3^2+(8-4)^2}\)
= 5Ω

b) Phase difference, Φ = tan^-1\(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
= tan^-1\(\left(\frac{4-8}{3}\right)\) = -53.1°

c) The power dissipated in the circuit is
P = I²R

Therefore, P = (40A)² × 3Ω = 4800W
= 4.8 kW

d) Power factor = cos Φ = cos 53.1° = 0.6.

Question 9.
Suppose the frequency of the source in the previous example can be varied,
(a) What is the frequency of the source at which resonance occurs ?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition.
Solution:
(a) The frequency at which the resonance occurs is
ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}}\)
= 222.1 rad/s
v_r = \(\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14}\) Hz = 35.4Hz

b) The impedance Z at resonant condition is equal to the resistance
Z = R = 3Ω
The rms current at resonance is ,
as V = \(\frac{v_{\mathrm{m}}}{\sqrt{2}}\)
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}\) = 66.7 A
The power dissipated at resonance is
P = I² × R = (66.7)² × 3 = 13.35 kW
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 8.

Question 10.
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ?
Solution:
The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

Question 11.
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Solution:
Let q₀ be the initial charge tin a capacitor. Let the charged capacitor be connected to an inductor of inductance L. this LC circuit will sustain an oscillation with frequency
\(\omega\left(2 \pi v=\frac{1}{\sqrt{\mathrm{LC}}}\right)\)
At an instant t, charge q on the capacitor and the current i are given by :
q(t) = q₀ cos ωt
i(t) = -q₀ co sin ωt
Energy stored in the capacitor at time ‘t’ is
U_E = \(\frac{1}{2}\) C V² = \(\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) cos² (ωt)
Energy stored in the inductor at time ‘t’ is
U_M = \(\frac{1}{2}\) L i²
= \(\frac{1}{2}\) L q₀² ω²sin² (ωt)
= \(\frac{Q_0^2}{2 C} \sin ^2(\omega t)\) [∵  ω = 1/ \(\sqrt{L C}\)]
Sum of energies
U_E + U_M = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) [cos² ωt + sin²ωt) = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\)
This sum is constant in time as q₀ and C, both are time-independent.

Students get through AP Inter 2nd Year Physics Important Questions 9th Lesson Electromagnetic Induction which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 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 ∝ \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

Question 4.
State Lenz’s law.
Answer:
The direction of induced e.m.f (or) current is such that it opposes the cause which produces 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) B/υ.

Question 6.
What are Eddy currents ? [A.P. Mar. 15]
Answer:
Eddy currents (or) Focault 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 = -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 veiority υ 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 Flemings left hand rule)
Electromagnetic force (ε) = \(\frac{W}{Q}\)
ε = \(\frac{\mathrm{Bqυl}}{\mathrm{q}}\) ⇒ ε = Blυ ……………. (3)

Question 2.
Describe the ways in which Eddy currents are used to advantage. [A.P. Mar. 17, 16; A.P. & T.S. Mar. 15]
Answer:
Eddy currents are used to advantage in
i) 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.

ii) Induction Motor: Eddy currents are used to rotate the short circuited motor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current.

iii) 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.

iv) 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.

v) 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.
Answer:
Consider two solenoids as shown in figure. The length of primary coil be l and area 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 \(\frac{\mathrm{N}_1}{l}\) I × A ……………. (2)
The same magnetic flux is linked with the secondary coil.
∴ Total magnetic flux linked with secondary = μ₀\(\frac{\mathrm{N}_1 \mathrm{i}}{l}\) × A × N₂ ……………. (3)
If M be mutual inductance of the two coils, the total flux linked with the secondary is Mi.
∴ Mi = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{iA}}{l}\) ……………… (4)
M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}\) …………….. (5) (∵ A = πr²)
(or) M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2\left(\pi \mathrm{r}^2\right)}{l}\) ………….. (6) (∵μ_r = \(\frac{\mu}{\mu_0}\))

Problems

Question 1.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? (Note that 1 G = 10^-4 T.)
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 2.
Number of turns in a coil are 100. When a current of 5A is flowing through the coil, the magnetic flux is 10^-6Wb. Find the self induction. [Board Model Paper]
Solution:
Self inductance, L = \(\frac{n \phi}{\mathrm{i}}\)
number of turns, n = 100; magnetic flux, Φ = 10^-6Wb; Current, i = 5A
L = \(\frac{100 \times 10^{-6}}{5}\) = 20 × 10^-6 = 20 µH
∴ Self inductance, L = 20 µH

Question 3.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit. [Mar. 16 (T.S.) 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 = \(\frac{200}{50}\) = 4 H.

Question 4.
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 ? [T.S. Mar. 17]
Solution:
Given, mutual inductance of coil M = 1.5 H
Current change in coil dI = 20 – 0 = 20 A
Time taken in change dt = 0.5s, Induced emf in the coil e M = \(M \frac{d I}{d t}=\frac{d \phi}{d t}\)
dΦ = M.dI = 1.5 × 20, dΦ = 30 Wb,
Thus the change of flux linkage is 30 Wb.

Question 5.
A jet plane is travelling towards west at a speed of 1800 km/K 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 = 3.1V
Thus, the voltage difference developed between the ends is 3.1 V.

Textual Examples

Question 1.
(a) What would you do to obtain a large deflection of the galvanometer ?
(b) How would you demonstrate the presence of an induced current in the absence of a gal-vanometer ?
Solution:
a) To obtain a large deflection, one or more of the following steps can be taken :

1. Use a rod made of soft iron inside the coil C₂.
2. Connect the coil to a powerful battery, and
3. Move the arrangement rapidly towards the test coil C₁.

b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.

Question 2.
A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time interval.
Solution:
The angle θ made by the area vector of the loop with the magnetic field is 45°.
From eq. Φ_B = B.A. = BA cosθ the initial magnetic flux is Φ = BA cosθ
= \(\frac{0.1 \times 10^{-2}}{\sqrt{2}}\) Wb
Final flux, Φ_min = 0
The change in flux is brought about in 0.70 s. From Eq. ε = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\), the magnitude of the induced emf is given by
ε = \(\frac{\left|\Delta \phi_{\mathrm{B}}\right|}{\Delta \mathrm{t}}=\frac{|(\phi-0)|}{\Delta \mathrm{t}}=\frac{10^{-3}}{\sqrt{2} \times 0.7}\) = 1.0 mV
And the magnitude of the current is
I = \(\frac{\varepsilon}{\mathrm{R}}=\frac{10^{-3} \cdot \mathrm{V}}{0.5 \Omega}\) = 2 mA.
Note that the earth’s magnetic field also produces a flux through the loop.

Question 3.
A circular coil of radius 10 cm, 1500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. .Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T.
Solution:
Initial flux through the coil,
Φ_B(initial) = BA C0S θ
= 3.0 × 10^-5 × (π × 10^-2) × cos 0°
= 3π × 10^-7 Wb.
Final flux after the rotation,
Φ_B(final) = 3.0 × 10^-5 × (π × 10^-2) × cos 180°
= -3π × 10^-7 Wb.
Therefore, estimated value of the induced emf is,
ε = N\(\frac{\Delta \phi}{\Delta t}\) = 500 × (6π × 10^-7)/0.25
= 3.8 × 10^-3 V
I = ε/R = 1.9. × 10^-3 A.
Note that the magnitudes of ε and I are the estimated values.

Question 4.
The following figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

Solution:

1. The magnetic flux through the rectangular loop abed increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux.
2. Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacd, so as to oppose the change in flux.
3. As the magnetic flux decreases due to motion of the irregular shaped loop abed out of the region of magnetic field, the induced current flows along edabe, so as to oppose change in flux.

Question 5.
a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets ?
b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) When it is partially outside the plates of the capacitor ? The electric field is normal to the plane of the loop.
c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region as in the figure, to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region ? The field is normal to the loops.

d) Predict the polarity of the capacitor in the situation described by the following figure.

Solution:
a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.

b) No current is induced in either case. Current can not be induced by changing the electric flux.

c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Question 6.
A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring as in figure. A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring ?

Solution:
Method I : As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.
Using equation (ε = – Bl \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = BlV), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
dε = Bυ dr. Hence,
ε = \(\int \mathrm{d} \varepsilon=\int_0^{\mathrm{R}} \mathrm{B} v \mathrm{dr}=\int_0^{\mathrm{R}} \mathrm{B} \omega \mathrm{rdr}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
Note that we have used υ = ωr. This gives
ε = \(\frac{1}{2}\) × 1.0 × 2π × 50 × (1²) = 157 V.

Method II: To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by
πR² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\), R²θ
where R is the radius of the circle. Hence, the induced emf is
e = B × \(\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{1}{2} \mathrm{R}^2 \theta\right]=\frac{1}{2} \mathrm{BR}^2 \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
[Note: \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = ω = 2πv]
This expression is identical to the – expression obtained by Method I and we get the same value of ε.

Question 7.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? Note that 1 G = 10^-4 T.
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 8.
Refer to fig. The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed o. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Solution:
Let us first consider the forward motion from x = 0 to x = 2b. The flux Φ_B linked with the circuit SPQR is
Φ_B = Blx 0 ≤ x < b
= Blυ b ≤ x < 2b
The induced emf is,
ε = – \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
= -Blυ 0 ≤ x < b
= 0 b ≤ x < 2b.
When the induced emf is non-zero, the current I is (in magnitude)
I = \(\frac{\mathrm{B} l v}{\mathrm{r}}\)

The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude
F = \(\frac{\mathrm{B}^2 l^2 v}{\mathrm{r}}\) 0 ≤ x < b
= 0 b ≤ x < 2b
The Joule heating loss is
P_j = I²r
= \(\frac{\mathrm{B}^2 l^2 v^2}{\mathrm{r}}\)   0 ≤ x < b
= 0   b ≤ x < 2b
One obtains similar expressions for the inward motion from x = 2b to x = 0.

Question 9.
Two concentric circular coils, one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.
Solution:
Let a current I₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = μ₀I₂/2r₂. Since the other co-axially placed coil has a very small radius. B₂ may be considered constant over its cross-sectional area. Hence,
Φ₂ = \(\pi \mathrm{r}_1^2 \mathrm{~B}_2\)
= \(\frac{\mu_0 \pi \mathrm{r}_1^2}{2 \mathrm{r}_2} \mathrm{I}_2\)
= M₁₂I₂
Thus,
M₁₂ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
From Equation M₁₂ = M₂₁ = M
M₁₂ = M₂₁ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area \(\pi r_1^2\). However, we can accept this value because r₁ << r₂.

Question 10.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.
(b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor ?
Solution:
a) From Equation ε = – L \(\frac{\mathrm{dI}}{\mathrm{dT}}\),
the magnetic energy is
U_B = \(\frac{1}{2}\) LI²
= \(\frac{1}{2} \mathrm{~L}\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
(since B = μ₀nI, for a solenoid)
= \(\frac{1}{2}\) (μ₀n²Al) \(\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
[from Equation L = μ₀n²Al]
= \(\frac{1}{2 \mu_0}\) B²Al

b) The magnetic energy per unit volume is,
u_B = \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{V}}\)
[where Vis volume that contains flux]
= \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{Al}}=\frac{\mathrm{B}}{2 \mu_0}\) ……………….. (1)
We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.
u_E = \(\frac{1}{2}\) ε₀E² ………………. (2)
In both the cases energy is proportional to the square of the field strength.

Question 11.
Kamla peddles a stationary bicycle, the pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil ?
Solution:
Here f = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T.
Employing Equation ε = NBA ω sin ωt
ε₀ = NBA (2πv)
= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5
= 0.314 V
The maximum voltage is 0.314 V

Students get through AP Inter 2nd Year Physics Important Questions 8th Lesson Magnetism and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 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.
Do you find two magnetic field lines intersecting? Why?
Answer:
Two magnetic field lines never intersect. If they intersect, at the point of intersection the field can have two directions. This is impossible. So, two field lines never intersect.

Question 3.
What happens to the compass needles at the Earth’s poles ? [T.S. Mar. 17; IPE 2014]
Answer:
At the Earth poles, the magnetic field lines are converging or diverging,vertically so that the horizontal component is negligible. Hence, the compass needle can point along any direction.

Question 4.
What do you understand by the ‘magnetisation’ of a sample ? Give its SI unit. [IPE 2016 (AP)]
Answer:
Magnetisation (M) of a sample is equal to its net magnetic moment per unit volume.
Magnetisation, M = \(\frac{m_{n e t}}{V}\), SI unit of magnetisation is Am^-1.

Question 5.
What is the magnetic moment associated with a solenoid ? [IPE 2016 (TS)]
Answer:
The magnitude of magnetic moment of the solenoid is, M = n × 2l × i × πa²
where, ‘n’ is number of turns, ‘2l’ is length of the solenoid, ‘i’ is current passing through coil, and ‘a’ is area of cross section of solenoid.

Question 6.
What are the units of magnetic moment, magnetic induction and magnetic field ? [IPE 2016 (AP), (TS)]
Answer:
The unit of magnetic moment (M) is ampere-meter² (Am²).
The unit of magnetic induction (B) is tesla (T) or N/A-m.
The unit of magnetic field (B) is tesla.

Question 7.
Magnetic lines form continuous closed loops. Why ? [T.S. 2017; IPE 2016(AP)]
Answer:
Magnetic lines are imaginary lines. Within the magnet, they move from south pole to north pole and outside the magnet they move from north pole to south pole. Hence, magnetic lines form continuous closed loops.

Question 8.
Define magnetic declination. [IPE 2016 (TS)]
Answer:
Magnetic declination at a place is the angle between magnetic meridian and geographic meridian at that place.

Question 9.
Define magnetic inclination or angle of dip. [A.P. Mar. 17; A.P. & T.S. 2015 (TS), 14]
Answer:
Magnetic inclination at a place is the angle between direction of total strength of earth’s magnetic field and horizontal line in magnetic meridian.

Question 10.
Classify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [T.S. 2015, 2016 (TS); A.P. Mar. 17]
Answer:
Manganese …………. Paramagnetic
Cobalt ………….. Ferromagnetic
Nickel …………….. Ferromagnetic
Bismuth ……………. Diamagnetic
Oxygen ………………. Paramagnetic
Copper …………………. Diamagnetic

Question 11.
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, F₁d₁² = F₂ d₂²
Fd² = 2 F d₂²
⇒ d₂² = \(\frac{\mathrm{d}^2}{2}\)
d₂ = \(\frac{\mathrm{d}}{\sqrt{2}}\)

Question 12.
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 = \(\frac{\mu_0 \mathrm{n} \mathrm{Ia}^2}{2 \mathrm{r}}\), 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 13.
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 solenoid is given by B = \(\frac{\mu_0 \mathrm{nI} \mathrm{a}^2(2 l)}{2 \mathrm{r}^3}\)
As I, a², 2l and r are constants, B ∝ n
⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \Rightarrow \frac{\mathrm{B}_2}{\mathrm{~B}}=\frac{2 \mathrm{n}}{\mathrm{n}}\)
∴ B₂ = 2B

Question 14.
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 Am² along the axis of solenoid.

Short Answer Questions

Question 1.
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 perpe-ndicular direction to the magnetic field.
c) When they are kept in a non-uniform magnetic field, they move 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 (χ) is low and negative.
E.g.: Copper, Silver, Water, Gold, Antimony, Bismuth, Mercury, Quartz Diamond etc.

Paramagnetic substances
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 magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they are kept in a non-uniform magnetic field„they move 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 (χ) is small and positive.
E.g.: Aluminium, Magnesium, Tungsten, Platinum, Mang-anese, liquid oxygen, Ferric chloride, Cupric chloride etc.

Ferromagnetic 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 are kept in a non-uniform magnetic field they move from the regions of lesser (magnetic field) strength to the regions of stronger (magnetic field) strength.
d) The relative permeability is much greater than 1. μ_r >> 1 and positive.
e) The susceptibility (χ) is high and positive.
E.g.: Iron, Cobalt, Nickel, Gadolinium and their alloys.

Question 2.
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 HE. 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 3.
Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility.
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.
   Suspectibility χ = \(\frac{\text { Intensity of magnetisation, } \mathrm{I}}{\text { Applied magnetic field, } \mathrm{H}}\)
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 χ :
   a) Suppose that material is placed in a magnetic field of intensity H. Let I be the intensity of magnetisation acquired by it.
   b) Then the magnetic induction within the material is
   B = μ₀H + μ₀I ⇒ \(\frac{\mathrm{B}}{\mathrm{H}}\) = μ₀[1 + \(\frac{\mathrm{I}}{\mathrm{H}}\)]
   ⇒ μ = μ₀[1 + χ] ⇒ \(\frac{\mu}{\mu_0}\) = 1 + χ
   ∴ μ_r = 1 + χ [∵μ_r = \(\frac{\mu}{\mu_0}\)]
5. Negative susceptibility (χ) of diamagnetic elements are Bismuth (-1.66 × 10^-5) and copper (-9.8 × 10^-6).
6. Positive susceptibility of paramagnetic 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 4.
Derive an expression for magnetic field induction on the equatorial line of a barmagnet. [Board Model Paper]
Answer:
At a point on equatorial line: Let us consider a point ‘P’ at a distance ‘d’ on the equatorial line from the centre of a bar magnet.

