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

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

Very Short Answer Questions

Question 1.

A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field ?

Answer:

The nature of the magnetic field’ is uniform, magnetic dipole (bar magnet) experiences a net force (or torque).

Question 2.

What happens to compass needles at the Earth’s poles ?

Answer:

At the poles, earths field is exactly vertical. So, the campass needles free to rotate in a horizontal plane only, it may point out in any direction.

Question 3.

What do you understand by the ‘magnetization’ of a sample ? (A.P. Mar. ’16)

Answer:

When a magnetic sample is placed in a magnetic field, their magnetic moments are add up in the direction of magnetic field. Hence the sample get a net magnetic moment (m_{net} ≠ 0)

Magnetisation is defined as the net magnetic moment per unit volume i.e., M = \(\frac{m_{\text {net }}}{V}\)

Question 4.

What is the magnetic moment associated with a solenoid?

Answer:

Magnetic dipole moment in a solinoid m = NIA

Where ‘N’ is the number of turns in the loop, ‘I’ the current and A the area vector.

Question 5.

What are the units of magnetic moment, magnetic induction and magnetic filed ?

Answer:

Unit of

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

Question 6.

Magnetic lines form continuous closed loops. Why ? (A.P. Mar. 16)

Answer:

Magnetic lines of force always start from north pole and forming curved path, enter south pole and travel to north pole inside the magnet. Thus lines of force are forming closed loops.

Question 7.

Define magnetic declination. (Mar. ’14)

Answer:

Magnetic Declinatin (D) : The angle between the true geographic north and the north shown by a compass needle is called magnetic declination or simply declination (D).

Question 8.

Define magnetic inclination or angle of dip. (A.P. & T.S. Mar. ’15)

Answer:

Inclination or Dip (I) : The angle which the total intensity of earth’s magnetic field makes with the horizontal at any place is called inclination (I).

Question 9.

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

Answer:

Ferromagnetic materials → Cobalt, Nickel.

Paramagnetic materials → Oxygen, Manganese

Diamagnetic materials → Bismuth, Copper

Short Answer Questions

Question 1.

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

Answer:

Expression for the axial field of a solenoid :

- Consider a solinoid of length ‘2l’ and radius ‘a’ having ‘n’ turns per unit length.
- Let ‘I’ be the current in the solenoid.
- We have to calculate magnetic field at any point P on the axis of solenoid, where OP = r.

- Consider a small element of thickness dx of the solenoid, at a distance ‘x’ from ‘O’.
- Number of turns in the element = ndx.
- Magnitude of magnetic field at P due to this current element is dB = \(\frac{\mu_0 \mathrm{ndx} \cdot \mathrm{Ia}^2}{2\left[(\mathrm{r}-\mathrm{x})^2+\mathrm{a}^2\right]^{3 / 2}}\)
- If P lies at a very large distance from 0, i.e., r > > a and r > > x, then [(r – x)
^{2}+ a^{2}]^{3/2}= r^{3}.

⇒ dB = \(\frac{\mu_0 \mathrm{ndx} \mid \mathrm{a}^2}{2 \mathrm{r}^3}\) - To get total magnetic field, integrating the above equation between the limits from X = -l to X = +l.

- The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa
^{2}).

B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\) - Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 2.

The force between two magnet poles separated by a distance’d’ in air is ‘F. At what distance between them does the force become doubled ?

Answer:

Force between two magnetic poles, F_{1} = F;

Distance between two magnetic poles, d_{1} = d

Force between two magnetic poles increased by double F_{2} = 2F

Distance between two magnetic poles, d_{2} = ?

From Coulombs law, \(\mathrm{F}_1 \mathrm{~d}_1^2\) = \(\mathrm{F}_2 \mathrm{~d}_2^2\)

Fd^{2} = 2 F \(\mathrm{d}_2^2\)

⇒ \(\mathrm{d}_2^2\) = \(\frac{\mathrm{d}^2}{2}\)

∴ d_{2} = \(\frac{\mathrm{d}}{\sqrt{2}}\)

Question 3.

Compare the properties of para, dia and ferromagnetic substances.

Answer:

Diamagnetic substances

a) When these materials placed in a magnetic field, they are magnetised feebly in the opposite direction to the applied external field.

b) When a rod of diamagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the perpendicular direction to the magnetic field.

c) When they kept in a non uniform magnetic field, they moves from the region of greater field strength to the region of less field strength.

d) The relative permeability is less than 1. μ_{r} < 1 and negative.

e) The susceptibility \((\chi)\) is low and negative.

E.g. : Copper, Silver, Water, Gold, Antimony, Bismuth, Mercury Quartz Diamond etc.

Paramagnetic Substance

a) When these materials placed in a magnetic field, they are magnetised feebly in the direction of the applied magnetic field.

b) When a rod of paramagnetic material is suspended freely in a uniform magneticfield, it comes to rest in the direction of the applied magnetic field.

c) When they kept in a non-uniform magnetic field, they moves from the region of less field strength to the region of greater field strength.

d) The relative permeability is greater than 1. μ_{r} > 1 and positive.

e) The susceptibility \((\chi)\) is small and positive.

E.g. : Aluminium, Magnetiam, Tungsten, Platinum, Manganese, liquid oxygen, Ferric chloride, Cupric chloride etc.

Fêrromagnetic substances

a) When these materials placed in a magnetic field, they are magnetised strongly in the direction of the applied external field.

b) When a rod of ferromagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.

c) When they kept in a non-uniform magnetic field they moves from the regions of lesser (magnetic field) strength to the regions of stronger (magnetic field) strength.

d) The relative permeability is much greater than μ_{r} >> 1 and positive.

e) The stisceptibility \((\chi)\) is high and positive.

