Class 12th Physics Chapter-5 Solutions

Question 5.1

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

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Solution

The torque on a magnetic dipole in a uniform magnetic field is given by:

$$\tau =mB\sin \theta$$Given:

• $\tau =4.5\times {10}^{-2}\text{ J}$

• $B=0.25\text{ T}$

• $\theta ={30}^{\circ }$, so $\sin {30}^{\circ }=\frac{1}{2}$

$$m=\frac{\tau }{B\sin \theta }=\frac{4.5\times {10}^{-2}}{0.25\times \frac{1}{2}}=\frac{4.5\times {10}^{-2}}{0.125}=0.36$$

Answer:

$$\boxed{{\displaystyle m=0.36{\text{J T}}^{-1}}}\text{or}\boxed{{\displaystyle 0.36{\text{A m}}^2}}$$

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Question 5.2

A short bar magnet of magnetic moment

$$m=0.32{\text{J T}}^{-1}$$

is placed in a uniform magnetic field

$$B=0.15\text{ T}\mathrm{.}$$

If the bar is free to rotate in the plane of the field:

(a) Which orientation corresponds to stable equilibrium?
(b) Which orientation corresponds to unstable equilibrium?
What is the potential energy of the magnet in each case?

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Concept Used

The magnetic potential energy of a magnetic dipole in a uniform magnetic field is:

$$U=-\vec{m}\cdot \vec{B}=-mB\cos \theta$$

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.

(a) Stable Equilibrium

• Stable equilibrium occurs when potential energy is minimum

• This happens when:

  $$\theta =0^{\circ }(\vec{m}\parallel \vec{B})$$

Potential Energy:

$$U_{\text{stable}}=-mB\cos 0^{\circ }=-mB$$

$$U_{\text{stable}}=-(0.32)(0.15)=-0.048\text{ J}$$

(b) Unstable Equilibrium

• Unstable equilibrium occurs when potential energy is maximum

• This happens when:

  $$\theta ={180}^{\circ }(\vec{m}\text{ antiparallel to }\vec{B})$$

Potential Energy:

$$U_{\text{unstable}}=-mB\cos {180}^{\circ }=+mB$$

$$U_{\text{unstable}}=+(0.32)(0.15)=+0.048\text{ J}$$

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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.

(a) Explain the sense in which the solenoid acts like a bar magnet.
(b) What is its associated magnetic moment?

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(a) Solenoid as a Bar Magnet

A current-carrying solenoid produces a magnetic field pattern similar to that of a bar magnet:

• One end of the solenoid behaves like a north pole

• The other end behaves like a south pole

• Magnetic field lines emerge from one end and enter the other, forming closed loops

• The polarity of the solenoid is given by the right-hand thumb rule:

  • Curl fingers in the direction of current → thumb gives the direction of the north pole

Hence, a solenoid acts like a magnetic dipole, just like a bar magnet.

(b) Magnetic Moment of the Solenoid

The magnetic moment of a solenoid is:

$$m=NIA$$

Given:

• $N=800$
  

• $I=3.0\text{ A}$

• $A=2.5\times {10}^{-4}{\text{m}}^2$

Calculation:

$$m=800\times 3.0\times 2.5\times {10}^{-4}$$

$$m=6000\times {10}^{-4}=0.6{\text{A m}}^2$$

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Question 5.4

If the solenoid in the previous exercise is free to turn about the vertical direction and a uniform horizontal magnetic field of

$$B=0.25\text{ T}$$

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?

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Given (from Exercise 5.3)

• Magnetic moment of the solenoid:

$$m=0.6{\text{J T}}^{-1}$$

• Magnetic field:

$$B=0.25\text{ T}$$

• Angle:

$$\theta ={30}^{\circ },\sin {30}^{\circ }=\frac{1}{2}$$

Formula Used

Torque on a magnetic dipole in a uniform magnetic field:

$$\tau =mB\sin \theta$$

Calculation

$$\tau =0.6\times 0.25\times \frac{1}{2}$$

$$\tau =0.075\text{ N m}$$

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Question 5.5

A bar magnet of magnetic moment

$$m=1.5{\text{J T}}^{-1}$$

lies aligned with the direction of a uniform magnetic field

$$B=0.22\text{ T}\mathrm{.}$$

(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)?