Qeustion 5.
What do you understand by “hysteresis” ? How does this propertry influence the choice of materials used in different appliances where electromagnets are used ?
Answer:

1. Cycle of magnetisation : When a ferromagnetic 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 :
   a) In fig, a closed curve ABCDEFA in H – I plane, called hysteris loop is shown in fig.
   
   b) When ferromagnetic specimen is slowly magnetised, I increases with H.
   c) Part OA of the curve shows that I increases with H.
   d) At point A, the value I becomes constant is called saturation value.
   e) At B, I has some value while H is zero.
   f) 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.

Question 6.
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 in terms 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 = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 m}{r^3}\)
6. The total magnetic field, at a point P due to solenoid is given by
   B = \(\frac{\mu_0 \mathrm{n} \mathrm{I}}{2} \frac{\mathrm{a}^2}{\mathrm{r}^3}(2 l)=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{n}(2 l) \mathrm{I} \pi \mathrm{a}^2}{\mathrm{r}^3}\)
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 7.
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 i 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 τ = m × B
4. In magnitude τ = mB sin θ.
   Here τ is restoring torque and θ is the angle between m and B.
5. Therefore, in equilibrium i \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – mB sinθ. Negative sign with mB sin0 implies that restoring torque is in opposition to deflecting torque.
   
6. For small values of o in radians, we approximate sinθ ≃ θ and get \(i \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) ≃ – mBθ
   \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2} \approx \frac{-\mathrm{mB}}{\mathcal{j}} \theta\) …………….. (1)
   This represents a simple harmonic motion. .
7. From defination of simple harmonic motion, we have \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – ω²θ …………… (2)
   From equation (I) and (II), we get ⇒ ω² = \(\frac{\mathrm{mB}}{\mathcal{J}}\)
   ∴ ω = \(\sqrt{\frac{\mathrm{mB}}{\mathcal{J}}}\)
8. Therefore, the time period is T = \(=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathcal{J}}{\mathrm{mB}}}\)

Qeustion 8.
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 barmagnet, held horizontally at A and which is set into angular oscillations in the Earth’s 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 nqt remain vertical component (B₁ sin θ₁). It can have only horizontal component (B₁ cosθ₁).
5. The time period of a bar magnet in uniform magnetic field is given by T = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_{\mathrm{H}}}}\)
   
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 = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_1 \cos \theta_1}}\) …………… (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 earth’s 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 = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_2 \cos \theta_2}}\) ………………… (2)
11. Dividing equation (1) by equation (2), we get \(\frac{T_1}{T_2}=\sqrt{\frac{\mathrm{mB}_2 \cos \theta_2}{\mathrm{mB}_1 \cos \theta_1}}\)
    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, \(\frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\mathrm{T}_2^2 \cos \theta_2}{\mathrm{~T}_1^2 \cos \theta_1}\)
14. By knowing T₁, T₂ and θ₁, θ₂ at different places A and B, we can find the ratio of resultant magnetic fields.

Question 9.
Obtain Gauss’ Law for magnetism and explain it.
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 ∆S of this surface as shown in figure.
4. Magnetic flux through this area element is defined as ∆Φ_B = B. ∆S. Then the net flux Φ_B, is,
   Φ_B = \(\sum_{\text {all }} \Delta \phi_B=\sum_{\text {all }} \text { B. } \Delta \mathrm{S}=0\)
5. If the area elements are really small, we can rewrite this equation as
   Φ_B = \(\oint\)B.ds = 0 …………………. (I)
   
6. Comparing this equation with Gauss’s law of electrostatics i.e., electric flux through a closed surface S is given by
   Φ_E = \(\oint \text { E. } \Delta S=\frac{q}{\varepsilon_0}\) …………….. (II) 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_E = \(\int_S\) B . dS = 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 10.
What are ferromagnetic materials ? Give examples. What happens to a ferromagnetic material at curie temperature ? [IPE 2015 (TS)]
Answer:
Ferro magnetic substances : (a) These are strongly attracted by magnet, (b) Susceptibility is large, positive and temperature dependent, (c) Relative permeability, μ_r > > 1 (d) Atoms have permanent dipole moments which are organised in domains. Ex: Iron, Cobalt, Nickel

Curie temperature : The temperature above which a ferro magnetic substance changes in to para magnetic substance changes in to para magnetic substance is called curie temperature.

Problems

Question 1.
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 = Bin A cos θ = 0.3 × 0.5 × 20 × 800× 10^-6 × cos 0°
∴ τ = 2.4 10^-3 Nm

Question 2.
In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression the magnetic moment (p) 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.

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

Qeustion 3.
A bar magnet of length 0.1m and with a magnetic moment of 5Am² is placed in a uniform a magnetic field of intensity 0.4T, with its axis making an angle of 60° with the field. What is the torque on the magnet ?
Answer:
Given, 2l = 0.1m; m = 5A – m²; B = 0.4T; θ = 60°.
Torque, T = mB sin θ = 5 × 0.4 × sin 60° = 2 × \(\frac{\sqrt{3}}{2}\)
∴ T = 1.732 N – m

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 = 3200 Am^-1

Qeustion 5.
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^-5 T
∴ B_E = 5.2 × 10^-5 T

Question 6.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that H_E = 0.26 G. From Fig., we have

The earth’s magnetic field, B_E, its horizontal and vertical components. H_E and Z_E. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Qeustion 7.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m².
Solution:
From Eq.

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment. .
Solution:
The equatorial magnetic field is,
We are given that B_E ~ 0.4 G = 4 × 10^-5 T. For r, we take the radius of the earth 6.4 × 10⁶ m.
Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 10² × (6.4 × 10⁶)³ (μ₀/4π = 10^-7)
= 1.05 × 10²³ A m²
This is close to the value 8 × 10²² A m² quoted in geomagnetic texts.

Textual Examples

Question 1.
In Fig, the magnetic needle has magnetic moment 6.7 × 10^-2 Am² and moment of inertia i = 7.5 × 10^-6 kg m². It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field ?
Solution:
The time period of oscillation is, :
T = \(\frac{6.70}{10}\) = 0.67 s

The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet.
From Eq. B =
= \(\frac{4 \times(3.14)^2 \times 7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times(.067)^2}\)
= 0.01 T

Question 2.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet ?
(b) What is the work done in moving it from its most stable to most unstable position ?
(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.
Solution:
a) From Eq., τ = m B sin θ, θ = 30°, hence sin θ = 1/2.
Thus, 0.016 = m × (800 × 10^-4 T) × (1/2)
m = 160 × 2/800 =0.40 Am²

b) From Eq . U_m = -m. B, the most stable position is θ = 0° and the most unstable position is q = 180°. Work done is given by
W = U_m (θ = 180°) – U_m (θ = 0°)
= 2 m B = 2 × 0.40 × 800 × 10^-4 = 0.064 J

c) From Eq., m_s = NIA. From part (a), m_s = 0.40 Am²
0.40 = 1000 × I × 2 × 10^-4
I = 0.40 × 10⁴/(1000 × 2) = 2A

Question 3.
a) What happens if a bar magnet is cut into two pieces :
(i) transverse to its length,
(ii) along its length ?
b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why ?
c) Must every magnetic configuration have a north pole and a south pole ? What about the field due to a toroid ?
d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised ? If only one is magnetised how does one ascertain which one ? (Use nothing else but the bars A and B].
Solution:
a) In either case, one gets two magnets, each with a north and south pole.

b) No force if the field is uniform. The iron nail experiences a non-uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. Then net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.

c) Not necessarily. True only if the source of the field has a net nonzero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.

d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is magnet, pick up one, (say, A) and lower one of its end first one of the ends of the other (say, B) and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised.

Question 4.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m², the same as in Example – 2.
Solution:
From Eq.

Question 5.
Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.

a) In which configuration the system is not in equilibrium ?
b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium ?
c) Which configuration corresponds to the lowest potential energy among all the configurations shown ?
Solution:
Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression. *
B_P = –\(\frac{\mu_0}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on the normal bisector)
B_P = \(\frac{\mu_0 2}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on axis)
where mp is the magnetic moment of the dipole P.
Equilibrium is stable when m_Q is parallel to B_P, and unstable when it is anti-parallel to B_P. For instance for the configuration Q₃ for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q₃ is stable. Thus,
a) PQ₁ and PQ₂
b) (i) PQ₃, PQ₆ (stable); (ii) PQ₅, PQ₄ (unstable)
c) PQ₆.

Question 6.
Many of the diagrams given in Fig. show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.

Solution:

a) Wrong: Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor.

b) Wrong: Magnetic field lines (like electric lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges.

c) Right: Magnetic lines are completely confined within a toroid: Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field.,

d) Wrong: Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops.

e) Right: These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the next flux of the field is zero.

f) Wrong: These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [(e) and (f)] should be carefully grasped.

g) Wrong: Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.

Question 7.
a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field line’s also represent the lines of force on a moving charged particle at every point ?
b) Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why ?
c) If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified ?
d) Does a bar magnet exert a torque on itself due to its own field ? Does one element of a current – carrying wire exert a force on another element of the same wire ?
e) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero ?
Solution:
a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force.

b) If field lines were entirely confined between two ends of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no ‘ends’.

c) Gauss’s law of magnetism states that the flux of,B thrugh any closed surface is always
zero \(\int_s B \cdot d s\) = o.
If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, \(\int_s B \cdot d s\) = μ₀q_m
where q_m is the (monopole) magnetic charge enclosed by S.]

d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero).

e) Yes. The average of the cahrge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment.
Solution:
The equatorial magnetic field is,
B_E = \(\frac{\mu_0 \mathrm{~m}}{4 \pi \mathrm{r}^3}\)
We are given that B_E ~ 0.4 G = 4 × 10^-5 T. For r, we take the radius of the earth 6.4 × 10⁶ m. Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 10² × (6.4 × 10⁶)³ (μ₀/4π = 10^-7)
= 1.05 × 10²³ A m²
This is close to the value 8 × 10²² A m² quoted in geomagnetic texts.

Question 9.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that H_E = 0.26 G. From Fig., we have

The earth’s magnetic field, B_E, its horizontal and vertical components. H_E and Z_E. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
B_E = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Question 10.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current I_m.
Solution:
a) The field H is dependent of the material of the core, and is
H = nI = 1000 × 2.0 = 2 × 10³ A/m

b) The magnetic field B is given by
B = μ_rμ₀H
= 400 × 4π × 10^-7 (N/A³) × 2 × 10³ (A/m) = 1.0 T

c) Magnetisation is given by
M = (B – μ₀ H)/μ₀
= (μ_rμ₀H – μ₀H)/μ₀ = (μ_r – 1) H = 399 × H ≃ 8 × 10⁵ A/m

d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_rn₀ (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794A.

Question 11.
A domain in ferromagnetic iron is in the form of a cube of side length 1 μm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm³. Assume that each iron atom has a dipole moment of 9.27 × 10^-24 A m³.
Solution:
The volume of the cubic domain is
V = (10^-6 m)³ = 10^-18 m³ = 10^-12 cm³
Its mass is volume × density = 7.9 g cm^-3 × 10^-12 cm³ = 7.9 × 10^-12 g
It is given that Afagadro number (6.023 × 10²³) of iron atoms have a mass of 55g. Hence,the number of atoms in the domain is
N = \(\frac{7.9 \times 10^{-12} \times 6.023 \times 10^{23}}{55}\)
= 8.65 × 10¹⁰ atoms
The maximum possible dipole moment m_max is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus,
m_max = (8.65 × 10¹⁰) × (9-27 × 10^-24)
= 8.0 × 10^-13 Am²
The consequent magnetisation is
M_max = m_max/DomainVolume :
= 8.0 × 10^-13 Am²/10^-18 m³
= 8.0 × 10⁵ Am^-1.

Students get through AP Inter 2nd Year Physics Important Questions 7th Lesson Moving Charges and Magnetism which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 7th Lesson Moving Charges and Magnetism

Very Short Answer Questions

Question 1.
What is the importance of Oersted’s experiment? [T.S. Mar. 17]
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.
Answer:
Ampere’s law: The line integral of the intensity of magnetic induction around a closed path is equal to g0 times the total current enclosed in it.
∴ \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} l\) = μ₀ i.
Biot – Savart’s laws: Biot – Savart’s law states that the intensity of magnetic induction (dB) due to a small element is directly proportional to the
i) current (i)
ii) length of the element (dZ)
iii) sine angle between radius vector (r) and dl and inversely proportional to the square of the point from current element.
∴dB ∝ \(\frac{\mathrm{i} \mathrm{dl} \sin \theta}{\mathrm{r}^2}\)
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} \mathrm{d} l \sin \theta}{\mathrm{r}^2}\)

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 = \(\frac{\mu_0 \mathrm{ni} \mathrm{r}^2}{2\left(\mathrm{r}^2+\mathrm{x}^2\right)^{3 / 2}}\)
2. At the centre of the coil B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 4.
A circular coil of radius T 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 = π 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 sinθ
2. If θ = 90°,F_Max = BiL
   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. 17; 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 RD 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^-9 A.

Question 10.
How do you convert a moving coil galvanometer into an ammeter ?
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. Mar. 16, 15, 14; A.P. Mar. 16]
Answer:
A high resistance is connected in series to the moving coil galvanometer, then it converts to voltmeter.

R = \(\frac{\mathrm{v}}{\mathrm{i}_g}\) – 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 vaccum?
Answer:
Speed of light in vaccum (C) = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Here μ₀ = m₀ = permeability in vaccum
ε₀ = permittivity in vaccum.

Question 13.
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic Held be set up in such a manner that the loop turns about the vertical axis ?
Answer:
Torque (τ) = \(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=\mathrm{i} \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) (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 itself.

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 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 the magnetic flux through it will assume a circular shape with its plane normal to the field.

Question 16.
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil ?
Solution:
Since the coil is tightly wound we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is (From Eq.),
B₀ = \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \hat{\mathrm{i}}\)
B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 10^2 \times 1}{2 \times 10^{-1}}\) = 2π × 10^-4 = 6.28 × 10^-4 T

Question 17.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ?
Solution:
The number of turns per unit length is,
n = \(\frac{500}{0.5}\) = 1000 turns / m
The length l = 0.5m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a.
Hence, we can use the long solenoid formula, namely, Eq. (B = μ₀nI)
B = μ₀ n I
= 4π × 10^-7 × 10³ × 5 = 6.28 × 10^-3 T

Question 18.
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^-4 T
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 19.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T ?
Solution:
According to the question
I = 8 A, 6 = 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

Short Answer Questions

Question 1.
State and explain Biot-Savart’s law.
Answer:
Consider a very small element of length dl of a conductor carrying current (i). Magnetic induction due to small element at a point P distance r from the element.
Magnetic induction (dB) is directly proportional to i) current (i) ii) Length of the element (di) iii) sine angle between r and dl and inversely proportional to the square of the distance from small element to point P.

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.
\(\oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d} l}=\oint \mathrm{B} \mathrm{d} l \cos \theta\)
= \(\mathrm{B} \oint \mathrm{d} l\) = B × 2π
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) × 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.

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
= \(\mathrm{B} \oint \mathrm{d} l\)
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = B (2πr) ………………. (1)
According to Ampere’s laws
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i ……………. (2)
From equations (1) and (2), B (2πr) = μ₀i
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

Question 4.
Derive an expression for the magnetic induction at the centre of a current carrying circular coil using Biot-Savart’s law.
Answer:
Consider a circular coil of radius r and carry a current! Consider a small element ‘dl’. Let O is the centre of the coil. By using Biot – Savart’s law,

dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
Here angle \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\) is 90° (i.e., θ = 90°)
dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{idl}}{\mathrm{r}^2}\) …………… (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)

Question 5.
Derive an expression for the magnetic induction of a point on the axis of a current carrying circular coil using Biot-Savart’s law.
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 = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\) ……………….. (1)
(∵ θ = 90° Angle between \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\))
dB can be resolved into two components dB cosθ and dB sinθ. If we consider another 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

Question 6.
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 \((\vec{v} \times \vec{B})\)
Consider electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.

