## Magnetism And Matter

### Exercises

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

## Show Answer

**Answer**

Magnetic field strength, $B=0.25 \mathrm{~T}$

Torque on the bar magnet, $T=4.5 \times 10^{-2} \mathrm{~J}$

Angle between the bar magnet and the external magnetic field, $\theta=30^{\circ}$

Torque is related to magnetic moment $(M)$ as:

$T=M B \sin \theta$

$\therefore M=\frac{T}{B \sin \theta}$

$$ =\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 \mathrm{~J} \mathrm{~T}^{-1} $$

Hence, the magnetic moment of the magnet is $0.36 \mathrm{~J} \mathrm{~T}^{-1}$.

**5.2** A short bar magnet of magnetic moment $\mathrm{m}=0.32 \mathrm{JT}^{-1}$ is placed in a uniform magnetic field of $0.15 \mathrm{~T}$. 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?

## Show Answer

**Answer**

Moment of the bar magnet, $M=0.32 \mathrm{~J} \mathrm{~T}^{-1}$

External magnetic field, $B=0.15 \mathrm{~T}$

(a)The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $\theta$, between the bar magnet and the magnetic field is $0^{\circ}$.

Potential energy of the system $=-M B \cos \theta$

$=-0.32 \times 0.15 \cos 0^{\circ}$

$=-4.8 \times 10^{-2} \mathrm{~J}$

(b)The bar magnet is oriented $180^{\circ}$ to the magnetic field. Hence, it is in unstable equilibrium.

$\theta=180^{\circ}$

Potential energy $=-M B \cos \theta$

$=-0.32 \times 0.15 \cos 180^{\circ}$

$=4.8 \times 10^{-2} \mathrm{~J}$

**5.3** A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4} \mathrm{~m}^{2}$ 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?

## Show Answer

**Answer**

Number of turns in the solenoid, $n=800$

Area of cross-section, $A=2.5 \times 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $I=3.0 \mathrm{~A}$

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

$M=n I A$ $=800 \times 3 \times 2.5 \times 10^{-4}$

$=0.6 \mathrm{~J} \mathrm{~T}^{-1}$

**5.4** If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25 \mathrm{~T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of applied field?

## Show Answer

**Answer**

Magnetic field strength, $B=0.25 \mathrm{~T}$

Magnetic moment, $M=0.6 \mathrm{~T}^{-1}$

The angle $\theta$, between the axis of the solenoid and the direction of the applied field is $30^{\circ}$.

Therefore, the torque acting on the solenoid is given as:

$$ \begin{aligned} \tau & =M B \sin \theta \\ & =0.6 \times 0.25 \sin 30^{\circ} \\ & =7.5 \times 10^{-2} \mathrm{~J} \end{aligned} $$

**5.5** A bar magnet of magnetic moment $1.5 \mathrm{~J} \mathrm{~T}^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \mathrm{~T}$.

(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?

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

## Show Answer

**Answer**

(a)Magnetic moment, $M=1.5 \mathrm{~J} \mathrm{~T}^{-1}$

Magnetic field strength, $B=0.22 \mathrm{~T}$

(i)Initial angle between the axis and the magnetic field, $\theta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $\theta_{2}=90^{\circ}$

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

$$ \begin{aligned} W & =-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\ & =-1.5 \times 0.22\left(\cos 90^{\circ}-\cos 0^{\circ}\right) \\ & =-0.33(0-1) \\ & =0.33 \mathrm{~J} \end{aligned} $$

(ii) Initial angle between the axis and the magnetic field, $\theta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $\theta_{2}=180^{\circ}$

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

$$ \begin{aligned} W & =-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\ & =-1.5 \times 0.22\left(\cos 180-\cos 0^{\circ}\right) \\ & =-0.33(-1-1) \\ & =0.66 \mathrm{~J} \end{aligned} $$

(b) For case (i): $\theta=\theta_{2}=90^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=1.5 \times 0.22 \sin 90^{\circ}$

$=0.33 \mathrm{~J}$

$\underline{\text { For case (ii): }} \theta=\theta_{2}=180^{\circ}$

$\therefore$ Torque, $\tau=M B \sin \theta$

$=M B \sin 180^{\circ}=0 \mathrm{~J}$

**5.6** A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4} \mathrm{~m}^{2}$, carrying a current of $4.0 \mathrm{~A}$, 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 $7.5 \times 10^{-2} \mathrm{~T}$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?

## Show Answer

**Answer**

Number of turns on the solenoid, $n=2000$

Area of cross-section of the solenoid, $A=1.6 \times 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $I=4 \mathrm{~A}$

(a)The magnetic moment along the axis of the solenoid is calculated as:

$M=n A I$

$=2000 \times 1.6 \times 10^{-4} \times 4$

$=1.28 \mathrm{Am}^{2}$

(b)Magnetic field, $B=7.5 \times 10^{-2} \mathrm{~T}$

Angle between the magnetic field and the axis of the solenoid, $\theta=30^{\circ}$

Torque, $\tau=M B \sin \theta$

$=1.28 \times 7.5 \times 10^{-2} \sin 30^{\circ}$

$=4.8 \times 10^{-2} \mathrm{Nm}$

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is $4.8 \times 10^{-2} \mathrm{Nm}$.

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

## Show Answer

**Answer**

Magnetic moment of the bar magnet, $M=0.48 \mathrm{~J} \mathrm{~T}^{-1}$

Distance, $d=10 \mathrm{~cm}=0.1 \mathrm{~m}$

The magnetic field at distance $d$, from the centre of the magnet on the axis is given by the relation:

$$ B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}} $$

Where,

$$ \begin{aligned} & \mu_{0}=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1} \\ & \therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}} \\ & \quad=0.96 \times 10^{-4} \mathrm{~T}=0.96 \mathrm{G} \end{aligned} $$

The magnetic field is along the $\mathrm{S}-\mathrm{N}$ direction.

The magnetic field at a distance of $10 \mathrm{~cm}$ (i.e., $d=0.1 \mathrm{~m}$ ) on the equatorial line of the magnet is given as:

$$ \begin{aligned} B & =\frac{\mu_{0} \times M}{4 \pi \times d^{3}} \\ & =\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}} \\ & =0.48 \mathrm{G} \end{aligned} $$

The magnetic field is along the $\mathrm{N}-\mathrm{S}$ direction.

The new null points will be located $11.1 \mathrm{~cm}$ on the normal bisector.