E.g. : Iron, Cobalt, Nickel, Gadolinium and their alloys.

Question 4.

Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.

Answer:

The magnetic field of the earth at a point on its surface can be specified by the declination D, the angle of dip or the inclination I and the horizontal component of the earth’s field H_{E}. These are known as the elements of the earth’s magnetic field.

Explanation:

- The total magnetic field at P can be resolved into a horizontal component H
_{E}and a vertical component Z_{E}. - The angle that B
_{E}makes with H_{E}is the angle of dip, I. - Representing the vertical component by Z
_{E}, we have

Z_{E}= B_{E}Sin I

H_{E}= B_{E}Cos I

Which gives Tan I = \(\frac{\mathrm{Z}_{\mathrm{E}}}{\mathrm{H}_{\mathrm{E}}}\)

Question 5.

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

Answer:

- Retentivity : The value of magnetic induction \(\overrightarrow{\mathrm{B}}\) left in the specimen, when the magnetising force (H) is reduced to zero is called Retentivity or Remanence or Residual magnetism.
- Coercivity : To reduce the retentivity to zero, we have to apply a magnetising force in opposite direction. This value of magnetising force is called coercivity or coercive force.
- Hysterisis eurve: The curve represents the relation between magnetic induction \(\overrightarrow{\mathrm{B}}\) (or) intensity of magnetization (I) of a ferromagnetic material with magnetizing force (or) magnetic intensity (\(\overrightarrow{\mathrm{H}}\)) is called Hystersis curve.

- Hysterisis curve for soft Iron and steel is shown below.

The hysterisis loops of soft iron and steel reveal that- The retentivity of soft iron is greater than the retentivity of steel.
- Soft Iron is more strongly magnetised than steel.
- Coercivity of Soft Iron is less than coercivity of steel. It means soft Iron loses its magnetisation more rapidly than steel does.
- As area of I – H loop for soft Iron is smaller than the area of I – H loop for steel. Therefore hysterisis loss in case of soft Iron is smaller than the hysterisis loss in case of steel.

Question 6.

If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns ?

Answer:

For first circular coil; B_{1} = B, n_{1} = 1; I_{1} = I; a_{1} = \(\frac{\mathrm{L}}{2 \pi}\)

For second circular coil, B_{2} = ? n_{2} = 10; I_{2} = I; a_{2} = \(\frac{\mathrm{L}}{2 \pi}\)

As B = \(\frac{\mu_0 \mathrm{n} I \mathrm{a}^2}{2 \mathrm{r}}\), B ∝ n

\(\frac{\mathrm{B}_2}{\mathrm{~B}_1}\) = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\)

\(\frac{\mathrm{B}_2}{\mathrm{~B}}\) = \(\frac{10}{1}\)

∴ B_{2} = 10 B

Question 7.

If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic field at the axis of the solenoid change?

Answer:

B_{1} = B (say); n_{1} = n; n_{2} = 2n; B_{2} =?

Magnetic field at the centre of a solinoid is given by B = \(\frac{\mu_0 \mathrm{nI} \mathrm{a}^2(2 l)}{2 \mathrm{r}^3}\)

⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}_1}\) = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\) ⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}}\) = \(\frac{2 \mathrm{n}}{\mathrm{n}}\)

∴ B_{2} = 2 B

Long Answer Questions

Question 1.

Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop.

Answer:

Expression for the magnetic field at a point on the axis of a current carrying circular loop:

- Consider ‘O’ is the centre of a circular coil of one turn and radius ‘a’.
- Let P is a point at a distance r from the centre, along the axis of coil.
- The plane of the coil is ⊥
^{r}to the plane of paper. - Consider two elements AB and AB’ each of length dl which are diametrically opposite.
- Then, the magnetic fields at P due to these two elements will be dB and dB in the direction PM and PN respectively.
- These directions are ⊥
^{r}to the lines joining the mid-points of the elements with the point P - Resolve these fields into two components parallel (dB sin θ) and perpendicular (dB cos θ) to the axis of the coil.
- The dB cos θ components cancel one another and dB sin θ components are in the same direction and add up due to the symmetric elements of the circular coil.
- Therefore, the total magnetic field along the axis = B = \(\int \mathrm{dB}\) sin θ of the circular coil along PC —– (1)
- The magnetic field at ‘P’ due to current element of length ‘dl’ is

‘dB’ = \(\frac{\mu_0}{4 \pi} \frac{I d l \sin \phi}{\left(a^2+x^2\right)}\) = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{a}^2+\mathrm{r}^2\right)}\) —– (II) [∵ φ = 90°] - From equation (I) and (II), B = \(\int \frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l}{\left(\mathrm{a}^2+\mathrm{r}^2\right)} \sin \theta\)

From Δ^{le}OPE, sin θ = \(\frac{a}{\sqrt{\left(a^2+r^2\right)}}\)

⇒ B = \(\int \frac{\mu_0}{4 \pi} \frac{\mathrm{Id} l \mathrm{a}}{\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}}\) = \(\frac{\mu_0 \mathrm{Ia}}{4 \pi\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}} \int \mathrm{d} l\)

But \(\int \mathrm{d} l\) = Circumference of the coil = 2πa

∴ B = \(\frac{\mu_0 \text { I a }}{4 \pi\left(a^2+r^2\right)^{3 / 2}}\) × 2πa = \(\frac{\mu_0 \mathrm{Ia}^2}{2\left(\mathrm{a}^2+\mathrm{r}^2\right)^{3 / 2}}\), Tesla along the direction of PC. - If the coil contains N turns, then B = \(\frac{\mu_0 \text { NIa }^2}{2\left(a^2+r^2\right)^{3 / 2}}\)
- At the centre of the coil r = 0, B = \(\frac{\mu_0 \mathrm{NI} \mathrm{a}^2}{2 \mathrm{a}^3}\) = \(\frac{\mu_0 \mathrm{NI}}{2 \mathrm{a}}\)

Note: If r >> a, B = \(\frac{\mu_0 \mathrm{NI}\left(\pi \mathrm{a}^2\right)}{2 \pi \mathrm{r}^3}\) = \(\frac{\mu_0 \mathrm{NI}_{\mathrm{A}}}{2 \pi^3}\)

A = πa^{2}= Area of current loop.