E = E\(\hat{\mathrm{j}}\),B = B\(\hat{\mathrm{k}}\), v = υ\(\hat{\mathrm{i}}\)
F_E = qE\(\hat{\mathrm{j}}\), F_B =q(v\(\hat{\mathrm{i}}\) × B\(\hat{\mathrm{k}}\)) = – qvB\(\hat{\mathrm{j}}\)
∴ F = F_E + F_B
F = q (E – υB)\(\hat{\mathrm{j}}\)
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 selefct charged particles of a particular velocity. The crossed field E and B serve as a velocity selector.

Question 7.
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. Vaccum 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.

Question 8.
Derive an expression for the magnetic dipole moment of a revolving electron. [A.P. Mar. 16]
Solution:
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) = \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
The electric current (i) = \(\frac{\text { Charge }}{\text { Time }}\) = charge × frequency
i = e × \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
∴ Magentic dipole moment (M) = iA (∵ N = 1)
M = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\) × πr² (∵ A = πr²)
M = \(\frac{\mathrm{evr}}{2}\)

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 T, area of cross section ’A1, carrying a current T, 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 = \(\frac{N}{V}\))
∴ Current i = nq V_d A
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_d B sin θ) (nlA)
(nqV_dA) (lB 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 ‘ix’ 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 Bx at a distance ‘r’ from the conductor ‘AB’ can be written as B₁ = \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\)
If ‘F’ is forœ acting on ‘CD’ clue to magnetic induction ‘B₁‘ then
F_CD = i₂lB₁
Where l = length of the conductor
F_CD = i₂l \(\left(\frac{\mu_0 i_1}{2 \pi \mathrm{r}}\right)=\frac{\mu_0 i_1 i_2 l}{2 \pi r}\) ……………. (1)
The direction of the force can be determined by using Flemings left hand rule.
Similarly we can find the force acting on the A conductor AB due to magnetic induction B₂.
F_AB = i₁lB₂
∴ F_AB = i₁l \(\left(\frac{\mu_0 i_2}{2 \pi r}\right)\) ………….. (2) [∵B₂ = \(\left(\frac{\mu_0 i_2}{2 \pi r}\right)\)]
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 = \(\mathrm{i} \overline{\mathrm{b}} \times \overline{\mathrm{B}}\) acting upwards along the axis of suspension
Force on arm BC = \(\mathrm{i} \overline{\mathrm{b}} \times \overline{\mathrm{B}}\) 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) = ilB (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
τ = n i AB sin θ
If ‘Φ’ is the deflection of the coil, that is the angle between the plane of the coil and magnetic field B
τ = 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 within 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 (τ) = 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 9 is the deflection made by the coil.
When the coil is in equilibrium position
Deflecting torque = Restoring torque
B i A N = Cθ
i = \(\left(\frac{\mathrm{C}}{\mathrm{BAN}}\right) \theta\)
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 to an ammeter ? Why is the parallel resistance smaller that the galvanometer resistance ? 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 ?
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 Kirchhoffs I^st law, i = i_g + i_s
As ‘G’ and ‘S’ are parallel P.D. across
Galvanometer = P.D. across shunt
i_gG = i_sS
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 \(\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}\) = n ⇒ i_g = \(\frac{\mathrm{i}}{\mathrm{n}}\)
∴ 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.
Solution for the problem : Current in the circuit i = 1A;
Current through the galvanometer, i_g = 10^-6A
Shunt resistance, S = \(\frac{G}{n-1}=\frac{G}{\frac{i}{i_g}-1}=\frac{G}{10^6-1}=\frac{G}{99.999} \Omega\)

Question 4.
How can a galvanometer be converted to a voltmeter ? Why is the series resistance greater that the galvanometer resistance ? A moving coil galvanometer of resistance 5Ω can measure a current of 15mA. What is the series resistance required if it is to measure 1.5V ?
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 P.D. 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]
i_g = Current passing through the galvanometer
\(\frac{\mathrm{V}}{\mathrm{i}_{\mathrm{g}}}\) = R + G
R = \(\frac{\mathrm{V}}{\mathrm{i}_{\mathrm{g}}}\) – 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 = \(\left(\frac{V}{V_g}-1\right)\)
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.
Solution for the problem:

Problems

Question 1.
Two long and parallel straight wires A and B canying 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^-4 N
The force on A of length 10 cm is F¹ = F × 0.1 (∵ 1 m = 100 cm)
F¹ = 2 × 10^-4 × 0.1
F¹ = 2 × 10^-5 N.

Question 2.
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 ? [A.P. & T.S. Mar. 15]
Answer:
i₁ = i₂ = 10A
r = 1m
\(\frac{\mathrm{F}}{l}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 10 \times 10}{2 \pi \times 1}\)
\(\frac{\mathrm{F}}{l}\) = 2 × 10^-5 Nm^-1.

Question 3.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig.). What is the magnitude of the magnetic field ?

Solution:
From Eq F = Il × B we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity.
m g = I lB mg
B = \(\frac{\mathrm{mg}}{\mathrm{I} l}\)
= \(\frac{0.2 \times 9.8}{2 \times 1.5}\) = 0.65 T
Note that it would have been sufficient to specify mll, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10^-5 T and we have ignored it.

Question 4.
The horizontal component of the earth’s magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west;
(b) south to north ?
Solution:
F = Il × B
F = IlB sin θ
The force per unit length is
f = F/l = IB sinθ
a) When the current is flowing from east to west,
θ = 90°
Hence,
f = IB
= 1 × 3 × 10^-5 = 3 × 10^-5 Nm^-1
This is larger than the value 2 × 10^-7 Nm^-1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

b) When the current is flowing from south to north, .
θ = 0°
f = 0
Hence there is no force on the conductor.

Textual Examples

Question 1.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal niagnetic field B (Fig.). What is the magnitude of the magnetic field ?

Solution:
From Eq F = Il × B we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity, m g = I lB .
B = \(\frac{\mathrm{mg}}{\mathrm{I} l}\)
= \(\frac{0.2 \times 9.8}{2 \times 1.5}\) = 0.65 T
Note that it would have been sufficient to specify m/l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10^-5 T and we have ignored it.

Question 2.
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).

Solution:
The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along -z axis, (b) for a positive charge (proton) the force is along +z axis.

Question 3.
What is the radius of the path of ah electron (mass 9 × 10^-31 kg and charge 1.6 × 10^-19 C) moving at a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10^-4 T perpendicular to it ? What is its frequency ? Calculate its energy in keV. (1 eV = 1.6 × 10^-19 J).
Solution:
Using Eq. r = mυ/qb we find
r = mυ/(qB) = 9 × 10^-31 kg × 3 × 10⁷ m s^-1 / (1.6 × 10^-19 C × 6 × 10^-4 T)
= 26 × 10^-2 m = 26 cm
v = υ / (2 πr) = 2 × 10⁶ s^-1 = 2 × 10⁶ Hz = 2MHz.
E = (1/2)mυ² = (1/2) 9 × 10^-31 kg × 9 × 10¹⁴ m2/s = 40.5 × 10^-17 J
≈ 4 × 10^-16 J = 2.5 keV.

Question 4.
A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons ? If the radius of its does is 60 cm, what is the kinetic energy (in MeV) of the proton beam produced by the accelerator.
(e = 1.60 × 10^-19 C, m_p = 1.67 × 10^-27 kg, 1 MeV = 1.6 × 10^-13 J).
Solution:
The oscillator frequency should be same as proton’s cyclotron frequency.
Using Eqs. r = mυ/qb and ω = 2πυ = \(\frac{\mathrm{qB}}{\mathrm{M}}\) we have
B = 2π m υ/q = 6.3 × 1.67 × 10^-27 × 10⁷ / (1.6 × 10^-19) = 0.66 T
Final velocity of protons is
υ = r × 2π v = 0.6 m × 6.3 × 10⁷ = 3.78 × 10⁷ m/s.
E = 1/2 mv² = 1.67 × 10^-27 × 14.3 × 10¹⁴/ (2 × 1.6 × 10^-13) = 7 MeV

Question 5.
element ∆1 = ∆x\(\hat{\mathrm{i}}\) is placed at the origin and carries a large current I = 10 A (Fig.). Wat is the magnetic field on the y-axis at a distance of 0.5 m. ∆x = 1 cm.

Solution:
|dB| = \(\frac{\mu_0}{4 \pi} \frac{I \mathrm{dl} \sin \theta}{\mathrm{r}^2}\)
dl = ∆x = 10^-2 m, I = 10 A, r = 0.5 m = y, μ₀/4π = 10^-7 \(\frac{\mathrm{Tm}}{\mathrm{A}}\)
θ = 90°; sin θ = 1
|dB| = \(\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}\) = 4 × 10^-8 T
The direction of the field is in the +z-direction. This is so since,
dl × r = \(\Delta \mathrm{x} \hat{\mathrm{i}} \times \mathrm{y} \hat{\mathrm{j}}=\mathrm{y} \Delta \mathrm{x}(\hat{\mathrm{i}} \times \hat{\mathrm{j}}) \mathrm{y} \Delta \mathrm{x} \hat{\mathrm{k}}\)
We remind you of the following cyclic property of cross-products,
\(\hat{\mathrm{i}} \times \hat{\mathrm{j}}=\mathrm{k} ; \hat{\mathrm{j}} \times \hat{\mathrm{k}}=\hat{\mathrm{i}} ; \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}}\)
Note that the field is small in magnitude.

Question 6.
A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments ? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble ? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. (b) ?

Solution:
a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|.

b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 × 10^-4 T normal to the plane of the paper going into it.

c) Same magnitude of B but opposite in direction to that in (b).

Question 7.
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?
Solution:
Since the coil is tightly wound we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is (From Eq.),
B₀ = \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \hat{\mathrm{i}}\)
B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 10^2 \times 1}{2 \times 10^{-1}}\) = 2π × 10^-4 = 6.25 × 10^-4 T

Question 8.
Magnetic field due to a long current-carrying wire Oersted’s experiments showed that there is a magnetic field around a current-carrying wire. We determine the magnitude of magnetic field at some distance from a long striaght wire carrying a current I.
Solution:
The direction of the field is given by the right hand rule. The figure shows an element dl of the current-carrying wire. The point P, where the field is to be determined is at a perpendicular distance ‘S’ from the wire. The position vector of P from dl is r.

The magnitude dB of the magnetic field due to dl is given by Biot-Savart law to be

now from the figure S = r cosθ which gives l/r² = cos²θ/s²
and l’ = S tanθ which gives dl’ = S sec²θ dθ = S dθ/cos²θ
thus dB = \(\frac{\mu_0 I \cos \theta}{4 \pi S}\) dθ
we integrate this to get B at P If the wire is very long then the limits for 0 would be -π/2 to π/2
thus B = \(\frac{\mu_0 l}{2 \pi S}\) (emerging from the paper at P)

Question 9.
Find \(\oint\) B . dl for the paths shown in (a) and (b)

Solution:
a) Going around theKpath in the anticlockwise direction, I₁ is taken as positive while I₃ is negative. Currents I₂ and I₄ do not matter as they are NOT enclosed by the path.
\(\oint\) B . dl = μ₀(I₁ – I₃)
Note : Currents I₂ and I₄ create magnetic fields all around them and B due to them on any element of the path would be non-zero. However, the sum B.dZ due to them would be zero, b) Calculation of B. dl for the entire path can be broken up into two separate calculations, one covering all contributions from an anti-clockwise traversal around I₁ and the other covering all contributions from a clockwise traversal around I₃. Thus
\(\int_1 \mathrm{~B} . \mathrm{d} l\) = μ₀I₁ for all elements around I₁ traversed in an anti-clockwise direction
\(\int_2 \mathrm{~B} . \mathrm{d} l\) = μ₀I₃ for all elements around I₃ traversed in a clockwise direction; I₃ taken as positive because it is flowing into the plane. Thus the total \(\oint\) B.dl = μ₀ (I₁ – I₃)

Question 10.
Figure shows the circular cross-section of a long straight wire of radius a carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a (dashed inner circle) and r > a (dashed outer circle).

Solution:
a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop,
L = 2πr
I_e = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire B(2πr) = μ₀I
B = \(\frac{\mu_0 I}{2 \pi \mathrm{r}}\) ……………… (1)
b) Consider the case r < a. The Amperian loop is a circle labelled 1 . For this loop, taking the radius of the circle to be r,
L = 2 π r
Now the current enclosed I_e is not I (because r < a), but is less than this value. Since the current distribution is uniform, the current enclosed is,

Figure shows a plot of the magnitude of B with distance r from the centre (axis) of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section.
This example possesses the required symmetry so that Ampere’s law can be applied readily.

Question 11.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ?
Solution:
The number of turns per unit length is,
n = \(\frac{500}{0.5}\) = 1000 turns / m
The length l = 0.5m and radius r = 0.01 m. Thus, l/a = 50 i.e., l > >a.
Hence, we can use the long solenoid formula, namely, Eq. (B = μ₀nI)
B = μ₀ n I
= 4π × 10^-7 × 10³ × 5 = 6.28 × 10^-3 T

Question 12.
The horizontal componet of the earth’s magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west;
(b) south to north ?
Solution:
F = Il × B
F = IlB sin θ
The force per unit length is
f = F/l = IB sinθ
a) When the current is flowing from east to west,
θ = 90°
Hence,
f = IB
= 1 × 3 × 10^-5 = 3 × 10^-5 Nm^-1
This is larger than the value 2 × 10^-7 Nm^-1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

b) When the current is flowing from south to north,
θ = 0°
f = 0
Hence there is no force on the conductor.

Question 13.
A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2A. (a) What is the field at the centre of the coil ? (b) What is the magnetic moment of this coil ?
The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field, (c) What are the magnitudes of the torques on the coil in the initial and final position ? (d) What is the angular speed acquired by the coil when it has rotated by 90° ? The moment of inertia of the coil is 0.1 kg m². [A.P. Mar. 19]
Solution:
a) From B = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}\)
Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence,
B = \(\frac{4 \pi \times 10^{-7} \times 10^2 \times 3.2}{2 \times 10^{-1}}=\frac{4 \times 10^{-5} \times 10}{2 \times 10^{-1}}\) (using π × 3.2 = 10)
= 2 × 10^-3 T
The direction is given by the right-hand thumb rule.

b) The magnetic moment is given by
m = N I A = N I π r² = 100 × 3.2 × 3.14 × 10^-2 = 10 A m²
The direction is once again given by the right hand thumb rule.

c) τ = |m × B|
= mBsin θ
Initially, θ = 0. Thus initial torque τ_i = 0. Finally, θ = π/2 (or 90°).
Thus, final torque τ_f = m B = 10 × 2 = 20 N m.

d) From Newton’s second law.
I\(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) = mBsin θ
where I is the moment of inertia of the coil. From chain rule,
\(\frac{\mathrm{d} \omega}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \omega\)
Using this,
Iω dω = m B sinθ dθ
Integrating from θ = 0 to θ = π/2,
\(\mathrm{I} \int_0^{\omega \mathrm{I}} \omega \mathrm{d} \omega=\mathrm{mB} \int_0^{\pi / 2} \sin \theta d \theta\)
\(\mathrm{I} \frac{\omega_{\mathrm{I}}^2}{2}=-\left.\mathrm{mB}[\cos \theta]\right|_0 ^{\pi / 2}=\mathrm{mB}\)
ω_f = \(\left(\frac{2 \mathrm{mB}}{\mathrm{I}}\right)^{1 / 2}=\left(\frac{2 \times 20}{10^{-1}}\right)^{1 / 2}\) = 20 s^-1

Question 14.
a) 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 around it self (i.e. turns about the vertical axis).
Solution:
No, because that would require τ to the in the vertical direction. But τ = I A × B, and since A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B.

b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its. orientation of stable equilibrium ? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum.
Solution:
Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.

c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape ? What could be the sense of current in the loop and the direction of magnetic field ?
Solution:
It assumes circular shape with its plane normal to the field to maximize flux, since for a given perimeter, a circle encloses greater area than any other shape.

Question 15.
In the circuit (Fig.) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance R_G = 60.00 Ω; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance r_s = 0.02 Ω; (c) is an ideal ammeter with zero resistance ?