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Concepts Used

• Magnetic potential energy:

$$U=-mB\cos \theta$$

• Work done by external agent:

$$W=\mathrm{\Delta }U$$

• Torque on a magnetic dipole:

$$\tau =mB\sin \theta$$

Given Initial Condition

Initially, the magnet is aligned with the field:

$${\theta }_i=0^{\circ }$$

Initial potential energy:

$$U_i=-mB=-(1.5)(0.22)=-0.33\text{ J}$$

(a) Work Done by External Torque

(i) Magnet turned normal to the field

$$\theta ={90}^{\circ },\cos {90}^{\circ }=0$$

Final potential energy:

$$U_f=0$$

Work done:

$$W=U_f-U_i=0-(-0.33)=0.33\text{ J}$$

(ii) Magnet turned opposite to the field

$$\theta ={180}^{\circ },\cos {180}^{\circ }=-1$$

Final potential energy:

$$U_f=+mB=+0.33\text{ J}$$

Work done:

$$W=U_f-U_i=0.33-(-0.33)=0.66\text{ J}$$

(b) Torque on the Magnet

(i) At ${90}^{\circ }$:

$$\tau =mB\sin {90}^{\circ }=(1.5)(0.22)=0.33\text{ N m}$$

(ii) At ${180}^{\circ }$:

$$\tau =mB\sin {180}^{\circ }=0$$

Final Answer (Summary)

(a) Work Done

• Normal to the field (90°):

$$\boxed{{\displaystyle W=0.33\text{ J}}}$$

• Opposite to the field (180°):

$$\boxed{{\displaystyle W=0.66\text{ J}}}$$

(b) Torque

• At 90°:

$$\boxed{{\displaystyle \tau =0.33\text{ N m}}}$$

• At 180°:

$$\boxed{{\displaystyle \tau =0}}$$

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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?

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(a) Magnetic moment of the solenoid

For a solenoid:

$$m=NIA$$

Given:

• $N=2000$

• $I=4.0\text{ A}$

• $A=1.6\times {10}^{-4}{\text{m}}^2$

Calculation:

$$m=2000\times 4.0\times 1.6\times {10}^{-4}$$

$$m=1.28{\text{A m}}^2$$

$$\boxed{{\displaystyle m=1.28{\text{J T}}^{-1}}}$$

(b) Force and torque on the solenoid

Force on the solenoid

In a uniform magnetic field, the net force on a magnetic dipole is zero.

$$\boxed{{\displaystyle F=0}}$$

Torque on the solenoid

Torque on a magnetic dipole:

$$\tau =mB\sin \theta$$

Given:

• $m=1.28{\text{J T}}^{-1}$

• $B=7.5\times {10}^{-2}\text{ T}$

• $\theta ={30}^{\circ },\sin {30}^{\circ }=\frac{1}{2}$

Calculation:

$$\tau =1.28\times 7.5\times {10}^{-2}\times \frac{1}{2}$$

$$\tau =0.048\text{ N m}$$

Final Answer (Summary)

(a) Magnetic moment of the solenoid:

$$\boxed{{\displaystyle m=1.28{\text{J T}}^{-1}}}$$

(b)

• Force on the solenoid:

$$\boxed{{\displaystyle F=0}}$$

• Torque on the solenoid:

$$\boxed{{\displaystyle \tau =4.8\times {10}^{-2}\text{ N m}}}$$

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Question 5.7

A short bar magnet has a magnetic moment

$$m=0.48{\text{J T}}^{-1}$$

Find the direction and magnitude of the magnetic field produced by the magnet at a distance of

$$r=10\text{ cm}=0.10\text{ m}$$

from the centre of the magnet on:

(a) the axial line,
(b) the equatorial line (normal bisector) of the magnet.

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Given

• Magnetic moment: $m=0.48{\text{J T}}^{-1}$
  

• Distance: $r=0.10\text{ m}$

• Permeability of free space:

$$\frac{{\mu }_0}{4\pi }={10}^{-7}{\text{T m A}}^{-1}$$

(a) Magnetic field on the axis of the magnet

Formula:

$$B_{\text{axis}}=\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}$$

Calculation:

$$B_{\text{axis}}={10}^{-7}\times \frac{2\times 0.48}{(0.10{)}^3}$$

$$B_{\text{axis}}={10}^{-7}\times \frac{0.96}{{10}^{-3}}$$

$$B_{\text{axis}}=9.6\times {10}^{-5}\text{ T}$$

Direction:

Along the axis of the magnet, in the same direction as the magnetic moment (from south pole to north pole outside the magnet).

(b) Magnetic field on the equatorial line (normal bisector)

Formula:

$$B_{\text{equatorial}}=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}$$

Calculation:

$$B_{\text{equatorial}}={10}^{-7}\times \frac{0.48}{(0.10{)}^3}$$

$$B_{\text{equatorial}}=4.8\times {10}^{-5}\text{ T}$$

Direction:

Along the equatorial line, opposite to the direction of the magnetic moment.

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Final Answer (Summary)