Question 2.

Prove that a bar magnet and a solenoid produce similar fields.

Answer:

Bar magnet produce similar field of Solenoid:

- We know that the current loop acts as a magnetic dipole. According to Ampere’s all magnetic phenomena can be explained interms of circulating currents.
- Cutting a bar magnet is like a solenoid. We get two similar solenoids with weaker magnetic properties.
- The magnetic field lines remain continuous, emerging from one face of solenoid and entering into other face of solenoid.
- 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 – - The magnetic field at point P due to bar magnet in the form of solenoid is B = \(\frac{\mu_0}{4 \pi} \frac{2 m}{r^3}\)
- The total magnetic field, at a point P due to solenoid is given by
- The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa
^{2}).

∴ B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\) - Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 3.

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

Answer:

Expression for time period of oscillation :

- A small compass needle (magnetic dipole) of known magnetic moment m and moment of Inertia is placing in uniform magnetic field B and allowing it to oscillate in the magnetic field.
- This arrangement is shown in Figure.
- The torque on the needle is \(\tau\) = m × B
- In magnitude \(\tau\) = mB sin θ.

Here \(\tau\) is restoring torque and θ is the angle between m and B. - Therefore, in equilibrium

Negative sign with mB sin θ implies that restoring torque is in opposition to deflecting torque.

- For small values of θ in radians, we approximate sin θ ≈ θ and get

This represents a simple harmonic motion. - From defination of simple harmonic motion, we have \(\) = -ω
^{2}θ ——– (II)

From equation (1) and (II), we get ⇒ ω^{2}= \(\frac{\mathrm{mB}}{\mathcal{J}}\)

- Therefore, the time period is

Question 4.

A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T_{1} and T_{2} at two places, where the angles of dip are θ_{1} and θ_{2} respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.

Answer:

- Suppose, the resultant magnetic fields is to be compared at two places A and B.
- A bar magnet, held horizontally at A and which is set into angular oscillatins in the Earths magnetic field.
- Let time period of a bar magnet at place ‘A’ is T
_{1}and angular displacement or angle of dip is θ_{1}. - As the bar magnet is free to rotate horizontally, it does not remain vertical component (B
_{1}sin θ_{1}) It can have only horizontal component (B_{1}cos θ_{1}) - The time period of a bar magnet in uniform magnetic field is given by T = \(2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mB}_{\mathrm{H}}}}\)
- Now, in this case T = T
_{1}and B_{H}= B_{1}cosθ_{1}

- Therefore time period of a bar magnet at place ‘A’ is given by

T_{1}= \(2 \pi \sqrt{\frac{I}{m B_1 \cos \theta_1}}\) —– (1) Where I is moment of Inertia of a barmagnet and m is magnitude of magnetic moment. - Similarly, the same bar magnet is placed at B and which is set into angular oscillations in the earths magnetic field.
- Let time period of a bar magnet at place B is T
_{2}and angle of dip is θ_{2}. - Since horizontal component of earths field at B is B
_{H}= B_{2}cos θ_{2}, time period,

T_{2}= \(2 \pi \sqrt{\frac{1}{\mathrm{mB}_2 \cos \theta_2}}\) —— (2) - Dividing equation (1) by equation (2), we get \(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\) = \(\sqrt{\frac{\mathrm{mB}_2 \cos \theta_2}{\mathrm{mB}_1 \cos \theta_1}}\)

Squaring on both sides, we have \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}\) = \(\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}\) - But B
_{1}= μ_{0}H_{1}and B_{2}= μ_{0}H_{2}

\(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}\) = \(\frac{\mu_0 \mathrm{H}_2 \cos \theta_2}{\mu_0 \mathrm{H}_1 \cos \theta_1}\) - Therefore, \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = \(\frac{\mathrm{T}_2^2 \cos \theta_2}{\mathrm{~T}_1^2 \cos \theta_1}\)

H2 T?cosOi . - By knowing T
_{1}, T_{2}and θ_{1}, θ_{2}at different places A and B, we can find the ratio of resultant magnetic fields.

Question 5.

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

Answer:

- 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.

- The susceptibility of a material represents its ability to get magnetism.
- Susceptibility is a dimension less quantity.
- Relation between μ
_{r}and \(\chi\):- Suppose that material is placed in a magnetic field of intensity H. Let I be the intensity of magnetisation acquired by it.
- Then the magnetic induction with in the material is

B = μ_{0}H + μ_{0}I ⇒ \(\frac{\mathrm{B}}{\mathrm{H}}\) = μ_{0}[1 + \(\frac{\mathrm{I}}{\mathrm{H}}\)]

⇒ μ = μ_{0}[1 + \(\chi\)] ⇒ \(\frac{\mu}{\mu_0}\) = 1 + \(\chi\)

μ_{r}= 1 + \(\chi\) [∵ μ_{r}= \(\frac{\mu}{\mu_0}\)]

- Negative susceptibility (y) of diamagnetic elements4are Bismuth (-1.66 × 10
^{-5}) and copper (-9.8 × 10^{-6}). - Positive susceptibility of Ferromagnetic elements are Aluminium (2.3 × 10
^{-5}) and oxygen at STP (2.1 × 10^{-6}). - Large and positive susceptibility of Ferromagnetic elements are Cobalt and Nickel.