Solution:
a) Total resistance in the circuit is,
R_G + 3 = 63 Ω. Hence I = 3/63 = 0.048 A.

b) Resistance of the galvanometer converted to an ammeter is,
\(\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}_{\mathrm{s}}}=\frac{60 \Omega \times 0.02 \Omega}{(60+0.02) \Omega}\) = 0.02 Ω
Total resistance in the circuit is,
0.02 Ω + 3 Ω = 3.02 Ω. Hence I = 3/3.02 = 0.99 A.

c) For the ideal ammeter with zero resistance,
1 = 3/3 = 1.00 A

Students get through AP Inter 2nd Year Physics Important Questions 6th Lesson Current Electricity which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 6th Lesson Current Electricity

Short Answer Questions

Question 1.
Derive an expression for the effective resistance when three resistors are connected in
(i) series
(ii) parallel.
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 Ohm’s 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 = \(\frac{V}{R_P}\)]
   ∴ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}\)

Question 2.
On what factors does the resistance of a conductor depend ? Define electric resistance and write its S.I unit. How does the resistance of a conductor vary if (a) Conductor is stretched to 4 times of its length (b) Temperature of conductor is increased ? [Board Model Paper]
Answer:
Dependence of Resistance on geometry :

1. Resistance is directly proportional to length of the conductor (R ∝ l)
2. Resistance is inversely proportional to area of cross-section (R ∝ 1/A)
   i.e., R ∝ l/A ⇒ R = ρl / A; ρ is called specific resistance or resistivity of the material.

Dependence of Resistance on temperature : R₁ = R₁[1 + α(∆T)] & α = \(\frac{\Delta R / R}{\Delta T}\)
α is called temperature coefficient of resistance and its SI unit: K^-1.

Definition of electrical resistance (R) : The ratio between the potential difference (V) across the ends of the conductor and the current (i) flowing through the conductor is called its electrical resistance, R = V/i; S.I unit of electrical resistance is ohm (Ω)
a) Resistance of the conductor, R = ρ \(\frac{l}{\mathrm{~A}}\) = ρ \(\frac{l^2}{V}\) ⇒ R ∝ l² (∵ ρ and V constant)
⇒ R₁/R₂ = (l₁/l₁)² ∴ R₁/R₂ = (l/4l)
⇒ R₂ = 16R
Hence resistance of the conductor increases by 16 times.
b) The value of the electrical resistance of a conductor increases with increase of temperature. Because, temperature coefficient of resistance is positive for metals.

Long Answer Questions

Question 1.
State Kirchhoffs law for an electrical network. Using th&se laws deduce the condition for balance in a Wheatstone bridge. [T.S. Mar. 16; 15; Mar. 14]
Answer:

1. Kirchhoffs 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 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 Kirchhoff s second law to the closed path ADBA,
-I₁R₁ – I_gG + I₂R₂ = 0
or
⇒ I₁R₁ + I_gG = I₂R₂ …………….. (3)
applying kirchhoft’s 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) and (8)
\(\frac{I_1 R_1}{I_3 R_3}=\frac{I_2 R_2}{I_4 R_4} \Rightarrow \frac{R_1}{R_3}=\frac{R_2}{R_4}\) [∵ I₁ = I₃ & I₂ = I₄]
∴ Wheatstone’s Bridge principle : R₄ = R₃ × \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 2.
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. [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. \(\frac{(1)}{(2)} \Rightarrow \frac{\varepsilon_1}{\varepsilon_2}=\frac{l_1}{l_2}\)

Question 3.
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. 17, 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. ε ∝ l ⇒ ε = 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 (I) through the resitance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
   Then V = Φ l₂ ………………… (2)
6. \(\frac{(1)}{(2)} \Rightarrow \frac{\varepsilon}{V}=\frac{l_1}{l_2}\) ……………. (3)
7. But ε = I (r + R) and V = IR. This gives
   \(\frac{\varepsilon}{V}=\frac{(r+R)}{R}\)
   \(\frac{l_1}{l_2}=\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = \(\mathrm{R}\left(\frac{l_1}{l_2}-1\right)\)

Question 4.
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 ∝ \(\frac{1}{R}\)

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

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 = 10 Ω,
l₁ = l
l₂ = 3l, R₂ ?

R₁ = \(\frac{\rho}{\mathrm{V}} l_1^2\)
R₂ = \(\frac{\rho}{\mathrm{V}} l_2^2\)
\(\frac{\dot{\mathrm{R}}_2}{\mathrm{R}_1}=\left(\frac{l_2}{l_1}\right)^2 \Rightarrow \frac{\mathrm{R}_2}{10}=\left(\frac{3 l}{l}\right)^2\)
∴ 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 ? [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 = \(\frac{2 R \times 2 R}{2 R+2 R}\)
∴ R_P = R.

Question 3.
Three resistors 3Ω, 6Ω and 9Ω are connected to a battery. In which of them will the power of 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 = \(\frac{18}{11}\) Ω
∴ Dissipated power in parallel,
P_P ∝ \(\frac{1}{R_P}\) ⇒ 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 = 18 Ω
∴ Dissipated power in series, P_S ∝ R_S ⇒ P_S ∝ 18
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 4.
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 = 2.7 Ω, t₂ = 100°C, α = ?
α = \(\frac{R_2-R_1}{R_1 \mathrm{t}_2-\mathrm{R}_2 \mathrm{t}_1}=\frac{2.7-2.1}{2.1 \times 100-2.7 \times 27.5}=\frac{0.6}{210-74.25}=\frac{0.6}{135.75}\)
∴Temperature coefficient of resistivity
α = 0.443 × 10^-2/°C

Question 5.
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 ? [A.P. Mar. 15
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 = \(\frac{\varepsilon_1 \times l_2}{l_1}=\frac{1.25 \times 63}{35}\) = 2.25 V

Question 6.
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. [T.S. Mar. 17]
Solution:
Circuit diagram for the given data is shown below.

Given, E = 2.5 V; R = 45Ω;
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 7.
The balancing point in meter bridge experiment is obtained at 30 cm from the left. If the right gap contains 3.5 £2, what is the resistance in the left gap? [BMP]
Solution:
\(\frac{x}{R}=\frac{I_1}{100-I_1} \Rightarrow \frac{x}{3.5}=\frac{30}{70} \Rightarrow x=\frac{30 \times 3.5}{70}=1.5\)
∴ x = 1.5 Ω
The resistance in the left gap in meter bridge is, x = 1.5 Ω

Question 8.
The storage battery of a car has an emf of 12 V If the internal resistance of the battery is 0.4Q, what is the maximum current that can be drawn from the battery ?
Solution:
Here E = 12 V, r = 0.4Ω
Maximum Current, I_max = \(\frac{E}{r}=\frac{12}{0.4}\) = 30 A

Question 9.
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 ? [T.S. Mar. 15]
Solution:
Here E = 10 V, r = 3Ω,1 = 0.5 A, R = ?, V = ?
I = \(\frac{E}{(R+r)}\) or (R + r) = \(\frac{\mathrm{E}}{\mathrm{I}}=\frac{10}{0.5}\)= 20 or
R = 20 – 3 = 17Ω
Terminal voltage V = IR = 0.5 × 17 = 8.5 Ω.

Question 10.
If the balancing point in a meter bridge from the left is 60 cm, compare the resistance in the left and right gaps of the meter bridge.
Solution:
\(\frac{R_1}{R_2}=\frac{I_1}{100-I_1}=\frac{60}{100-60}=\frac{60}{40}=\frac{3}{2} \Rightarrow \frac{R_1}{R_2}=\frac{3}{2}\)

Question 11.
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell that balances against a length of 180cm of the potentiometer wire. [A.P. Mar. 17, 16]
Solution:
Length of potentiometer wire L = 5m
Potential difference V = 6 Volt
Potential gradient Φ = \(\frac{V}{L}=\frac{6}{5}\) = 1.2V/m
Balancing length l = 180cm
= 1.80m
Emf of the cell E = Φl
= 1.2 × 1.8 = 2.16V.

Students get through AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance

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. The electric potential at a point is defined as the amount of work done 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 work done 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 = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
4. This amount of 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 it’s axis makes an angle θ with E.
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   
   Torque τ = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}} \) ⇒ BC = 2a sinθ
   ∴ Torque τ = 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 = τdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂
   workdone W = \(\int_{\theta_1}^{\theta_2}\) PE sinθdθ = 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 cos₁.
   In vector form U = –\(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. [A.P. Mar. 16; T.S. 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 point x , E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   \(\int {\mathrm{dV}}=\int_{\mathrm{d}}^0-\mathrm{Edx}=\int_{\mathrm{d}}^0 \frac{-\sigma}{\varepsilon_0} \mathrm{dx}=\frac{\sigma \mathrm{d}}{\varepsilon_0}\)
6. Capacitance of the capacitor C = \(\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{A} \sigma}{\frac{\sigma \mathrm{d}}{\varepsilon_0}}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\) Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field.
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 is the external field strength and E_i is the electric field strength induced, then the net field, \(\overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_0+\overrightarrow{\mathrm{E}}_{\mathrm{i}}\)
   (E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Question 6.
Define electric potential. Derive an 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,
   V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{BP}}\right]\)
   ∴ V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}\right]\) [∵ BP = NP]
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 = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{MP}}\right]\) [∵ AP = MP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}-\frac{\mathrm{q}}{\mathrm{MP}}\right]\) ………… (1)
9. In △le ONB, ON = OB cos0 = a cosθ; ∴NP = OP – ON = r – a cosθ ………………… (2)
10. In △le AMO, OM = AO cos0 = a cosθ; ∴ MP = MO + OP = r + a cosθ ………………. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
13. (a) Electric potential on the axial line of dipole:
    
    • 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]
    • 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]

Qeustion 7.
What is series combination of capacitors. Derive the formula for equivalent capacitance in series combination. [A.P.& 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. P.D’s across the capacitors is not equal.

Consider three capacitors of capacitances 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)
RD across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
RD across III^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across IIII^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1),V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}=\mathrm{Q}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)
\(\frac{\mathrm{V}}{\mathrm{Q}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\) [∵ \(\frac{1}{C}=\frac{V}{Q}\)
For ‘n’ number of capacitors, the effective capacitance
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}+\ldots+\frac{1}{\mathrm{C}_{\mathrm{n}}}\)

Question 8.
What is parallel combination of capacitors. Derive the formula for equivalent capacitance in parallel combination. [T.S. Mar. 17; A.P.& T.S. Mar. 15]
Answer:
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 P.D’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal.
Consider three capacitors of capacitance C₁, C₂ and C₃ are connected in parallel across a P.D 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 Q₃ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{\mathrm{Q}}{\mathrm{V}}\) = 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 9.
Derive an expression for the energy stored in a capacitor.
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 = \(\frac{Q^2}{2 C}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}=\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Qeustion 10.
What is the energy stored when the space between the plates is filled with dielectric.
a) With charging battery disconnected ?
b) With charging battery connected in the circuit ?
Answer:
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 disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge remains constant.
Capacity increases by’K’ times.
New capacity C’ = \(\frac{Q}{V}\) = \(\frac{\frac{Q}{V}}{K}\) = K\(\frac{Q}{V}\) = KC [V’ = \(\frac{V}{K}\); C = \(\frac{Q}{V}\)]
∴ C’ = KC
Energy stored U’ = \(\frac{1}{2}\) QV = \(\frac{1}{2}\) Q \(\frac{V}{K}\) = \(\frac{U}{K}\)
U’ = \(\frac{U}{K}\)
∴ 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’ = \(\frac{1}{2}\) QV; = \(\frac{1}{2}\) (KQ) V = KU
U = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7 C located 9 cm away,
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Qeustion 2.
Two charges 3 × 10^-8 C and -2 × 10^-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Thke the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Question 3.
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when die slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 4.
Four charges are arranged at the corners of a square ABCD of side d. as shown in fig. Find the work required to put together this arrangement, (b) A charge q₀ is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A and then the charges -q, +q, and -q are brought to B, C and D, respectively. The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
2. Work needed to bring -q to B when + q is at A. This is given by (charge at B) × (electrostatic potential at B due to change +q at A) = -q × \(\left(\frac{q}{4 \pi \varepsilon_0 d}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q₀ to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7μC and -2μC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each ‘ other?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

Question 6.
There is a uniform electric field in the XOY plane represented by (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) 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,
\(\overrightarrow{\mathrm{E}}\) = (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) Vm^-1
Electric potential at the origin = 200V
Position vector \(\mathrm{d} \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}})\) m
We know that,

dV = \(-\vec{E} \cdot d \vec{r}=-(40 \hat{i}+30 \hat{j}) \cdot(2 \hat{i}+\hat{j})\)
V_p – V₀ = -(80 + 30) = -110Volt. ’
V_p = V₀ – 110 = (200 – 110) Volt = 90 Volt
∴ Potential at point P, V_p = 90Volt.

Question 7.
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 ratio 2 : 1.
That means r_max = 2 and r_min = 1

Question 8.
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 ?
Solution:
(For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V₁² = \(\frac{1}{2}\) CV² …………………. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}=\frac{1}{4}\)CV²; Let potential difference across the capacitor = V₂

Then, U₂ = \(\frac{1}{2}\) C₂V₂²
⇒ \(\frac{1}{4}\) CV² = \(\frac{1}{2}\) × 2C × V₂²
⇒ V₂² = \(\frac{\mathrm{V}^2}{4}\)
∴ V₂ = \(\frac{V}{2}\)

Question 9.
Three Capacitors each of capaitance 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) Resultant capacitance in series combination \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \Rightarrow \frac{1}{C_S}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)
C_S = 3pF

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

Question 10.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. [A.P. Mar. 17]
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 = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₂ = C₃V = 4 × 100 = 400pC

Textual Examples

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7C located 9 cm away.
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Question 2.
Two charges 3 × 10^-8 C and -2 × 10^-8C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-\mathrm{x}) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm.
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x. = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the claculation required choosing potential to be zero at infinity.

Question 3.
Figure (a) and (b) shows the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference V_p – V_Q; V_B – V_A.
(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.
(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.
(d) Give the sign of the work done by the external agency in moving a small negative charge from B And A.
(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A ?
Solution:
(a) As V ∝ \(\frac{1}{\mathrm{r}}\), V_p > V_Q. Thus, (V_p – V_Q) is positive. Also VB is less negative than V_A. Thus, V_B > V_A or (V_B – V_A) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly. (PE.)_A > (P.E.)_B and hence sign of potential energy differences is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the eletric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going form B to A.

Question 4.
Four charges are arranged at the comets of a square ABCD of side d. as shown in fig. 5.15.(a) Find the work required to put together this arrangement, (b) A charge q0 is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges -q, +q, and -q are brought to B, C and D, respectively The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elesewhere: this is zero.
2. Work needed to bring -q to B when +q is at A. This is given by (charge at B) x (electrostatic .potential at B due to charge +q at A) = -q × \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{~d}}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7µC and -2µC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each other ?
c) Suppose that the same system of charges is now placed in an external electric field E = A(1/r²); A = 9 × 10⁵ C m^-2. What would the electrostatic energy of the configuration be ?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
q₁V(r₁) + q₂V(r₂) = A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
q₁V(r₁) + q₂V(r₂) + \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r_{12}}\)
= A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
= 70 – 20 – 0.7 = 49.3J

Question 6.
A molecule of a substance has a perma-nent electric dipole moment of magnitude 10^-29C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10⁶ V m^-1. The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.
Solution:
Here, dipole moment of each molecules = 10^-29C m
As 1 mole of the substance contains 6 × 10²³ molecules,
total dipole moment of all the molecules, p = 6 × 10²³ × 10^-29 Cm
= 6 × 10^-6 C m
Initial potential energy, U_t = -pE cos θ
=-6 × 10^-6 × 10⁶ cos 0° = -6 J
Final potential energy (when θ = 60°),
U_f = -6 × 10^-6 × 10⁶ × cos 60° = – 3J
Change in potential energy
= -3 J – (-6j) = 3 J
So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.

Question 7.
(a) A comb run through one’s dry hair attracts small bits of paper. Why ? What happens if the hair is wet or if it is a rainy day ? (Remember, a paper does not conduct electricity.)
(b) Ordinary rubber is an insulator. But special rubber types of aircraft are made slightly conducting. Why is this necessary ?
(c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?
(d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why ?
Solution:
(a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction, if the hair is wet, or if it is a rainy day. friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper.

(b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire.

(c) Reason similar to (b).

(d) Current passes only when there is difference in potential.

