CBSE · NCERT · Class 12 Physics · Chapter 5

NCERT Solutions: Class 12 Physics Chapter 5 - Magnetism and Matter

6 textbook Q&A6 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Magnetism and Matter, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercises 6
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1Exercises6 questions
Q.5.1A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?v
Solution

The torque on a magnetic dipole is $\tau=mB\sin\theta$. Hence $m=\tau/(B\sin\theta)=4.5\times10^{-2}/[0.25\sin30^\circ]=0.36\,\text{J T}^{-1}$.

Answer:

$0.36\,\text{J T}^{-1}$.

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Q.5.2A short bar magnet of magnetic moment m = 0.32 JT–1 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?v
Solution

Potential energy of a magnetic dipole in a field is $U=-\mathbf{m}\cdot\mathbf{B}=-mB\cos\theta$. For stable equilibrium, $\theta=0^\circ$, so $U=-mB=-(0.32)(0.15)=-4.8\times10^{-2}\,\text{J}$. For unstable equilibrium, $\theta=180^\circ$, so $U=+mB=+4.8\times10^{-2}\,\text{J}$.

Answer:

(a) Stable equilibrium occurs when $\mathbf{m}$ is parallel to $\mathbf{B}$; $U=-4.8\times10^{-2}\,\text{J}$. (b) Unstable equilibrium occurs when $\mathbf{m}$ is antiparallel to $\mathbf{B}$; $U=+4.8\times10^{-2}\,\text{J}$.

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Q.5.3A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2 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?v
Solution

A current-carrying solenoid produces a magnetic field similar to that of a bar magnet: one end behaves as a north pole and the other as a south pole, with field lines forming closed loops. The magnetic moment is $m=NIA=800(3.0)(2.5\times10^{-4})=0.60\,\text{A m}^2$.

Answer:

The solenoid behaves like a bar magnet with north and south poles at its ends; its magnetic moment is $0.60\,\text{A m}^2$.

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Q.5.5A bar magnet of magnetic moment 1.5 J T–1 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: (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)?v
Solution

Potential energy is $U=-mB\cos\theta$. Initially, $\theta=0^\circ$, so $U_i=-mB=-(1.5)(0.22)=-0.33\,\text{J}$. For normal alignment, $\theta=90^\circ$, $U=0$, so work required is $0-(-0.33)=0.33\,\text{J}$. For opposite alignment, $\theta=180^\circ$, $U=+0.33\,\text{J}$, so work required is $0.33-(-0.33)=0.66\,\text{J}$. Torque magnitude is $mB\sin\theta$; it is $0.33\,\text{N m}$ at $90^\circ$ and zero at $180^\circ$.

Answer:

(a) Work required is $0.33\,\text{J}$ for normal alignment and $0.66\,\text{J}$ for opposite alignment. (b) Torque is $0.33\,\text{N m}$ in case (i) and zero in case (ii).

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Q.5.6A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, 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 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?v
Solution

Magnetic moment is $m=NIA=2000(4.0)(1.6\times10^{-4})=1.28\,\text{A m}^2$. A magnetic dipole in a uniform magnetic field experiences no net force. The torque is $\tau=mB\sin\theta=1.28(7.5\times10^{-2})\sin30^\circ=4.8\times10^{-2}\,\text{N m}$.

Answer:

(a) $1.28\,\text{A m}^2$. (b) Net force is zero; torque is $4.8\times10^{-2}\,\text{N m}$.

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Q.5.7A 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 equatorial lines (normal bisector) of the magnet.v
Solution

For a short magnet, axial field is $B_a=(\mu_0/4\pi)(2m/r^3)$ and equatorial field has magnitude $B_e=(\mu_0/4\pi)(m/r^3)$, directed opposite to $\mathbf{m}$. With $m=0.48\,\text{J T}^{-1}$ and $r=0.10\,\text{m}$, $B_a=10^{-7}(0.96)/(10^{-3})=9.6\times10^{-5}\,\text{T}$ and $B_e=10^{-7}(0.48)/(10^{-3})=4.8\times10^{-5}\,\text{T}$.

Answer:

(a) $9.6\times10^{-5}\,\text{T}$ along the magnetic moment on the axis. (b) $4.8\times10^{-5}\,\text{T}$ opposite to the magnetic moment on the equatorial line.

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