Question 6.

Obtain Gauss Law for magnetism and explain it. (T.S. Mar. ’19)

Answer:

Gauss law for Magnetism:

- According to Gauss’s law for magnetism, the net magnetic flux (ϕ
_{B}) through any closed surface is always zero. - 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.
- Suppose a closed surface S is held in a uniform magnetic field B. Consider a small vector area element ΔS of this surface as shown in figure.
- Magnetic flux through this area element is defined as

Δϕ_{B}= B. ΔS. Then the net flux ϕ_{B}, is,

- If the area elements are really small, we can rewrite this equation as

- Comparing this equation with Gauss’s law of electrostatics i.e., electric flux through a closed surface S is given by

Where q is the electric charge enclosed by the surface. - 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. - The fact that ϕ
_{B}= 0 indicates that the simplest magnetic element is a dipole or current loop. - The isolated magnetic poles, called magnetic monopoles are not known to exist.
- All magnetic phenomena can be explained interms of an arrangement of magnetic dipoles and /or current loops.
- Thus corresponding to equation (II) of Gauss’s theorem in electrostatics, we can visualize equation (I) as

ϕ_{B}= \(\int_{S_{,}} B . d S\) = µ_{0}(m) + µ_{0}(-m) = 0 where m is strength of N-pole and -m is strength of S -pole of same magnet. - The net magnetic flux through any closed surface is zero.

Question 7.

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

Answer:

- Cycle of magnetisation : When a fen or magnetic specimen is slowly magnetised, the intensity of magnetisation varies with magnetic field through a cycle is called cycle of magnetisation.
- 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.
- Retentivity : The value of I for which H = 0 is called retentivity or residual magnetism.
- Coercivity: The value of magnetising force required to reduce I is zero in reverse direction of H is called coercive force or coercivity.
- 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.
- Explanation of hysterisis loop or curve :
- In fig, a closed curve ABCDEFA in H – I plane, called hysterisis loop is shown in fig.
- When ferromagnetic specimen is slowly magnetised, I increases with H.
- Part OA of the curve shows that I increases with H.

- At point A, the value I becomes constant is called saturation value.
- At B, I has some value while H is zero.
- In fig. BO represents retentivity and OC represents coercivity.

- Uses : The properties of hysterisis curve, i.e., saturation, retentivity, coercivity and hysterisis loss help us to choose the material for specific purpose.
- Permanent magnets : A permanent magnet should have both large retentivity and large coercivity. Permanent magnets are used in galvanometers, voltmeres, ammeters, etc.
- 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.
- Transformer cores, Dynamocore, Chokes, Telephone diaphragms: The core material should have high initial permeability, low hysterisis loss and high specific resistance to reduce eddy currents. Soft iron is the best suited material.

Problems

Question 1.

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

Answer:

For Rectangular LOOP PQRS :

Length PR = QS = l

breadth PQ = RS = b

Current = i

Magnetic field Induction = B

Angle made by normal to plane of coil with B = θ

Forces on conductor PR and SQ, F = Bil sin θ

Force on conductor PQ and RS, F = 0

Torque on a rectangular coil, T – F × ⊥^{r} distnace (b) ⇒ \(\tau\) – Bil sin θ (b)

∴ \(\tau\) = BiA sinθ[∵ A = l × b]

If the loop has n turns, then \(\tau\) = B in A sin θ.

Question 2.

A coil of 20 turns has an area of 800 mm^{2} 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 mm^{2} = 800 × 10^{-6} m^{2}; i = 0.5A; B = 0.3T; θ = 0°.

When the plane parallel to the field,

T = B in A cos θ = 0.3 × 0.5 × 20 × 800 × 10^{-6} × cos 0°

\(\tau\) = 2.4 × 10^{-3} Nm

Question 3.

In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression the magnetic moment (μ) of the electron in a Hydrogen atom in terms of its angular momentum L.

Answer:

Consider an electron of charge e, moves with constant speed v in a circular orbit of radius ‘r’ in Hydrogen atom as shown in fig.

Time period of orbiting electron,

The current constitute by revolving electron in circular motion around a nucleus, I = \(\frac{\mathrm{e}}{\mathrm{T}}\).

orbital magnetic moment, μ = IA = I (πr^{2})

⇒ μ = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\left(\pi \mathrm{r}^2\right)=\frac{\mathrm{evr}}{2}\)

μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}(\mathrm{mvr})\) [∵ Multiplying and dividing with ‘m’ on right side]

∴ μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}} \mathrm{~L}\) where L = mvr = angular momentum.

Question 4.

A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid ?

Answer:

l = 22.5 cm = 22.5 × 10^{-2} m = \(\frac{45}{2}\) × 10^{-2}m

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 × 10^{2} × 2

∴ H = 3200Am^{-1}

Question 5.

A bar magnet of length 0.1 m and with a magnetic moment of 5Am^{2} is placed in a uniform amagnetic field n of intensity 0.4T, with its axis making an angle of 60° with the field. What is the torque on the magnet ? (Mar. ’14)

Answer:

Given, 2l = 0.1m; m = 5A – m^{2}; B = OAT; θ = 60°.