Question 8.
A slab of material of dielectric constant K has the same area as the plates of a parallelplate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 9.
A network of four 10μF capacitors is connected to a 500V supply, as shown in Fig. Determine
(a) the equivalent capacitance of the network and
(b) the charge on each capacitor, (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Solution:
(a) In the given network, C₁, C₂ and C₃ are connected in series. The effective capacitance C’ of these three capacitors isgivgftby
\(\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
For C₁ = C₂ = C₃ = 10 μF. C = (10/3) μF. The network has C and C₄ connected in parallel. Thus, the equivalent capacitance C of the network is
C = C’ + C₄ = (\(\frac{10}{3}\) + 10) μF = 13.3 μF

(b) Clearly, from the figure, the charge on each of the capacitors, C₁, C₂ and C₃ is the same, say Q. Let the charge on C₄ be Q. Now, since the potential difference c across AB is Q/C₁ across BC is Q/C₂. across CD is Q/C₃, we have
\(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}\) = 500 V
Also Q’/C₄ = 500V.
This gives for the given value of the capacitances,
Q = 500 V × \(\frac{10}{3}\) μF = 1.7 × 10^-3C and
Q’ = 500 V × 10μF = 5.0 × 10^-3C

Question 10.
(a) A 900pF capacitor is charged by 100 V battery [Fig.a]. How much electrostatic energy is stored by the capacitor ?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic y energy stored by the system ?

Solution:
The charge on the capacitor is
Q = CV = 900 × 10^-12F × 100 V
= 9 × 10^-8C
The energy stored by the capacitor is
= (1/2) CV² = (1/2) QV
= (1/2) × 9 × 10^-8C × 100 V
= 4.5 × 10^-6 J

(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V. The charge on each capacitor is then Q’ = CV’. By charge conservation, Q’ = Q/2. This implies V’ = V/2. The total energy of the system is
= 2 × \(\frac{1}{2}\)Q’V’ = \(\frac{1}{4}\)QV = 2.25 × 10⁶J
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone ?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

Students get through AP Inter 2nd Year Physics Important Questions 4th Lesson Electric Charges and Fields which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 4th Lesson Electric Charges and Fields

Very Short Answer Questions

Question 1.
What is meant by the statement ‘charge is quantized’? [IPE 2015 (TS)]
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 like a charged body. Hence repulsion is the sure test of electrification.

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

Question 4.
What happens to the weight of a body when it is charged positively ?
Answer:
When a body is 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 [Board Model Paper]
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.
[∵ F₂ = \(\frac{F_1 d_1^2}{\left(\frac{d_1}{2}\right)^2}\) = 4 F₁]

b) When distance is doubled to half, force increases by four times.
[∵ F₂ = \(\frac{F_1 d^2}{\left|2 d_1\right|^2}\) = \(\frac{1}{4}\) F₁]

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. [IPE 2015 (TS)]
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 = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\). 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 = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) perpendicular to 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 = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Question 14.
A proton and an α-particle are released in a uniform electric field. Find the ratio of (a) forces experienced by them (b) accelerations gained by each.
Answer:
a) As F = Eq, F ∝ q, ⇒ \(\frac{F_p}{F_\alpha}=\frac{Q_p}{Q_\alpha}=\frac{1}{2}\)
∴ \(\frac{F_p}{F_\alpha}=\frac{1}{2}\)

b) As a = \(\frac{E Q}{m}\) ⇒ a ∝ \(\frac{Q}{m}\) ⇒ \(\frac{a_p}{a_\alpha}=\frac{e_p}{Q_\alpha} \times \frac{m_\alpha}{m_p}=\frac{1}{2} \times \frac{4}{1}=\frac{2}{1}\)
∴ \(\frac{a_p}{a_\alpha}=\frac{2}{1}\)

Question 15.
The electric field in a region is given by \(\bar{E}=a \bar{i}+b \bar{j}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
Answer:
Electric field, \(\bar{E}=a \bar{i}+b \bar{j}\)
Flux passing through square area, Φ = \(\overline{\mathrm{E}} \cdot \overline{\mathrm{A}}=(\mathrm{a} \overline{\mathrm{i}}+\mathrm{b} \overline{\mathrm{i}}) \cdot\left(\mathrm{L}^2 \overline{\mathrm{i}}\right)\) = aL² Wb

Question 16.
A hollow sphere of radius ‘r’ has a unifrom charge density ‘σ’. It is kept in a cube of edge 3r such that the center of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
Answer:
Charge on the hallow sphere, q = σ × 4πr².
The flux through a single face of the cube, Φ¹ = \(\frac{f}{6}=\frac{1}{6} \cdot \frac{Q}{\varepsilon_0}=\frac{\sigma \times 4 \pi \mathrm{r}^2}{6 \varepsilon_0}=\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}\)

Question 17.
Consider a uniform electric field . What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the YZ plane ?
Answer:
Given, (field is along positive x-axis)
Surface area of square, S = (10 × 10^-2)(10 × 10^-2) = 10^-2m².
When plane of the square is parallel to yz-plane its area vector points towards 4-ve x-axis.
So θ = 0°.
∴ Flux through square, Φ = EScosθ = 3 × 10³ × 10^-2 × cos0° => <(> = 30 NC^-1m².

Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity. [T.S. Mar. 17; Mar. 14]
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 ∝ \(\frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\)
∴ F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\) 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.
In a medium, F_m = \(\frac{1}{4 \pi \varepsilon} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{1}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}}} \times \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\) [∵ ε = ε₀ε_r]
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. [A.P. Mar. 16]
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 P.
3. Force acting on q₀ due to q is F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qq}_0}{\mathrm{r}^2}\)
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 = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}\)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 θ 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, τ = ⊥^r distance × magnitude of one of force
   ∴ τ = (2a sin θ)qE = 2aqE sin θ = PE sin θ
7. In vector form, \(\vec{\tau}=\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole. [A.P. Mar. 17; T.S. Mar. 16]
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.
   

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. 15]
Answer:
Electric field intensity on equatorial 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 equatorial 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 I lel 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 = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{2 \mathrm{aq}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{3 / 2}}\)
7. If r >> a, then a² can be neglected in comparison to r². Then E
   E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
   In vector form E = \(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and explain its importance. [T.S. Mar. 15]
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, Φ = \(\oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{\mathrm{q}}{\varepsilon_0}\)
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.
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., Φ = \(\oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{\mathrm{q}}{\varepsilon_0}\)

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 ABÇD 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 \(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}=\int\) Eds = E(S + S) = 2ES ……………….. (1)
6. Consider cylindrical surface of AB and CD. Let their areas are say dS₃ and dS₄. These surfaces are ⊥^r to electric intensity \(\overrightarrow{\mathrm{E}}\).
7. So angle between \(\overrightarrow{\mathrm{E}}\) and d\(\overrightarrow{S_3}\) or dS₄ is 90°. Total flux through these surfaces is zero.
   Since \(\oint_S\) E.dS = 0.
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 = \(\frac{\sigma S}{\varepsilon_0}\) [∵Q = σ × S]
9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \(\frac{\sigma}{2 \varepsilon_0}\).

Textual Examples

Question 1.
How can you charge a metal sphere positively without touching it ?
Solution:
Figure (a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig. 4.5 (d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

Question 2.
If 10⁹ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? .
Solution:
In one second electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10^-19 × 10⁹C = 1.6 × 10^-10 C. The time required o accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.6 × 10^-10 C/s) = 6.25 × 10⁹ s = 6.25 × 10⁹ ÷ (365 × 24 × 3600) years = 198 years. Thus to collect a charge of one coulomb, from a body from which 10⁹ electrons move, out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.

It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about 2.5 × 10²⁴ electrons.

Question 3.
How much positive and negative charge is there in a cup of.water ?
Solution:
Let us assume that the mass of one cup of water is 250 g. The molecular mass of. water is 18g. Thus, one mole (= 6.02 × 10²³ molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 10²³.

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10^-19 C = 1.34 × 10⁷C.

Question 4.
Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square depen-dence on the distance between the charges/masses, (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a protron and (ii) for two protons (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= 10^-10 m) apart ?
(m_p = 1.67 × 10^-27 kg, m³ = 9.11 × 10^-31 kg).
Solution:
a) i) The electric force between an electron and a proton at a distance r apart is :
F_e = –\(\frac{\mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\)
Where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is :
F_G = -G\(\frac{\mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\)
Where m_p and m_e are the masses of a proton and an electron respectively.
\(\left|\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{G}}}\right|=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}\) = 2.4 × 10³⁹

ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is :
\(\left|\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{G}}}\right|=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{G} \mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}\) = 1.3 × 10³⁶
However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces between two protons inside a nucleus (distance between two protons
is ~ 10^-15m inside a nucleus) are F_e ~ 230 N whereas F_G ~ 1.9 × 10^-34 N.
The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however the masses of an electron and a proton are different. Thus, the magnitude of force is
|F| = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}^2}\) = 8.987 × 10⁹ Nm²/C² × (1.6 × 10^-19C)² / (10^-10 m)²
= 2.3 × 10^-8N
Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is a = 2.3 × 10^-8 N/9.11 × 10^-31 kg = 2.5 × 10²² m/s²
Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.
The value for acceleration of the proton is
a = 2.3 × 10^-8 N/1.67 × 10^-27 kg = 1.4 × 10⁹ m/s².

Question 5.
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. (c). What is the expected repulsion of A on the basis of Coulomb’s law ? Spheres A and C and spheres B and D have identical sizes. Ignore the size of A and B in comparison to the separation between their centres.

Solution:
Let the original charge on sphere A be q and that on B be q’. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}\)
Neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
F’ = \(\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}\) = F
Thus the electrostatic force on A, due to B, remains unaltered.

Question 6.
Consider three charges q₁, q₂, q₃ each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. ?

Solution:
In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC,
AD = AC cos 30° = \(\left(\frac{\sqrt{3}}{2}\right) l\) and the distance AO of the centroid O from A is
(2/3) AD = \(\left(\frac{1}{\sqrt{3}}\right) l\).
By symmetry AO = BO = CO.
Thus,
Force F₁ on Q due to charge q at A = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along AO
Force F₂ on Q due to charge q at B = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along BO
Force F₃ on Q due to charge q at C = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along CO
The resultant of forces F₂ and F₃ is \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along OA, by the parallelogram law. Therefore, the total force on Q = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}(\hat{\mathrm{r}}-\hat{\mathrm{r}})\) = 0, where \(\hat{\mathrm{r}}\) is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60° about O.

Question 7.
Consider the charges q, q and -q placed at the vertices of an equilateral triangle, as shown in Fig. What is the force on each charge ?

Solution:
The forces acting on charge q at A due to charges q at B and -q at C are F₁₂ along BA and F₁₃ along AC respectively, as shown in Fig. By the parallelogram law, the total force F₁ on the charge q at A is given by
F₁ = F \(\hat{\mathrm{r}}_1\) where \(\hat{\mathrm{r}}_1\) is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude
F = \(\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 l^2}\)
The total force F₂ on charge q at B is thus F₂ = F \(\hat{\mathrm{r}}_2\), where \(\hat{\mathrm{r}}_2\) is a unit vector along AC.
Similarly the total force on charge -q at C is F₃ = \(\sqrt{3} \mathrm{~F} \hat{\mathrm{n}}\), where \(\hat{\mathrm{n}}\) is the unit vector along the direction bisecting the ∠BCA.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
F₁ + F₂ + F₃ = 0
The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise.

Question 8.
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N C^-1 (Fig. a). The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance (Fig. b) Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Solution:
In Fig. (a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude f the electric field. The acceleration of the electron is a_e = eE/m_e.
Where m_e is the mass of the electron.
Starting from rest, the time required by the electron to fall through a distance h is given by
t_e = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{a}_{\mathrm{e}}}}=\sqrt{\frac{2 \mathrm{hm_{ \textrm {e } }}}{\mathrm{eE}}}\)
For e = 1.6 × 10^-19C, m_e = 9.11 × 10^-31 kg.
E = 2.0 × 10⁴ NC^-1, h 1.5 × 10^-2 m.
t_e = 2.9 × 10^-9 s
In Fig. (b), the field is downward and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is
a_p = eE/m_p
Where mp is the mass of the proton ; m_p = 1.67 × 10^-27 kg. The time of fall for the proton is
t_p = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{a}_{\mathrm{p}}}}=\sqrt{\frac{2 h \mathrm{~m}_{\mathrm{p}}}{\mathrm{eE}}}\) = 1.3 × 10^-7 s
Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of, free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field
a_p = \(\frac{\mathrm{eE}}{\mathrm{m}_{\mathrm{p}}}=\frac{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(2.0 \times 10^4 \mathrm{NC}^{-1}\right)}{1.67 \times 10^{-27} \mathrm{~kg}}\)
= 1.9 × 10¹² ms^-2
Which is enormous compared to the value of g (9.8 ms^-2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.

Question 9.
Two point charges q₁ and q₂, of magnitude +10^-8 C and -10^-8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig.

Solution:
The electric field vector E_1A at A due to the positive charge qx points towards the right and has a magnitude
E_1A = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.05 \mathrm{~m})^2}\) = 3.6 × 10⁴ NC^-1
The electric field vector E_2A at A due to the negative charge q₂ points towards the right and has the same magnitude. Hence the magnitude of the total electric field E_A at A is
E_A = E_1A + E_2A = 7.2 × 10⁴ NC^-1
E_A is directed toward the right.
The electric field vector E_1B at B due to the positive charge q₁ points towards the left and has a magnitude.
E_1B = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.05 \mathrm{~m})^2}\) = 3.6 × 10⁴ NC^-1
The electric field vector E_2B at B due to the negative charge q₂ points towards the right and has a magnitude.
E_2B = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.15 \mathrm{~m})^2}\) = 4 × 10⁴ NC^-1
The magnitude of the total electric field at B is E_B = E_1B – E_2B = 3.2 × 10⁴ NC^-1
E_B is directed towards the left.
The magnitude of each electric field vector at point C, due to charge q₁ and q₂ is
E_1C = E_2C = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.10 \mathrm{~m})^2}\) = 9 × 10³ NC^-1
The directions in which these two vectors point are indicated in Fig. The resultant of these two vectors is
E_C = E₁ cos \(\frac{\pi}{3}\) + E₂ cos \(\frac{\pi}{3}\) = 9 × 10³ NC^-1
E_C points towards the right.

Question 10.
Two charges 10 μC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. (a) and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. (b).

Solution:
a) Field at P due to charge +10 μC
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15-0.25)^2 \times 10^{-4} \mathrm{~m}^2}\)
= 4.13 × 10⁶ NC^-1 along BP
Field at P due to charge -10 μC
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15+0.25)^2 \times 10^{-4} \mathrm{~m}^2}\)
= 3.86 × 10⁶ NC^-1 along PA
The resultant electric field at P due to the two charges at A and B is 2.7 × 10⁵ NC^-1 along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can’expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ±q,.2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude.
E = \(\frac{2 \mathrm{p}}{4 \pi \varepsilon_0 \mathrm{r}^3}\) (r/a > > 1)
Where p = 2aq is the magnitude of the dipole moment.
The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from -q to q). Here, p = 10^-5 × C ; 5 × 10^-3 m = 5 × 10^-8 C m
Therefore,
E = \(\frac{2 \times 5 \times 10^{-8} \mathrm{Cm}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15)^3 \times 10^{-6} \mathrm{~m}^3}\)
= 2.6 × 10⁵ N C^-1
Along the dipolemoment direction AB, which is close to the result obtained earlier.

b) Field at Q due to charge +10 μC at B
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{\left(15^2+(0.25)^2\right] \times 10^{-4} \mathrm{~m}^2} \mathrm{x}\)
= 3.99 × 10⁶ N C^-1 along BQ
Field at Q due to charge – 10 μC at A
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{\left[15^2+(0.25)^2\right] \times 10^{-4} \mathrm{~m}^2}\)
= 3.99 × 10⁶ × N C^-1 along QA ‘
Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is
= 2 × \(\frac{0.25}{\sqrt{15^2+(0.25)^2}}\) × 3.99 × 10⁶NC^-1 along BA
= 1.33 × 10⁵ N C^-1 along BA.
As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole :
E = \(\frac{\mathrm{p}}{4 \pi \varepsilon_0 \mathrm{r}^3}\) (r/a > > 1)
= \(\frac{5 \times 10^{-8} \mathrm{Cm}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15)^3 \times 10^{-6} \mathrm{~m}^3}\)
= 1.33 × 10⁵ N C^-1
The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.

Question 11.
The electric field components in Fig. are E_x = ax^1/2, E_y = E_z = 0, in which a = 800 N/C m^1/2. Calculate (a) the flux through the cube and (b) the charge within the cube. Assume that a = 0.1 m.

