Chapter 5 Magnetism and Matter Class 12 Physics

Question 5.1

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

Formula

Torque $\tau = m B \sin\theta$, where $m$ is magnetic moment.

Calculation

$$m = \frac{\tau}{B \sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5}$$ $$m = \frac{4.5 \times 10^{-2}}{0.125} = 0.36 \text{ J/T}$$

Magnetic Moment $m = 0.36 \text{ J/T}$

Question 5.2

A short bar magnet of magnetic moment $0.32 \text{ J/T}$ is placed in a uniform magnetic field of 0.15 T. 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?

Concept

Potential Energy $U = -mB \cos\theta$.

(a) Stable Equilibrium

$\theta = 0^\circ$ (Moment parallel to Field).

$$U = -mB \cos 0^\circ = -0.32 \times 0.15 \times 1 = -4.8 \times 10^{-2} \text{ J}$$

(b) Unstable Equilibrium

$\theta = 180^\circ$ (Moment antiparallel to Field).

$$U = -mB \cos 180^\circ = -0.32 \times 0.15 \times (-1) = +4.8 \times 10^{-2} \text{ J}$$

Question 5.3

A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4} \text{ m}^2$ carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Explanation

A current-carrying solenoid behaves like a bar magnet because its magnetic field lines resemble those of a bar magnet. One end acts as the North Pole and the other as the South Pole, depending on the direction of current flow (Right-Hand Grip Rule).

Calculation

$$m = N I A = 800 \times 3.0 \times 2.5 \times 10^{-4}$$ $$m = 2400 \times 2.5 \times 10^{-4} = 6000 \times 10^{-4} = 0.6 \text{ J/T}$$

Magnetic Moment = 0.6 J/T along the axis.

Question 5.4

If the solenoid in Exercise 5.3 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 $30^\circ$ with the direction of the applied field?

Calculation

Given $m = 0.6 \text{ J/T}$ (from Q5.3), $B = 0.25 \text{ T}$, $\theta = 30^\circ$.

$$\tau = m B \sin\theta = 0.6 \times 0.25 \times \sin 30^\circ$$ $$\tau = 0.15 \times 0.5 = 0.075 \text{ N m}$$

Result: $7.5 \times 10^{-2} \text{ N m}$

Question 5.5

A bar magnet of magnetic moment $1.5 \text{ J/T}$ lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?
(b) What is the torque on the magnet in cases (i) normal to field, (ii) opposite to field?

(a) Work Done

Initial angle $\theta_1 = 0^\circ$. Final angle $\theta_2 = 90^\circ$.

$$W = -mB(\cos\theta_2 – \cos\theta_1) = -1.5 \times 0.22 \times (\cos 90^\circ – \cos 0^\circ)$$ $$W = -0.33 \times (0 – 1) = 0.33 \text{ J}$$

(b) Torque

(i) Normal ($\theta = 90^\circ$): $\tau = mB \sin 90^\circ = 1.5 \times 0.22 \times 1 = 0.33 \text{ N m}$.

(ii) Opposite ($\theta = 180^\circ$): $\tau = mB \sin 180^\circ = 1.5 \times 0.22 \times 0 = 0$.

Question 5.6

A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4} \text{ m}^2$, 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?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2} \text{ T}$ is set up at an angle of $30^\circ$ with the axis of the solenoid?

(a) Magnetic Moment

$$m = N I A = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 1.28 \text{ J/T}$$

(b) Force and Torque

Force: Since the field is uniform, the net magnetic force on a dipole (solenoid) is Zero.

Torque:

$$\tau = m B \sin\theta = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^\circ$$ $$\tau = 9.6 \times 10^{-2} \times 0.5 = 4.8 \times 10^{-2} \text{ N m}$$

Question 5.7

A short bar magnet has a magnetic moment of $0.48 \text{ J/T}$. 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 equatorial lines (normal bisector) of the magnet.

Concept

$d = 10 \text{ cm} = 0.1 \text{ m}$. Since it is a short magnet ($l \ll d$), we use simplified formulas.

[Image of Magnetic field lines due to a bar magnet]

(a) Axial Point

Direction: Along the magnetic moment (S to N).

$$B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} = \frac{10^{-7} \times 2 \times 0.48}{(0.1)^3} = \frac{0.96 \times 10^{-7}}{10^{-3}} = 0.96 \times 10^{-4} \text{ T}$$

(b) Equatorial Point

Direction: Opposite to magnetic moment (N to S).

$$B_{eq} = \frac{\mu_0}{4\pi} \frac{m}{d^3} = \frac{1}{2} B_{axial} = 0.48 \times 10^{-4} \text{ T}$$

Question 5.8

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. 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 of 14 cm from the centre?