Torque, T = mB sin θ = 5 × 0.4 × sin 60° = 2 × \(\frac{\sqrt{3}}{2}\)

∴ T = 1.732 N-m

Question 6.

If the Earth’s magnetic field at the equator is about 4 × 10^{5}T, What is its approximate magnetic dipole moment ?

Answer:

Given, B_{E} = 4 × 10^{-5} T; r = 6.4 × 10^{6}m; m = ?

B_{E} = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^3}\)

4 × 10^{-5} = \(\frac{4 \pi \times 10^{-7}}{4 \pi}\) × \(\frac{\mathrm{m}}{\left(6.4 \times 10^6\right)^3}\)

m = 4 × 10^{2} × (6.4 × 10^{6})^{3}

∴ m = 1.05 × 10^{23} Am^{2} ≈ 1 × 10^{23} Am^{2}

Question 7.

The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10^{-5}T 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^{-5}T;

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 8.

A solenoid, of insulated wire, is wound on a core with relative permeability 400. If the number of turns per metre is 1000 and the solenoid carries a current of 2A, calculate H, B and the magnetisation M.

Answer:

Given, µ_{r} = 400, I = 2A, n = 1000

H = nI = 1000 × 2 = 2 × 10^{3} A/m

B = µ_{r}µ_{0} H = 400 × 4π × 10^{-7} × 2 × 10^{3} = 1.0 T Magnetisation M = (µ_{r} – 1) H = (400 – 1)H = 399 × 2 × 10^{3}

∴ M \(\simeq\) 8 × 10^{5} A/m

Textual Exercises

Question 1.

Answer the following questions regarding earth’s magnetism :

a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

Answer:

The three independent quantities used to specify earth’s magnetic field are, Magnetic declination (θ), Magnetic dip (δ) and Horizontal component of earth’s field (H).

b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain ?

Answer:

Yes, we expect greater dip angle in Britian, because it is located close to North pole; δ = 70° in Britain.

c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground ?

Answer:

As Melbourne is situated in Southern hemisphere where north pole of earth’s magnetic field lies, therefore, magnetic lines of foece seem to come out of the ground.

d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north ro south pole ?

Answer:

At the poles, earth’s field is exactly vertical. As the compass needle is free to rotate in a horizontal plane only, it may point out in any direction.

e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10^{22} J T^{-1} located at its centre. Check the order of magnitude of this number in some way.

Answer:

Here, M = 8 × 10^{22} J T^{-1}.

Let us calculate magnetic field intensity at magnetic line of short mangetic dipole for which, d = R = radius of earth 6,400 km = 6.4 × 10^{6} m.

B = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}^3}\) = 10^{-7} × \(\frac{8 \times 10^{22}}{\left(6.4 \times 10^6\right)^3}\) = 0.31 × 10^{-4}T.

= 0.31 gauss

This value is in good approximation with observed values of earth’s magnetic field.

f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all ?

Answer:

The earth’s magnetic field is only approximately a dipole field. Therefore, local N-S poles may exist oriented in different directions. This is possible due to deposits of magnetised minerals.

Question 2.

Answer the following questions :

a) The earth’s magnetic field varies from point to point in space. Does it also change with time ? If so, on what time scale does it change appreciably ?

Answer:

Yes, earth’s field undergoes a change with time. For example, daily changes, annual changes secular changes with period of the order of 960 years and irregular changes like magnetic storms. Time scale for appreciable change is roughly a few hundred years.

b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why ?

Answer:

The earth’s core does contain iron but in the molten form only. This is not ferromagnetic and hence it cannot be treated as a source of earth’s magnetism.

c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents ?

Answer:

One of the possibilities is radioactivity in the interior of the earth. But it is not certain.

d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past ?

Answer:

Earth’s magnetic field gets recorded weakly in certain rocks during their solidification. An analysis of these rocks may reveal the history of earth’s magnetism.

e) The earth’s field departs from its dipole shape substantially at large, distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

Answer:

The earth’s magnetic field gets modified by the field produced by motion of ions in earth’s ionosphere.

f) Interstellar space has an extremely weak magnetic field of the order of 10^{-12} T. Can such a weak field be of any significant consequence? Explain.

[Note : Exercise 2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.)

Answer:

When a charged particle moves in a magnetic field, it is deflected along a circular path such that BeV = \(\frac{\mathrm{mV}^2}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{mV}}{\mathrm{Be}}\)

When B is low, r is high i.e., radius of curvature of path is very large. Therefore, over the gigantic inter stellar distance, the deflection of charged particles becomes less noticeable.

Question 3.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a troque of magnitude equal to 4.5 × 10^{-2} J. What is the magnitude of magnetic moment of the magnet?

Answer:

Here θ = 30°, B = 0.25 T, r = 4.5 × 10^{-2} J, M =?

As r = mB sin θ ∴ m = \(\frac{r}{B \sin \theta}\) = \(\frac{4.5 \times 10^{-2}}{0.25 \sin 30^{\circ}}\) = 0.36JT^{-1}

Question 4.

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

Answer:

Here m = 0.32JT^{-1}, B = 0.15T

- In stable equilibrium, the bar magnet is aligned along the magnetic field, i.e., θ = 0°.

Potentiâl Energy = -mB cos θ° = -032 × 0.15 × 0 = -4.8 × 10^{-2}J. - In unstable equilibrium the magnet is so oriented that magnetic moment is at 180° to the magnetic field i.e., θ = 180°.

Potential Energy = -mB cos 180° = – 0.32 × 0.15 (-1) = 4.8 × 10^{-2}J.

Question 5.