Solution:
a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and ∆S is ± π/2. Therefore, the flux Φ = E. ∆S is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is E_L = x^1/2 = αa^1/2
(x = a at the left face).
The magnitude of electric field at the right face is E_R = αx^1/2 = α(2a)^1/2
(x = 2a at the right face).
The corresponding fluxes are
Φ_L = E_L . ∆S = ∆SE_L . \(\hat{n}_L\) = E_L ∆S cos θ = -E_L ∆S, since θ = 180°
= -E_La²
Φ_R = E_R . ∆S = E_R ∆S cos θ = E_R ∆S, since θ = 0°
= E_Ra²
Net flux through the cube.
= Φ_R + Φ_L = E_Ra² – E_La² = a² (E_R – E_L) = αa² [(2a)^1/2 – a^1/2]
= αa^5/2 (\(\sqrt{2}\) – 1)
= 800 (0.1)^5/2 (\(\sqrt{2}\) – 1)
= 1.05 N m² C^-1

b) We can use Gauss’s law to find the total charge q inside the cube.
We have f = \(\frac{\mathrm{q}}{\varepsilon_0}\) or q = Φε₀. Therefore,
q = 1.05 × 8.854 × 10^-12 C = 9.27 × 10^-27 C.

Question 12.
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 \(\hat{\mathrm{i}}\) N/C for x > 0 and E = -200 \(\hat{\mathrm{i}}\) N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = – 10 cm (Fig.),
(a) What is the net outward flux through each flat face ?
(b) What is the flux through the side of the cylinder ?
(c) What is the net outward flux through the cylinder ?
(d) What is the net charge inside the cylinder ?
Solution:
a) We can see from the figure that on the left face E and ∆S are parallel. Therefore, the outward flux is
Φ_L = E. ∆S = -200 \(\hat{\mathrm{i}}\) . ∆S
= +200 ∆S, since \(\hat{\mathrm{i}}\) . ∆S = – ∆S
= +200 × π(0.05)² = + 1.57 Nm²C^-1
On the right face, E and AS are parallel and therefore
Φ_R = E. ∆S = +1.57 Nm²C^-1.

b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E. ∆S = 0. Therefore, the flux out of the side of the cylinder is zero.

c) Net outward flux through the cylinder Φ = 1.57 + 1.57 + 0 = 3.14 Nm²C^-1.

d) The net charge within the cylinder can be found by using Gauss’s law which gives
q = ε₀Φ
= 3.14 × 8.854 × 10^-12 C
= 2.78 × 10^-11 C

Question 13.
An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus ?
Solution:
The charge distribution for this model of the atom is as shown in Fig. The total negative charge in the uniform spherical charge distribution of radius R must be -Ze, since the atom (nucleus of charge Z e + negative charge) is neutral.

This immediately gives us the negative charge density p, since we must have
\(\frac{4 \pi \mathrm{R}^3}{3}\) ρ = 0 – Ze or ρ = – \(\frac{3 \mathrm{Ze}}{4 \pi \mathrm{R}^3}\)
To find the electric field E(r) at a point P which is a distance r away from the nucleus, we use Gauss’s law. Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance, no matter what the direction of r. Its direction is along (or opposite to) the radius vector r from the origin to the point P. The obviouis Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely r < R and r > R.

i) r < R : The electric flux Φ enclosed by the spherical surface is Φ = E(r) × 4πr²
Where E(r) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface.
The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r,
i.e., q = Ze + \(\frac{4 \pi \mathrm{r}^3}{3}\) ρ
Substituting for the charge density p obtained earlier, we have
q = Ze – Ze\(\frac{\mathrm{r}^3}{\mathrm{R}^3}\)
Gauss’s law then gives,
The electric field is directed radially outward.

ii) r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law,
E(r) × 4 π r² = 0 or E(r) = 0 ; r > R
At r = R, both cases give the same result: E = 0.

Students get through AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics

Short Answer Questions

Question 1.
Explain Doppler effect in light. Distinguish between red shift and blue shift. [T.S. Mar. 16]
Answer:
Doppler effect in light : The change in the apparent frequency of light, due to relative motion between source of light and observer. This phenomenon is called Doppler effect.

The apparent frequency of light increases when the distance between observer and source of light is decreasing and the apparent frequency of light decreases, if the distance between source of light and observer increasing.
Doppler shift can be expressed as = \(\frac{\Delta v}{v}=\frac{-v_{\text {radial }}}{c}\)
Applications of Doppler effect in light:

1. It is used in measuring the speed of a star and speed of galaxies.
2. Measuring the speed of rotation of the sun.

Red shift: The apparent increase in wave length in the middle of the visible region of the spectrum moves towards the red end of the spectrum is called red shift.

Blue shift: When waves are received from a source moving towards the observer, there is an apparent decrease in wave length, this is called blue shift.

Question 2.
Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity. [A.P. Mar. 16; T.S. Mar. 15]
Answer:
Let y₁ and y₂ be the displacements of the two waves having same amplitude a and <|> is the phase difference between them.

y₁ = a sin ωt …………… (1)
y₂ = a sin (ωt + Φ) …………….. (2)
The resultant displacement y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ)
y = a sin ωt + a sin ωt cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt (a sin Φ)
Let R cos θ = a(1+ cos Φ) ……………….. (4)
R sin θ = a sin Φ ………………. (5)
y = R sin ωt. cos θ + R cos ωt. sin θ
y = R sin (ωt + θ) ………………… (6)
where R is the resultant amplitude at P, squaring equations (4) and (5), then adding
R² [cos² θ + sin2 θ] = a²[l + cos² Φ + 2 cos Φ + sin² Φ]
R² [1] = a² [1 + 1 + 2 cos Φ]
I = R² = 2a² [1 + cos Φ] = 2a² × 2 cos² \(\frac{\phi}{2}\);
I = 4a² cos² \(\frac{\phi}{2}\) …………. (7)

i) Minimum intensity (I_max)
cos² \(\frac{\phi}{2}\) = 1
Φ = 2nπ Where n = 0, 1, 2, 3 ……….
Φ = 0, 2π, 4π, 6π
∴ I_max = 4a².

ii) Minimum intensity (I_min)
cos² \(\frac{\theta}{2}\) = 1
Φ = (2n + 1)π Where n = 0, 1, 2, 3 ……….
Φ = π, 3π, 5π, 7π …………..
∴ I_min = 0

Question 3.
Does the principle of conservation of energy hold for interference and diffraction phenomena? Explain briefly. [Mar. 14]
Answer:
Yes, law of conservation of energy is obeyed. In case of constructive interference, intensity becomes maximum. Hence bright fringes are formed on the screen where as in the case of destructive interference, intensity becomes minimum. Hence dark fringes are formed on the screen.

This establishes that in the interference and diffraction pattern, the intensity of light is simply being redistributed i.e., energy is being transferred from dark fringe to bright fringe. No energy is being created (or) destroyed in the process. Hence energy is redistributed.

Thus the principle of conservation of energy is being obeyed in the process of interference and diffraction.

Question 4.
How do you determine the resolving power of your eye ? [A.P. Mar. 17]
Answer:
Make black strips of equal width separated by white strips. All the black strips having same width, while the width of white strips should increase from left to right.

Now watch the pattern with one eye. By moving away (or) closer to the wall, find the position where you can just see some two black strips as separate strips.

All black strips to the left of this strips would merge into one another and would not be distinguishable on the other hand, the black strips to the right of this would be more and more clearly visible.

Note the width d of the white strips and measure the distance D of the wall from eye.
Then resolution of your eye = \(\frac{\mathrm{d}}{\mathrm{D}}\)

Question 5.
Explain polarisation of light by reflection and arrive at Brewster’s law from it.
Answer:
Polarisation of light by reflection : When unpolarized light is incident on the boundary of a denser medium, at a particular angle of incidence the reflected light is completely plane polarised. This incident angle is called Brewster’s angle (i_B).

Brewster’s law:
When light is incident on a transparent surface at Brewster s angle, then reflected and refracted rays are at right angles to each other.
From snell’s law, refractive index,
∴ i_B + r = \(\frac{\pi}{2}\) ⇒ r = \(\frac{\pi}{2}\) – i_B

n = \(\frac{\sin \mathrm{i}_B}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_B}{\sin \left(\frac{\pi}{2}-\mathrm{i}_B\right)}=\frac{\sin \mathrm{i}_B}{\cos \mathrm{i}_B}\) = tan i_B
n = tani_B, This is known as Brewster’s law.
Brewster’s law – statement: The refractive index of a denser medium is equal to tangent of the polarising angle.

Question 6.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. [T.S. Mar. 17; Mar. 12]
Answer:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ.
Where θ is the angle between pass axes P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (\(\frac{\pi}{2}\) – θ)
Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ . cos²(\(\frac{\pi}{2}\) – θ)
= I₀cos²θ . sin²θ
I = \(\frac{\mathrm{I}_0}{4}\) sin² 2θ
∴ The transmitted intensity will be maximum when θ = \(\frac{\pi}{4}\).

Long Answer Questions

Question 1.
Distinguish between Coherent and Incoherent addition of waves. Develop the theory of constructive interferences.
Answer:
Coherent sources : The two sources which maintain zero (or) any constant phase relation between themselves are known as Coherent sources.

Incoherent sources : If the phase difference changes with time, the two sources are known as incoherent sources.
Theory of constructive and destructive interference :
Let the waves of two coherent sources be
y₁ = a sin ωt ………………….. (1)
y₂ = a sin (ωt + Φ) …………….. (2)
where a is amplitude and Φ is the phase difference between two displacements.
According to superposition principle, y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ) = a sin ωt + a sin ω cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt [a sin Φ] ………………….. (3)
Let A cos θ = a(1 + cos Φ], ………………. (4)
A sin θ = a sin Φ ……………… (5)
Substituting equations (4) and (5) in equation (3)
y = A sin ωt. cos θ + A cos ωt sin θ
y = A sin (ωt + θ) …………………… (6)
Where A is resultant amplitude. Squaring equations (4) and (5), then adding
A²[cos² θ + sin² θ] = a²[1 + cos² Φ + 2 cos Φ + sin² Φ]
A² [1] = a² [1 + 1 + 2 cos Φ]
I = A² = 2a² [1 + cos Φ]
I = 2a² × 2 cos² \(\frac{\phi}{2}\)
I = 4a² cos² \(\frac{\phi}{2}\)
I = 4I₀ cos² \(\frac{\phi}{2}\) …………….. (7) [∵ I₀ = a²]

Case (i) For constructive interference: Intensity should be maximum.
cos \(\frac{\phi}{2}\) = 1 ⇒ Φ = 2nπ
Where n = 0, 1, 2, 3 ⇒ Φ = 0, 2π, 4π, 6π ………….. I_max = 4I₀
Case (ii) For destructive interference: Intensity should be minimum
i.e., cos Φ = 0 ⇒ Φ = (2n + 1) π ; where π = 0, 1, 2, 3 ; Φ = π, 3π, 5π ⇒ I_min = 0.

Question 2.
Describe Young’s experiment for observing interference and hence arrive at the expression for ‘fringe width’.
Answer:
Interference: The modification of intensity obtained by the super position of two (or) more light waves is called interference.
Description :

1. Thomas Young experimentally observed the phenomenon of interference of light using two coherent sources.
2. A small pin hole ‘S’ illuminated by monochromatic source of light which produces a spherical wave.
3. S₁ and S₂ are two narrow pin holes equidistant from S.
4. Screen is placed at a distance D.
5. The points at which any two crests (or) any two troughs are superimposed, constructive interference takes place bright fringe will be observed on the screen.
6. The points at which crest of one wave and trough of another wave are superimposed, destructive interference takes place dark fringe will be observed on the screen.
7. Thus on the screen alternately bright and dark frings are observed.

Expression for fringe width :
1. It is the distance between two successive bright (or) dark fringes, denoted by β.

The path difference (δ) = d sin θ
2. If θ is very small then from figure sin θ ≈ tan θ = \(\frac{x}{D}\)

3. For bright fringes path difference S₂P – S₁P = nλ
∴ d sin θ = nλ
d × \(\frac{x}{D}\) = nλ
x = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) ……………. (1) where n = 0, 1, 2, 3 …………….

This equation represents the position of bright fringe.
When n = 0, x₀ = 0.
n = 1, x₁ = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) and n = 2, x₂ = \(\frac{2\lambda \mathrm{D}}{\mathrm{d}}\)
The distance between any two consecutive bright fringes is
x₂ – x₁ = \(\frac{2 \lambda D}{d}-\frac{\lambda D}{d} \Rightarrow \beta=\frac{\lambda D}{d}\) ………… (2)

4. For dark fringes path difference S₂P – S₁P = (2n + 1) \(\frac{\lambda}{2}\) ∴ d sin θ = (2n + 1) \(\frac{\lambda}{2}\)
d × \(\frac{x}{D}\) = (2n + 1) \(\frac{\lambda}{2}\) ⇒ x = \(\frac{(2 n+1) \lambda D}{2 d}\) ………….. (3) where n = 0, 1, 2, 3…………
This equation (3) represents, position of dark fringe.
When n = 0, x₀ = \(\frac{\lambda D}{2 d}\) ⇒ n = 1, x₁ = \(\frac{3 \lambda D}{2 d}\); n = 2, x₂ = \(\frac{5 \lambda D}{2 d}\) ………….
The distance between any two consecutive dark fringes is x₂ – x₁ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}-\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}=\frac{5 \lambda \mathrm{D}-3 \lambda \mathrm{D}}{2 \mathrm{~d}}\)
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) …………………….. (4)
Hence fringe width is same for bright and dark fringes.

Problems

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ? .
Solution:
Since vλ = c, \(\frac{\Delta v}{v}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Question 3.
What is the Brewster angle for air to glass transition ? (Refractive index of glass = 1.5.)
Solution:
Here, i_p = ? μ = 1.5; As tan i_p = μ = 1.5
∴ i_p = tan^-1 (1.5); i_p = 56.3

Question 4.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?
Solution:
Let I₁ = I₂ = I. If Φ is phase difference between the two light waves, then resultant intensity,
I_R = I₁ + I₂ + \(2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}\) . cos Φ
When path difference = λ, Phase difference Φ = 0°
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) . cos 0° = 4I = k
When path difference = \(\frac{\lambda}{3}\), phase difference Φ = \(\frac{2 \pi}{3}\) rad
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) cos \(\frac{2 \pi}{3}\) ⇒ I’_R = 2I + 2I(\(\frac{-1}{2}\)) = I = \(\frac{\mathrm{k}}{4}\)

Question 5.
Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then ∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
= 2.9 × 10^-7 radians.

Question 6.
In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m, β = 1.2 × 10^-2 m, n = 4
Since, β = \(\frac{\mathrm{D}}{\mathrm{d}}\) nλ ⇒ \(\frac{\beta}{n} \cdot \frac{d}{D}\) ⇒ λ = \(\frac{1.2 \times 10^{-2} \times 2.8 \times 10^{-2}}{4 \times 1.4}\)
⇒ λ = 600 × 10^-9 m
⇒ λ = 600 nm.

Textual Examples

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ?
Solution:
Since vλ = c, \(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
(a) When monochromatic light is incident on a surface separating, two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why ?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave ?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Solution:
a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) and undergo forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.

b) No. energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing ah unit area per unit time.

Question 3.
Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when blue*: green light of wavelength 500 nm is used ?
Solution:
Fringe spacing = \(\frac{\mathrm{D} \lambda}{\mathrm{d}}=\frac{1 \times 5 \times 10^{-7}}{1 \times 10^{-3}}\) m
= 5 × 10^-4 m = 0.5 mm

Question 4.
What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations :
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochro-matic) source of shorter wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved Closer to the double-slit plane;
(e) the width of the source slit is increased;
(f) the monochromatic source is replaced by a source of white light ?
(In each operation, take all parameters, other than the one specified, to remain unchanged.)
Solution:
a) Angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits.

b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.

c) The separation of the fringes (and also angular separation) decreases. See however, the condition mentioned in (d) below.

d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition s/S < λ/d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.

e) Same as in (d). As the source slit which increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s/S ≤ λ/d is not satisfied, the interference pattern disappears.

f) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P = λ_b/2, where λ_b (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue.
Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 5.
In Textual Example 3, what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern ?
Solution:
We want aθ = λ, θ = \(\frac{\lambda}{\mathrm{a}}\)
10 \(\frac{\lambda}{\mathrm{d}}\) = 2\(\frac{\lambda}{\mathrm{a}}\), a = \(\frac{\mathrm{d}}{5}\) = 0.2 mm

Question 6.
Assume that light of wavelength 6000 Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a =100 inch = 254 cm.
Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then
∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
≈ 2.9 × 10^-7 radians.

Question 7.
For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm ?
Solution:
Z_F = \(\frac{\mathrm{a}^2}{\lambda}=\frac{\left(3 \times 10^{-3}\right)^2}{5 \times 10^{-7}}\) = 18 m

Question 8.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ?
Solution:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ,
where θ is the angle between pass axes of P₁ and P₂. Since P₂ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ cos² (\(\frac{\pi}{2}\) – θ)
= I₀ cos² θ sin² θ = (I₀/4) sin² 2θ
Therefore, the ‘transmitted intensity will be maximum when θ = π/4.