1. Finding Magnetic Moment

At null points on axis, field of magnet $B_{axial}$ balances Earth’s horizontal component $H_E$.

$$B_{axial} = H_E = 0.36 \text{ G}$$

2. Field on Normal Bisector

At the same distance $d$ on equatorial line, the magnet’s field is half the axial field.

$$B_{eq} = \frac{1}{2} B_{axial} = \frac{0.36}{2} = 0.18 \text{ G}$$

At this point, Earth’s field $H_E$ and magnet’s field $B_{eq}$ are in the same direction.

Total Field $B = B_{eq} + H_E = 0.18 + 0.36 = 0.54 \text{ G}$.

Result: 0.54 Gauss

Question 5.9

If the bar magnet in exercise 5.8 is turned around by $180^\circ$, where will the new null points be located?

Analysis

Initially, null points were on the axis (Axial Field = Earth’s Field). Turning $180^\circ$ reverses the magnet’s field direction. Now, null points will shift to the equatorial line.

Calculation

At new null points (distance $d’$), $B_{eq} = H_E$.

$$\frac{\mu_0}{4\pi} \frac{m}{d’^3} = H_E$$

We know from Q5.8: $H_E = \frac{\mu_0}{4\pi} \frac{2m}{d^3}$ (where $d=14$).

$$\frac{\mu_0}{4\pi} \frac{m}{d’^3} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} \implies \frac{1}{d’^3} = \frac{2}{d^3}$$ $$d’^3 = \frac{d^3}{2} \implies d’ = \frac{d}{\sqrt[3]{2}} = \frac{14}{1.26} \approx 11.1 \text{ cm}$$

Null points on equatorial line at 11.1 cm.

Question 5.10

A short bar magnet of magnetic moment $5.25 \times 10^{-2} \text{ J/T}$ 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 $45^\circ$ with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field is 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Condition for 45 Degrees

Resultant is at $45^\circ$ when the Magnet’s Field $B$ equals Earth’s Field $H_E = 0.42 \times 10^{-4} \text{ T}$.

(a) Normal Bisector

$$B_{eq} = \frac{\mu_0 m}{4\pi d^3} = H_E$$ $$d^3 = \frac{10^{-7} \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}} = \frac{5.25 \times 10^{-9}}{0.42 \times 10^{-4}} = 12.5 \times 10^{-5} = 125 \times 10^{-6}$$ $$d = \sqrt[3]{125} \times 10^{-2} = 5 \times 10^{-2} \text{ m} = 5 \text{ cm}$$

(b) Axis

$$B_{axial} = \frac{\mu_0 2m}{4\pi d^3} = H_E$$ $$d^3 = 2 \times (\text{Value from part a}) = 250 \times 10^{-6}$$ $$d = \sqrt[3]{250} \times 10^{-2} \approx 6.3 \times 10^{-2} \text{ m} = 6.3 \text{ cm}$$

Question 5.11

Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost 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?

(a) Cooling reduces random thermal motion, allowing magnetic dipoles to align better with the external field, increasing magnetisation.
(b) Diamagnetism arises from intrinsic electron orbital motion changes induced by the external field. This internal atomic mechanism is not affected by thermal agitation.
(c) Slightly less. Bismuth is diamagnetic ($\chi < 0$, $\mu_r < 1$), so it reduces the magnetic field inside the core.

Question 5.12

Answer the following questions:
(a) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(b) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. Why?

(a) No, permeability $\mu$ depends on $B$ (hysteresis). It is generally higher for lower fields and decreases at saturation.
(b) Ferromagnets have very high permeability ($\mu_r \gg 1$). Magnetic field lines tend to concentrate inside them and enter/exit normally to minimize reluctance, similar to how electric field lines are normal to conductors.

Question 5.13

Which material is better for making a Permanent Magnet and which for an Electromagnet? (Soft Iron vs Steel). Give reasons.

Permanent Magnet

Steel is better. It has high Retentivity (retains magnetism) and high Coercivity (hard to demagnetize).

Electromagnet

Soft Iron is better. It has high Permeability (easily magnetized), low Retentivity (loses magnetism quickly when current stops), and low Hysteresis Loss (energy efficient).

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