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^{-4} m^{2} 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} m^{2}, I = 3.0 A

A magnetic field develop along the axis of the solenoid. Therefore the current carrying solenoid behaves like a bar magnet m = N IA = 800 × 3.0 × 2.5 × 10^{-4}

= 0.6 JT^{-1} along the axis of solenoid.

Question 6.

If the solenoid in Exercise 5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field ?

Answer:

Here M = 0.6 JT^{-1} (from Question 5)

B = 0.25 T r = ? q = 30°

As r = m B sin θ’ ∴ r = 0.6 × 0.25 sin 30° = 0.075 N.m.

Question 7.

A bar magnet of magnetic moment 1.5 JT^{-1} lies aligned with the direction of a uniform magnetic field of 0.22T.

a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:

(i) normal to the field direction,

(ii) opposite to the field direction ?

Answer:

Here m = 1.5 JT^{-1}, B = 0.22 T, W = ?

a) Here θ_{1} = 0° (along the field)

θ_{2} = 90° (⊥ to the field)

As W = -mB (cos θ_{2} – cos θ_{1})

W =-1.5 × 0.22 (cos 90° – cos 0°) = -0.33 (0 – 1)

= 0.33 f

ii) Here θ_{1} = 0°, θ_{2} = 180°

W = -1.5 × 0.22 (cos 180° – cos 0°)

= – 0.33 (-1 – 1) = 0.66 J.

b) What is the torque on the magnet in cases (i) and (ii) ?

Answer:

Torque r = mB sin θ

i) Here θ = 90°, r = 1.5 × 0.22 sin 90° = 0.33 Nm

ii) Here θ = 180°, r = 1.5 × 0.22 sin 180° = 0

Question 8.

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^{-4} m^{2}, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

a) What is the magnetic moment associated with the solenoid ?

Answer:

N = 2000, A = 1.6 × 10^{-4} m^{2}, I = 4 amp, M = ?

As m = NIA

∴ M = 2000 × 4 × 1.6 × 10^{-4} = 1.28 JT^{-1}

b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^{-2} T is set up at an angle of 30° with the axis of the solenoid ?

Answer:

Net force on the solenoid = 0

Torque, r = m B sin θ = 1.28 × 7.5 × 10^{-2} sin 30°

= 1.28 × 7.5 × 10^{-2} × \(\frac{1}{2}\)

r = 4.8 × 10^{-2} Nm.

Question 9.

A circular coil of 76 turns and radius 10 cm. carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 8.0 × 10^{-2} T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^{-1}. What is the moment of inertia of the coil about its axis of rotation.

Answer:

Here n = 16, r = 10 cm = 0.1 m, I = 0.75A, B = 5.0 × 10^{-2}T

v = 2.0 s^{-1}, I = ?

Question 10.

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer:

Here θ = 22° 1 + 0.35 G, R = ?

As H = R cos θ

R = H/cos θ = \(\frac{0.35}{\cos 22^{\circ}}\) = \(\frac{0.35}{0.9272}\) = 0.38 G

Question 11.

At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Answer:

Here declination θ = 12° west, dip θ = 60°

H = 0.16 gauss = 0.16 × 10^{-4} tesla. R = ?

As H = R cos θ

R = \(\frac{\mathrm{H}}{\cos \theta}\) = \(\frac{0.16 \times 10^{-4}}{\cos 60^{\circ}}\)

R = \(\frac{0.16 \times 10^{-4}}{1 / 2}\) = 0.32 × 10^{-4}T

The earth’s field lies in a vertical plane 12° west of geographic meridian at an angle of 60° above the horizontal.

Question 12.

A short bar magnet has a magnetic moment of 0.48 J T^{-1}. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equitorial lines (normal bisector) of the magnet.

Answer:

Here M = 0.48 J T^{-1}, B = ?

d = 10 cm = 0.1 m

a) On the axis of the magnet

B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3}\) = 10^{-7} × \(\frac{2 \times 0.48}{(0.1)^3}\)

= 0.96 × 10^{T} along S – N direction

b) On the equitorial line of the magnet

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

Question 13.

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point- (i.e., 14 cm) from the centre of the magnet ? (At null distants, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

As null points are on the axis of the magnet, therefore

B_{1} = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^3}\) = H

On the equitorial line of magnet at same distance (d), field due to the magnet is

B_{2} = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}^3}\) = \(\frac{B_1}{2}\) = \(\frac{\mathrm{H}}{2}\)

∴ Total Magnetic field at this point on equitorial line is

B = B_{2} + H = H + \(\frac{\mathrm{H}}{2}\) = \(\frac{3}{2} \mathrm{H}\)

B = \(\frac{3}{2}\) × 0.36 = 0.54G

Question 14.

If the bar magnet in Exercise 13 is turned around by 180°, where will the new null points be located ?

Answer:

When the bar magnet is turned through 180°, neutral points would lie on equitorial line, so that

B_{2} = \(\frac{\mu_0}{4 \pi} \frac{M}{d_2^3}\) = H

On the equitorial line of magnet at same direction (d), field due to the magnet is

B_{2} = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}_2^3}\) = H

In the previous question

B_{1} = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}_1^3}\)

\(\mathrm{d}_2^3\) = \(\frac{d_1^3}{2}\) = \(\frac{(14)^3}{2}\)

d_{2} = \(\frac{14}{2^{1 / 3}}\) = 11.1 cm

Question 15.

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

Answer:

Here M = 5.25 × 10^{-2} J T^{-1} r = ?

Earth’s field \(\overrightarrow{\mathrm{B}}_{\mathrm{e}}\) = 0.42 G = 0.42 × 10^{-4} T

a) At a point P distant r on normal bisector, fig, field due to the magnet is

\(\overrightarrow{\mathrm{B}}_2\) = \(\frac{\mu_0}{4 \pi} \frac{M}{r^3}\) along PAI/NS.

The resultant field \(\overrightarrow{\mathrm{R}}\) will be inclined at 45° to the earth’s field along PQ only when

|\(\vec{B}_2\)| = |\(\overrightarrow{\mathrm{B}}_{\mathrm{e}}\)|

\(\frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3}\) = 0.42 × 10^{-4}

which gives, r = 0.05 m = 5 cm

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

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

|\(\overrightarrow{\mathrm{B}}_1\)| = |\(\overrightarrow{\mathrm{B}_{\mathrm{e}}}\)|

\(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{r}^3}\) = 0.42 × 10^{-4} which gives

r = 6.3 × 10^{-2} m = 6.3 cm

Additional Exercises

Question 1.

Answer the following Questions :

a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled ?

Answer:

This is because at lower temperatures, the tendency to disrupt the alignment of dipoles (due to magnetising field) decreases on account of reduced random thermal motion.

b) Why is diamagnetism, in contrast, almost independent of temperature ?

Answer:

In a diamagnetic sample, each molecule is not a magnetic dipole in itself. Therefore, random thermal motion of molecules does not affect the magnetism of the specimen. This is why diamagnetism is independent of temperature.

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

Answer:

As bismuth is diamagnetic, therefore, the field in the core will be slightly less than when the core is empty.

d) Is the permeability of a ferromagnetic material independent of the magnetic field ? If not, is it more for lower or higher fields ?

Answer:

No, permeability of a ferromagnetic material is not independent of magnetic field. As is clear from the hysteresis curve, μ is greater for lower fields.

e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why ?

Answer:

Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. The proof of this important fact is based on the boundary conditions of magnetic fields (B and H) at the interface of two media.

f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?

Answer:

The magnetic permeability of a ferromagnetic material μ > > 1. That is why the field lines meet this medium normally.

Question 2.

Answer the following questions :

a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

Answer:

Since, in a ferromagnetic substance the magnetic properties are due to alignment of domains, therefore on with drawing the magnetising field the original domain formation does not take place.

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

Answer:

Carbon steel piece, because heat lost per cycle is proportional to the area of the hysteresis loop.

c) A system displaying a hysterisis loop such as a ferromagnet, is a device for storing memory ?’ Explain the meaning of this statement.

Answer:

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

d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer ?

Answer:

Ceramics (specially treated barium iron oxides) also called ferrites.

e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Answer:

Surround the region by soft iron rings. Magnetic field lines will be drawn into the rings, and the enclosed space will be free of magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field.

Question 3.

A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable) ? (At neutral points, magnetic field due to a current – carrying cable is equal and opposite to the horizontal component of earth’s magnetic field).

Answer:

Here i = 2.5 amp

R = 0.33G = 0.33 × 10^{-4} T; θ = 0°

Horizontal component of earth’s field

H = R cos θ = 0.39 × 10^{-4} cos 35°

= 0.39 × 10^{-4} × 0.8192

= 3.9 × 10^{-5} tesla.

Vertical component of earth’s field.

H = R cos θ = 0.33 × 10^{-4} cos 0°

= 0.33 × 10^{-4} tesla.

Let the neutral points lie at a distance r from the cable

Strength of magnetic field on this line due to current in the cable = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

At neutral point,

\(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) = H

r = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{H}}\) = \(\frac{4 \pi \times 10^{-7} \times 2.5}{2 \pi \times 0.33 \times 10^{-4}}\) = 1.5 × 10^{-2}m

Hence neutral points lie on a straigt line parallel to the cable at a perpendicular distance of 1.5 cm above te plane of the paper.

Question 4.

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable ?

Answer:

There, no. of wires, n = 4, i = 1.0 amp

Earth’s field R = 0.39 G = 0.39 × 10^{-4} T

dip, θ = 35 declination θ = 0°

R_{1} = ?, R_{2} = ?

r = 4 cm each 4 × 10^{-2} m

Magnetic field at 4 cm due to currents in 4 wires

B = 4 × \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) = 4 × \(\frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 4 \times 10^{-2}}\)

Horizontal component of earth’s field

H = R cos θ = 0.39 × 10^{-4} cos 35°

= 0.39 × 10^{-4} × 0.8192 = 3.19 × 10^{-5}tesla

Vertical component of Earth’s field

V = R sin θ = 0.39 × 10^{-4} sin 35°

= 0.39 × 10^{-4} × 0.5736 = 2.2 × 10^{-5} tesla.

At point Q, 4 cm below the wire, horizontal component due to Earth’s field and field due to current are in opposite directions (fig.)

H_{1} = H – B

∴ H_{1} = 3.19 × 10^{-5} – 2 × 10^{-5}

= 1.19 × 10^{-5} tesla.

Hence R_{1} = \(\sqrt{\mathrm{H}_1^2+\mathrm{V}^2}\)

= \(\sqrt{\left(1.19 \times 10^{-5}\right)^2+\left(2.2 \times 10^{-5}\right)^2}\)

= 2.5 × 10^{-5} tesla.

At point P, 4 cm above the wire, horizontal component of Earth’s field and field due to current are in the same direction [fig.]

H_{2} = H + B = 3.19 × 10^{-5} + 2 × 10^{-5} = 5.19 × 10^{-5} T

R_{2} = \(\sqrt{\mathrm{H}_2^2+\mathrm{V}^2}\) = \(\sqrt{\left(5.19 \times 10^{-5}\right)^2+\left(2.2+10^{-5}\right)^2}\) = 5.54 × 10^{-5} tesla.

Question 5.

A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

a) Determine the horizontal component of the earth’s magnetic field at the location.

b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer:

a) Here n = 30, r = 12 cm = 12 × 10^{-2} m

i = 0.35 amp, H = ?

As is clear from fig. 11 the needle can point west to east only when H = B sin 45°

Where B = Magnetic field strength due to current in coil = \(\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{ni}}{\mathrm{r}}\)

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

Question 6.

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^{-2} T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field ?

Answer:

Here θ = 60°; B_{1} = 1.2 × 10^{-2} tesla

θ_{1} = 15°; θ_{2} = 60° – 15° = 45°

In equilibrium, torque due to two fields must balance i.e., r_{1} = r_{2}

MB_{1} sin θ_{1} = MB_{2} sin θ_{2}

B_{2} = \(\frac{B_1 \sin \theta_1}{\sin \theta_2}\) = \(\frac{1.2 \times 10^{-2} \sin 15^{\circ}}{\sin 45^{\circ}}\)

B_{2} = \(\frac{1.2 \times 10^{-2} \times 0.2588}{0.7071}\) = 4.4 × 10^{-3} tesla.

Question 7.

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_{e} = 9.11 × 10^{-31} kg).

[Note : Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field bn the motion of the electron beam from the electron gun to the screen in a TV set] .

Answer:

Here energy E = 18 KeV = 18 × 1.6 × 10^{-19}J

B = 0.40 G = 0.40 × 10^{-4} T

x = 30 cm = 0.3 m

As E = \(\frac{1}{2}\) mυ^{2} ∴ υ = \(\sqrt{\frac{2 E}{m}}\)

In a magnetic field electron beam is deflected along a circular arc of radius r, such that

BeV = \(\frac{m v^2}{r}\) or r = \(\frac{\mathrm{mv}}{\mathrm{Be}}\)

r = \(\frac{m}{B e} \sqrt{\frac{2 E}{m}}\) = \(\frac{1}{\mathrm{Be}} \sqrt{2 \mathrm{Em}}\) = 11.3 m

If y is the deflection at the end of the path it is clear from fig.

θ = \(\frac{\mathrm{x}}{\mathrm{r}}\) = \(\frac{\mathrm{y}}{\mathrm{x} / 2}\) = 2\(\frac{y}{x}\)

or y = \(\frac{x^2}{2 r}\) = \(\frac{0.30 \times(0.30)}{2 \times 11.3}\)m = 0.004m = 4mm

Question 8.

A sample of paramagnetic salt contains 2.0 × 10^{24} atomic dipoles each of dipole moment 1.5 × 10^{-23} J T^{-1}. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie’s law)

Answer:

Here no. of dipoles n = 2 × 10^{24}

Magnetic moment of each dipole M = 1.5 × 10^{-23} J T^{-1}.

Total dipole moment of sample = n × M = 2 × 10^{24} × 1.5 × 10^{-23} = 30

As saturation achieved is 15% therefore, effective dipole moment

M_{1} = \(\frac{15}{100}\) × 30 = 4.5 J T^{-1}; B_{1} = 0.64T, T_{1} = 4.2 k,

M_{2} = ?; B_{2} = 0.98 T,T_{2} = 2.8 k

Question 9.

A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Answer:

Here r = 15 cm = 15 × 10^{-2} m, N = 3500, M_{r} = 800

I = 1.2A, B =?

Number of turns, length, n = \(\frac{N}{2 \pi r}\) = \(\frac{3500}{2 \pi \times 15 \times 10^{-2}}\)

B = μ_{0}μ_{r}nI

= 4π × 10^{-7} × 800 × \(\frac{3500 \times 1.2}{2 \pi \times 15 \times 10^{-2}}\)

= 4.48T

Question 10.

The magnetic moment vectors μ_{s} and μ_{l} associated with the intrinsic spin angular momentum S and orbital angular momentum 1, respectively, of an electron are predicted

by quantum theory (and verified experimentally to a high accuracy) to be given by:

m_{s} = -(e/m) S,

μ_{l} = -(e/2m)l

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

Out of the two relations given only one is in accordance with classical physics. This is

\(\vec{\mu}_1\) = \(-\left(\frac{\mathrm{e}}{2 \mathrm{~m}}\right) \vec{l}\)

It follows from the definitions of μ_{1} and l.

μ = iA = \(\left(\frac{-\mathrm{e}}{\mathrm{T}}\right) \pi \mathrm{r}^2\)

l = mvr = \(\mathrm{m}\left(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\right) \mathrm{r}\)

Where r is the radius of the circular orbit, which the electron of mass m and charge (-e) completes in time T.

Divide (i) by (ii) \(\frac{\mu_1}{l}\) = \(\frac{-\mathrm{e}}{\mathrm{T}} \pi \mathrm{r}^2 \times \frac{\mathrm{T}}{\mathrm{m}^2 \pi \mathrm{r}^2}=\frac{-\mathrm{e}}{2 \mathrm{~m}}\)

\(\vec{\mu}_1\) = \(\left(\frac{-\mathrm{e}}{2 \mathrm{~m}}\right) \vec{l}\)

‘Clearly \(\vec{\mu}_1\) and \(\vec{l}\) will be antiparallel (both being normal to the plane of the orbit)

In contrast \(\frac{\mu_{\mathrm{S}}}{\mathrm{S}}\) = \(\frac{\mathrm{e}}{\mathrm{m}}\). It is obtained on the basis of quantum mechanics.