Question 9.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Students get through AP Inter 2nd Year Physics Important Questions 2nd Lesson Ray Optics and Optical Instruments which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 2nd Lesson Ray Optics and Optical Instruments

Very Short Answer Questions

Question 1.
What is optical density and how is it different from mass density?
Answer:
Optical density: Optical density is defined as the ratio of the speed of light in media. Mass density: Mass per unit volume is defined as mass density.
The mass density of an optically denser medium is less than that of an optically rarer medium.

Question 2.
What are the laws of reflection through curved mirrors?
Answer:

1. “The angle of reflection equals to the angle of incidence”.
2. “The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane”.

Question 3.
Define ‘power’ of a convex lens. What is its unit ? [A.P. Mar. 17, T.S. Mar. 16]
Answer:
Power of a lens : Power of a lens is defined as its bending ability and is measured as reciprocal of focal length in metre.
∴ Power of a lens = \(\frac{1}{f(\text { in metres) }}=\frac{100}{\mathrm{f}(\text { in cms) }}\)
Unit → Dioptre (D)

Problems

Question 1.
A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?
Answer:
f = 10 cm, υ = 35 cm
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{-u}\) (using sign convention)
\(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{35}-\frac{1}{10}\)
\(\frac{1}{u}=\frac{10-35}{35 \times 10}=\frac{-1}{14}\)
U = – 14 cm.
Distance of the object from the wall = 35 – 14 = 21cm.

Question 2.
A concave mirror produces an image of a long vertical pin, placed 40cm from the mirror, at the position of the object. Find the focal length of the mirror. [T.S. Mar. 17, 16]
Answer:
Give u = υ = 40cm
\(\frac{1}{\mathrm{f}}=\frac{1}{v}+\frac{1}{u}\)
\(\frac{1}{f}=\frac{1}{40}+\frac{1}{40}\)
\(\frac{1}{\mathrm{f}}=\frac{2}{40}\)
f = 20 cm.

Question 3.
A small angled prism of 4° deviates a ray through 2.48°. Find the refractive index of the prism.
Answer:
A = 4°, D_m = 2.48°
D_m = A (μ – 1)
μ – 1 = \(\frac{\mathrm{D}_{\mathrm{m}}}{\mathrm{A}}=\frac{2.48}{4}\) = 0.62
μ = 1 + 0.62
μ = 1.62

Question 4.
What is ‘dispersion’? Which colour gets relatively more dispersed ? [Mar. 14]
Answer:
Dispersion : The phenomenon of splitting of white light ipto its constituent colours, on passing through a prism is called dispersion of light.
The deviation is maximum for violet colour.

Question 5.
The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is 1/10 of its size ?
Solution:
f = 30 cm, h₁ = h, h₂ = \(\frac{\mathrm{h}}{10}\)

Question 6.
What is myopia ? How can it be corrected ? [T.S. Mar. 15]
Answer:
Myopia (or) Near sightedness :
The light from a distant object arriving at the eye-lens may get converged at a point infront of the retina. This type of defect is called myopia.
To correct this, we interpose a concave lens between the eye and the object.

Question 7.
What is hypermetropia ? How can it be corrected ? [A.P. Mar. 16]
Answer:
Hypermetropia (or) Farsightedness :
The light from a distant object arriving at the eye-lens may get converged at a point behind the retina. This type of defect is called Hypermetropia.
To correct this, we interpose a convex lens (Convergent lens) between the eye and the object.

Question 8.
Draw neat labelled ray diagram of simple microscope. [PE 2015 (A.P.)]
Answer:

u = object distance
D = distance of near point.

Short Answer Questions

Question 1.
Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal length. [A.P. Mar. 17]
Answer:
Focal length of concave mirror:
The distance between the focus F and the pole P of the mirror is called the focal length of the concave mirror.

Consider a ray AB parallel to principal axis incident on a concave mirror at B and is re-flected along BF. The line CB is normal to the mirror.
Let θ be the angle of incidence, ∠ABC = ∠BCP = θ
Draw BD ⊥ CP
In right angled ∆^le
Tan θ = \(\frac{\mathrm{BD}}{\mathrm{CD}}\) ……………. (1)
From ∆^le BFD, Tan 2θ = \(\frac{\mathrm{BD}}{\mathrm{FD}}\) ……………. (2)
Dividing eq (2) by eq (1), \(\frac{{Tan} 2 \theta}{{Tan} \theta}=\frac{\mathrm{CD}}{\mathrm{FD}}\) ………………….. (3)
If θ is very small, then tan θ ≈ θ and tan 2θ ≈ 2θ since the aperture of the lens is small
∴ The point B lies very close to p.
CD ≈ CP and FD ≈ FP
From eq (3), \(\frac{2 \theta}{\theta}=\frac{\mathrm{CP}}{\mathrm{FP}}=\frac{\mathrm{R}}{\mathrm{f}} \Rightarrow 2=\frac{\mathrm{R}}{\mathrm{f}}\)
R = 2f

Question 2.
Define critical angle. Explain total internal reflection using a neat diagram. [T.S. Mar. 15]
Answer:
Critical angle:
When light ray travelling from denser medium to rarer medium, then the angle of incidence for which angle of refraction in air is 90a is called critical angle.
C = sin^-1 \(\left(\frac{1}{\mu}\right)\)
Total internal reflection:
When a light ray travels from denser to rarer medium, the angle of incidence is greater than the critical angle, then it reflects into the same medium is called total internal reflection.

Explanation:
Consider an object in the denser medium. A ray OA incident on XY bends away from the normal. As the angle of incidence is increased, the angle of refraction goes on increasing. For certain angle of incidence, the refracted ray parallel to XY surface (r = 90°).

When the angle of incidence is further increased, the ray is not refracted but is totally reflected back in the denser medium. This phenomenon is called total internal reflection.

Question 3.
Explain the formation of a mirage. [A.P. Mar. 16]
Answer:
In a desert, the sand becomes very hot during the day time and it rapidly heats the layer of air which is in its contact. So density of air decreases. As a result the successive upward layers are denser than lower layers.

When a beam of light travelling from the top of a tree enters a rarer layer, it is refracted away from the normal. As a result at the surface of layers of air, each time the angle of incidence increases and ultimately a stage is reached, when the angle of incidence becomes greater than the critical angle between the two layers, the incident ray suffers total internal reflection.

So it appears as inverted image of the tree is formed and the same looks like a pool of water to the observer.

Question 4.
Explain the formation of a rainbow. [A.P. Mar. 15]
Answer:

Figure shows how sun light is broken into its segments in the process and a rainbow appears. The dispersion of the violet and the red rays after internal reflection in the drop is shown in figure.

The red rays emerge from the drops of water at one angle (43°) and the violet rays emerge at another angle (41°). The large number of water drops in the sky makes a rain-bow. The rainbow appears semicircular for an observer on earth.

Question 5.
Why does the setting sun appear red ? [T.S. Mar. 17, Mar. 14]
Answer:
As sunlight travels through the earths atmosphere, gets scattered by the large number of molecules present. This scattering of sun light is responsible for the colour of the sky, during sunrise and sunset etc.
The light of shorter wave length is scattered much more than light of larger wave-length. Scattering ∝ \(\frac{1}{\lambda^4}\).
Most of blue light is scattered, hence the bluish colour of sky predominates.

At sunset (or) sunrise, sun rays must pass through a larger atmospheric distance. More of the blue colour is scattered away only red colour which is least scattered ap-pears to come from sun. Hence it appears red.

Question 6.
With a neat labelled diagram explain the formation of image in a simple microscope. [T.S. Mar. 16, A.P. Mar. 15]
Answer:
Simple microscope: It consists a single short focus convex lens. It increases the visual angle to see an object clearly. It is also called magnifying glass (or) reading glass.

Working : The object is adjusted within the principal focus Of the convex lens to form the image at the near point. The image is formed on same side of the object and it is virtual, erect and magnified as shown in fig.

Magnifying power : The ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye is called magnifying power of a simple microscope.
It is denoted by ‘m’.
m = \(\frac{\alpha}{\beta} \simeq \frac{{Tan} \alpha}{{Tan} \beta}\)

Question 7.
What is the position of the object for a simple microscope ? What is the maximum magnifi-cation of a simple microscope for a realistic focal length ?
Answer:
When an object is placed between principal focus and optical centre of a convex lens, a virtual and erect image will be formed on the same side of the object.

Magnifying power: It is defined as the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the eye.
m = \(\frac{\alpha}{\beta} \simeq \frac{{Tan} \alpha}{{Tan} \beta}\)
From figure OJ = IJ’, ∠IO’G = α and ∠IO’J’ = β

The above equation can be written as
m = 1 + \(\frac{\mathrm{D}}{\mathrm{f}}\) …………… (2)
This shows that smaller the focal length of the lens, greater will be the magnifying power of microscope.

Long Answer Questions

Question 1.
Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.
Answer:
Description : It consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The distance of the object can be adjusted by
means of a rack and pinion arrangement.

Working: The object OJ is placed outside the principal focus of the objective and the real image is formed on the other side of it. The image I₁ G₁ is real, inverted and magnified.
This image acts as the object for the eyepiece. The position of the eyepiece is so adjusted that the image due to the objective is between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified.

Magnifying Power: It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point.

Imagining that the eye is at the optic centre, the angle subtended by the final image is a. When the object is imagined to be taken at near point it is represented by IJ’ and OJ – IJ’.
The angle made by I J’ at the eye is β. Then by the definition of magnifying power

Dividing and multiplying by I₁ G₁ on the right side, we get
m = \(\left(\frac{\mathrm{IG}}{\mathrm{I}_1 \mathrm{G}_1}\right)\left(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\right)\)
Magnifying power of the objective (m₀) = I₁ G₁ / OJ = Height of the image due to the objective / Height of its object.
Magnifying power of the eye piece (m_e) = IG/I₁G₁ = Height of the final image / Height of the object for the eyepiece.
∴ m = m₀ × m_e ……………… (1)
To find m0: In figure OJ O’ and I₁ G₁ O’ are similar triangles. \(\left(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\right)=\frac{\mathrm{O}^{\prime} \mathrm{I}_1}{\mathrm{O}^{\prime} \mathrm{O}}\)
Using sign convention, we find that O’I₁ = + v₀ and O’O = -u where v₀ is the image distance due to the objective and u is the object distance for the objective or the compound microscope. I₁G₁ is negative and OJ is positive.
∴ m₀ = \(\frac{\mathrm{v}_0}{\mathrm{U}}\) (∵ \(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\) = m₀)
To find m_e : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece.
∴ m_e = (1 + \(\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\))
Where f_e is the focal length of the eyepiece.
Substituting m0 and me in equation (1),
m = \(+\frac{\mathrm{v}_0}{\mathrm{u}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\)
When the object is very close to the principal focus F₀ of the objective, the image due to the objective becomes very close to the eyepiece,
u ≈ -f₀ and v₀ ≈ L
Where L is the length of the microscope. Then
m ≈ \(-\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)\)

Question 2.
a) Define Snell’s Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism.
b) A ray of light, after passing through a medium, meets the surface separating the medium from air at an angle of 45° and is just not refracted. What is the refractive index of the medium ?
Answer:
a) Snell’s law:
The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant, called the refractive index of the medium.
\(\frac{\sin i}{\sin r}\) = μ (constant).
Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is p. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = emergent ray.
Let angle of incidence = i₁
angle of emergence = i₂
angle of refraction = r₁
angle of refraction at R = r₂
After travelling through the prism it falls on AC and emerges as RS.

The D angle of deviation.
From the ∆ QRT
r₁ + r₂ + ∠T = 180° ……………….. (1)
From the quadrilateral AQTR
∠A + ∠T = 180°
∠T = 180° – A ………………. (2)
From the equations (1) and (2)
r₁ + r₂ + ∠T = 180° we get
r₁ + r₂ + 180° – A = 180°
r₁ + r₁ = A ………………. (3)
from the ∆ QUR
i₁ – r₁ + i₂ – r₂ + 180° – D = 180°
i₁ + i₂ = (r₁ + r₂) = D
i₁ + i₂ – A = D [∵r₁ + r₂ = A]
i₁ + i₂ = A + D ……………… (4)
Minimum deviation : Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation ‘δ’ as shown in the fig.

When D decreases the two angles i₁ and i₂ become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e, i ₁ = i₂.
As i₁ = i₂, r₁ = r₂
∴ i₁ = i₂ = i, r₁ = r₂ = r
substituting this in (1) and (2) we get
2r = A ⇒ r = A/2
i + i = A + δ ⇒ i = \(\frac{\mathrm{A}+\delta}{2}\)
According to Snell’s law μ =
μ =
Note :The minimum deviation depends on the refractive index of the prism material and the angle of the prism.

b) Given that i = C = 45°
μ = \(\frac{1}{\sin \mathrm{c}}\) ⇒ μ = \(\frac{1}{\sin 45^{\circ}}\)
μ = \(\frac{1}{1 / \sqrt{2}}=\sqrt{2}\)
μ = 1.414

Textual Examples

Question 1.
Suppose that the lower half of the concave mirror’s reflecting surface in figure is covered with an opaque (non- reflective) material. What effect will this have on the image of an object p’ iced in front of the mirror ?

Solution:
You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the re-fleeting surface has been reduced, the intensity of the image will be low (in this case, half).

Question 2.
A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror ?

Solution:
The ray diagram for the formation of the image of the phone is shown in fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B’C = BC. You can yourself realise why the image is distorted.

Question 3.
An object is placed at (i) 10 cm. (ii) 5 cm in front of a concave mirror of radius of curvature 15cm. Find the position, nature, and magnification of the image in each case.
Solution:
The focal length f = – 15/2 cm = – 7.5 cm
i) The object distance u = – 10 cm. Then
Eq. \(\frac{1}{v}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\) gives
\(\frac{1}{v}+\frac{1}{-10}=\frac{1}{-7.5}\)
or υ = \(\frac{10 \times 7.5}{-2.5}\) = -30 cm
The image is 30 cm from the mirror on the same side as the object.
Also, magnification
m = –\(\frac{v}{u}=-\frac{(-30)}{(-10)}\) = -3
The image is magnified, real and inverted.

ii) The object distance u = -5cm. Then
from Eq. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}+\frac{1}{-5}=\frac{1}{-7.5}\)
or υ = \(\frac{5 \times 7.5}{(7.5-5)}\) = 15 cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = – \(\frac{v}{u}=-\frac{15}{(-5)}\) = 3
The image is magnified, virtual and erect.

Question 4.
Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5ms^-1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29m, (c) 19 m and (d) 9 m away.
Solution:
From the mirror equation, Eq. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
we get
υ = \(\frac{\mathrm{fu}}{\mathrm{u}-\mathrm{f}}\)
For convex mirror, since R = 2m, f = 1 m.
Then for u = -39 m. v = \(\frac{(39) \times 1}{-39-1}\)
= \(\frac{39}{40}\)
Since the jogger moves at a constant speed of 5ms”1, after Is the position of the image u (for u = – 39 + 5 = – 34) is (34/35) m.
The shift in the position of image in 1 s is
\(\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}\)
Therefore, the average speed of the im-age when the jogger is between 39 m and 34 m from the mirror, is (1/280) ms^-1. Similarly, it can be seen that for u = – 29 m, -19 m and -9 m, the speed with which the image appears to move is \(\frac{1}{150}\)ms^-1, \(\frac{1}{60}\) ms^-1 and \(\frac{1}{10}\) ms^-1 respectively.

Question 5.
The earth takes 24 h to rotate once about its axis. How much time does the sun take to shift by 1° when viewed from the earth ?
Solution:
Time taken for 360° shift = 24h
Time taken for 1° shift = 24/360
h = 4 min.

Question 6.
Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed ?
Solution:
We use the relation given by Eq.
\(\frac{\mathrm{h}_2}{v}-\frac{\mathrm{h}}{\mathrm{u}}=\frac{\mathrm{h}_2-\mathrm{h}_1}{\mathrm{R}}\) Here
u = – 100 cm, υ = ?. R = + 20 cm, n₁ = 1, and n₂ = 1.5. We then have
\(\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}\)
or υ = + 100 cm

Question 7.
A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid ? Could the liquid be water ?
Answer:
The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n₁ = n₂. This gives 1/f = 0 or f → ∞. The lens in the liquid will act like a plane sheet of glass. No, the liquid is not water. It could be glycerine.

Question 8.
(i) If f = 0.5 m for a glass lens, what is the power of the lens ? (ii) The radii of curvature 6f the faces of a double convex lens are 10 cm and 15cm. Its focal length is 12 cm. What is the refractive index of glass ? (iii) A convex lens has 20 cm focal length in air. What is focal length in water ?
(Refractive index of air-water = 1.33. Refractive index for air – glass = 1.5.)
Solution:
i) Power = + 2 dioptre.

ii) Here, we have f = + 12 cm,
R₁ = + 10 cm, R₂ = – 15 cm.
Refractive index of air is taken as unity.
We use the lens formula of Eq. \(\frac{\mathrm{h}_1}{\mathrm{DB}}+\frac{\mathrm{h}_1}{\mathrm{DB}}=\frac{\mathrm{h}_1}{\mathrm{f}}\). The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac{1}{12}\) = (n – 1) (\(\frac{1}{10}\) – \(\frac{1}{-15}\))
This gives n = 1.5.

iii) For a glass lens in air, n₂ = 1.5, n: = 1, f = + 20cm. Hence, the lens formula gives.
\(\frac{1}{20}=0.5\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\)
For the same glass lens in water. n₂ = 1.5, n₁ = 1.33. Therefore,
\(\frac{1.33}{f}=(1.5-1.33)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\)
Combining the above two equations, we find f = + 78.2 cm.

Question 9.
Find the position of the image formed by the lens combination given in the fig.

Solution:
Image formed by the first lens
\(\frac{1}{v_1}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_1}\)
\(\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}\)
or v₁ = 15 cm
The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens. Though the image is reeil, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.
\(\frac{1}{v_3}-\frac{1}{10}=\frac{1}{-10}\) or v₂ ∞
The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.
\(\frac{1}{v_3}-\frac{1}{\mathrm{u}_3}=\frac{1}{\mathrm{f}_3} \text { or } \frac{1}{v_3}-\frac{1}{\infty}+\frac{1}{30}\)
or v₃ = 30 cm
The final image is formed 30 cm to the right of the third lens.

Question 10.
What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm ?
Solution:
The distance of normal vision is 25cm. So if a book is at u= – 25 cm. Its image should be formed at υ = – 50 cm. There-fore, the desired focal length given by
\(\frac{1}{\mathrm{f}}=\frac{1}{v}-\frac{1}{\mathrm{u}} \text { or } \frac{1}{\mathrm{f}}=\frac{1}{-50}-\frac{1}{-25}=\frac{1}{50}\)
or f = + 50 cm (convex lens).

Question 11.
a) The far point of a myopic person is 80 cm in front of the eye. What is ,. the power of the lens required to enable him to see very distant objects clearly ?
b) In what way does the corrective lens help the above person ? Does the lens magnify very distant objects ? Explain carefully.
c) The above person prefers to remove his spectacles while reading a book. Explain why?
Solution:
a) Solving as in the previous example, we find that the person should use a concave lens of focal length = – 80 cm. i.e:, of power = – 1.25 dioptres.

b) No. The concave lens, in fact, reduces the size of the object, but the angle subtended by the image (at the far point) at the eye. The eye is able to see distant objects not because the corrective lens magnifies the object, but because it brings the (i.e., it produces virtual image of the object) at the far point of the eye which then can be focussed by the eye-lens on the retina.

c) The myopic person may have a normal near point, i.e., about 25 cm (or even less). In order to read a book with the spectacles, such a person must keep the book at a distance greater than 25cm so that the image of the book by the concave lens is produced not closer than 25cm. The angular size of the book (dr its image) at the greater distance is evidently less than the angular size when the book is placed at 25 cm and no spectacles are needed. Hence, the person prefers to remove the spectacles while reading.

Question 12.
a) The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25cm from the eye ?
b) In what way does the corrective lens help the above person ? Does the lens magnify objects held near the eye ?
c) The above person prefers to remove the spectacles while looking at the sky. Explain why?
Solution:
a) u = – 25 cm, υ = – 75 cm
1/f = 1/25 – 1/75, i.e., f = 37.5 cm.
The corrective lens- needs to have a converging power of +2.67 dioptres.

b) The corrective lens produces a virtual image (at 75 cm) of an object at 25 cm. The angular size of this image is the same as that of the object. In this sense the lens does not magnify the object but merely brings the object to the near point of the hypermetric eye, which then gets focussed on the retina. However, the angular size is greater than that of the same object at the near point (75 cm) viewed without the spectacles.

c) A hypermetropic eye may have normal far point i.e., it may have enough converging power to focus parallel rays from infinity on the retina of the shortened eyeball. Wearing spectacles of converging lenses (used for near vision) will amount to more converging power than needed for parallel rays. Hence the person prefers not to use the spectacles for far objects.

Students get through AP Inter 2nd Year Physics Important Questions 1st Lesson Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 1st Lesson Waves

Very Short Answer Questions

Question 1.
Write the formula for the speed of sound in solids and gases.
Answer:
Speed of sound in solids,
V_s = \(\sqrt{Y / \rho}\) [y = Young’s modulus of solid, ρ = density of solid]
Speed of sound in gases,
V_s = \(\sqrt{\gamma P / \rho}\) [γP = Adiabatic Bulk modulus of gas, ρ = density of gas] .

Question 2.
What does a wave represent?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 3.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare factions are formed alternatively.

Question 4.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = \(\frac{2 \pi}{T}\); k = \(\frac{2 \pi}{\lambda}\)
Parameters:

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 5.
What is the principle of superposition of waves ? .
Answer:
When two or more waves are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.
If y₁, y₂, ……………… y_n be the individual displacements of the particles,then resultant displacement
y = y₁ + y₂ + ……………… + y_n.

Question 6.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 7.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary ?
Answer:
Phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary is radian or 180°.

Question 8.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transeverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 9.
What do you understand by the terms node’ and ‘antinode’ ?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes: The points at which the amplitude is maximum, are called antinodes.

Question 10.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is \(\frac{\lambda}{4}\).

Question 11.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 12.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics..
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 13.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + \(\frac{1}{2}\)) \(\frac{v}{2l}\) where n = 0, 1, 2, 3, ……….

Question 14.
If the air column in a long tube, closed at one end, is set in vibration, what harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n +1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3,

Question 15.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible ?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by no
v_n = \(\frac{\mathrm{nv}}{21}\)
where n = 1, 2, 3, ……………….

Question 16.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.
Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.
It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard, ∆υ = υ₁ ~ υ₂
Importance:
1. It can be used to tune musical instruments.
2. Beats are used to detect dangerous gases.
Explanation for tuning musical instruments with beats:
Musicians use. the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 4.
What is ‘Doppler effect’ ? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between .the observer and the source of sound is called doppler effect.
Examples:

1. The frequency of whistling, engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Sample Problem on Doppler effect:
Two trucks heading in opposite direction with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330m/s). After the two trucks have passed each other, what frequency does the driver of the second truckhear?
Answer:
Speed of first truck = 60 kmph
= 60 × \(\frac{5}{18}\) = 16.66 m/s;
Speed of second truck = 70 kmph 5
= 70 × \(\frac{5}{18}\) = 19.44 m/s
Frequency of horn of first truck = 400 Hz;
Velocity of sound, (V) =330 m/s
Frequency of sound heard by the driver of the second truck when approaching each other,
v¹ = \(\left(\frac{V+V_0}{V-V_s}\right) v=\left(\frac{330+19.44}{330-16.66}\right)\) × 400 = 446 Hz
Frequency of sound heard by the driver of the second truck when approaching each other,
V¹¹ = \(\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{v}=\left(\frac{330-19.44}{330+16.66}\right)\) × 400 = 358.5 Hz

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. [IPE]
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length λ and frequency v, travelling in opposite direction be given by

y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω = 2πv and k = \(\frac{2 \pi}{\lambda}\)
The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}\) ……………… etc, the amplitude = zero
These positions are known as “Nodes”.
If x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}\) …………… etc., the amplitude = maximum (2a)
These positions are called “Antinodes”.
If the string vibrates in ‘P’ segments and T is its length, then length of each segment = \(\frac{l}{\mathrm{P}}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{P}}=\frac{\lambda}{2} \Rightarrow \lambda=\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v \mathrm{P}}{2 l}\) ………………. (1)
If’ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is
v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) …………… (2)
From the Eqs (1) and (2) :
Harmonic frequency v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\Gamma}{\mu}}\)
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) …………….. (3)

Laws of Transverse Waves Along Stretched String:
Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when ‘T’ and ‘μ’ are constant. .

Second Law: When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ \(\sqrt{T}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.
JT .
Third Law: When the length (l) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and T are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. [A.P. 17; IPE 2015, 2016 (TS)]
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super-imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two anti nodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)
l = \(\frac{\lambda_1}{2}\) ⇒ λ₁ = 2l
fundamental frequency v₁ = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity,of sound in air, v₁ = \(\frac{v}{21}\) = v
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ₂ = \(\frac{21}{2}\)
If ‘v₂’ is frequency of second harmonic then v₂ = \(\frac{v}{\lambda_2}=\frac{v \times 2}{2 l}\) = 2v
v₂ = 2v ……………… (2)
Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)
λ₃ = \(\frac{2l}{3}\)
If ‘v₂‘ is frequency of third harmonic then
v₃ = \(\frac{v}{\lambda_3}=\frac{v \times 3}{2 l}\) = 3V
v₃ = 3v …………… (3)
Similarly we can find the remaining or higher harmonic frequencies i.e., v3, v4 etc., can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : V₁ : v₂ = 1 : 2 : 3

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. [IPE 2015, 2016(A.P.), (T.S) A.P. & T.S. Mar. 15]
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air
v₁ = \(\frac{v}{4 l}\) = v …………….. (1)
To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.
∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ₃‘ is wave length of third harmonic
λ₃ = \(\frac{4l}{3}\)
where ‘X3’ is wave length of third harmonic.
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = \(\frac{v}{\lambda_3}=\frac{3 v}{41}\)
v₃ = 3v …………….. (2)
Similarly the next overtone in the close pipe is only fifth harmonic, it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ₅)
∴ l = \(\frac{5 \lambda_5}{4}\) where ‘λ₅‘ is wave length of fifth harmonic.
λ₅ = \(\frac{4l}{5}\)
If ‘v₅‘ is frequency of fifth harmonic (second overtone)
v₅ = \(\frac{v}{\lambda_5}=\frac{5 v}{4 l}\)
v₅ = 5v …………….. (3)
∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [Mar. 17, BMP, 2016 (AP) Mar. 14, (TS)]
Answer:
Doppier effect: The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppier effect. When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = Listener
Let ‘S’ be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T = υ_s. T Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength λ’ = λ – υ_sT.
λ’ = λ – \(\frac{v_s}{v}\) [∵ υ = \(\frac{1}{T}\)]
= \(\frac{\lambda v-v_s}{v}=\frac{v-v_s}{v}\) [∵ υ = vλ]
If “v'” is apparent frequency heard by the listener then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is Velocity of sound in air
v’ = \(\frac{v . v}{v-v_S}\)
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency v’ = \(\frac{v . v}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity.

Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg / m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [IPE 2016 (T.S)
Answer:
v = 30 Hz; I = 0.6 m; μ = 0.05 kg m^-1
υ = ?; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N .

Question 2.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;
μ = \(\frac{\mathrm{M}}{1}=\frac{0.16 \times 10^{-3}}{0.4}\) = 0.4 × 10^-3 kg/m;
T = 70 N; v_n = \(\frac{\mathrm{P}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
v₁ = \(\frac{1}{21} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 \times 0.4} \sqrt{\frac{70}{0.4 \times 10^{-3}}}\) = 523 Hz
v₂ = 2v₁ = 2 × 523 = 1046 Hz
v₃ = 3v₁ = 3 × 523 = 1569 Hz

Question 3.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
l = 70 cm = 70 × 10^-2m; v = 331 m/s ;
v = ?
v = \(\frac{v}{4 l}=\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz.

Question 4.
A steel cable of diameter 3 cm is kept under a tension of lOkN. The density of steel is 7.8 g/ cm3. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = \(\frac{D}{2}=\frac{3}{2}\) cm
= \(\frac{3}{2}\) × 10^-2m;

Question 5.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle. [T.S. Mar. 17]
Solution:
When a whistling train away from rest observer,
v’ = \(\left[\frac{v}{v-v_s}\right] v\) ……………… (1)
When a whistling train away from rest observer,
v” = \(\left[\frac{v}{v+v_s}\right] v\) ……………… (2)
Here v’ = 219 Hz; v” = 184 Hz;
v = 340 m/s

Question 6.
A rocket is moving at a speed of 200 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+v_{\mathrm{S}}}{v}\right]^{-1}\) and not the approximate
v = v₀ [1 – \(\frac{v_s}{v}\)]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [1 – \(\frac{v_s}{v}\)]^-1
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v₀, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Textual Examples

Question 1.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both.
a) Motion of kink in a longitudinal spring produced by displacing one end of the spring sideways.
b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
c) Waves produced by a motorboat sailing in water.
d) Ultrasonic waves in air produced by a vibrating quartz crystal.
Solution:
a) Transverse and longitudinal
b) Longitudinal
c) Transverse and longitudinal
d) Longitudinal.

Question 2.
A wave travelling along a string is des-cribed by, y(x, t) = 0.005 sin (80.0 x – 3.01), in which the numerical constants are in SI untis (0.005 m, 80.0 rad m^-1, and 3.0 rad s^-1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ?
Solution:
On comparing this displacement equation with Eq. y (x, t)n = a sin(kx – ωt + Φ)
y(x, t) = a sin (kx – ωt).
We find
a) the amplitude of the wave is
0. 005 m = 5 mm.

b) the angular wave number k and angular frequency ω are k = 80.0 m^-1 and ω = 3.0 s^-1
We then relate the wavelength λ to k through Eq.
λ = \(\frac{2 \pi}{K}\)
= \(\frac{2 \pi}{80.0 \mathrm{~m}^{-1}}\) = 7.85 cm

c) Now we relate T to ω by the relation
T = \(\frac{2 \pi}{\omega}\)
= \(\frac{2 \pi}{3.0 \mathrm{~s}^{-1}}\)
= 2.09 s
and frequency, v = \(\frac{1}{T}\) = 0.48 Hz
The displacement y at x = 30.0 cm and time t= 20s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (-36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) ≃ 5 mm

Question 3.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the .wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? [A.P. Mar. 19]
Solution:
Mass per unit length of the wire,
μ = \(\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}\) = 6.9 × 10^-3 kg m^-1
Tension, T = 60 N
The speed of wave on the wire is given by
υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}\) = 93 m s^-1

Question 4.
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.
Solution:
We know that 1 mole of any gas occupies 22.4 litres at STP Therefore, density of air at STP is:
ρ₀ = (mass of one mole of air) / (Volume of one mole of air at STP)
= \(\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}\) = 1.29 kgm^-3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
υ = \(\left[\frac{1.01 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1.29 \mathrm{~kg} \mathrm{~m}^{-3}}\right]^{1 / 2}\) = 280 m s^-1

Question 5.
A pipe, 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source ? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s^-1.
Solution:
The first harmonic frequency is given by
v¹ = \(\frac{v}{\lambda_1}=\frac{v}{2 L}\) (open pipe)
Where L is the length of the pipe. The frequency of its n^th; harmonic is
v_n = \(\frac{n v}{2 L}\) for n = 1, 2, 3, ……………… (open pipe)
First few modes of an open pipe are shown in Fig.
For L = 30.0 cm. υ = 330 m s^-1
v_n = \(\frac{\mathrm{n} \times 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}\) = 550 s^-1

Clearly, for a source of frequency 1.1 kHz the air column will resonate at υ₂, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig.), the fundamental frequency is

and only to odd numbered harmonics are present :
v₃ = \(\frac{3 v}{4 L}/latex], v₅ = [latex]\frac{5 v}{4 L}\) and s0 0n.
For L = 30 cm and υ = 330 m s^-1;, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Question 6.
Two sitar strings A and B playing the note ‘Dha’ are slightly put of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Solution:
Increase in the tension of a string increases its frequency. It the original frequency of B (v_B) were greater than that of A(v_A) .further increase in v_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that v_B < v_A. Since v_A – v_B = 5 Hz, and v_a = 427 Hz, we get v_B = 422 Hz.

Question 7.
A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 ms^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+\mathrm{v}_{\mathrm{S}}}{\mathrm{v}}\right]^{-1}\) and not the approximate
v = v₀ \(\left[1-\frac{v_s}{v}\right]\)
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ \(\left(1-\frac{v_{\mathrm{S}}}{v}\right)^{-1}\)
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v₀, the frequency intercepted by the target and not v0, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz