Exercises — Magnetism and Matter

1. 12 marksNCERT Exercises, Chapter 5

   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 $4.5\times 10^{-2}$ J. What is the magnitude of the magnetic moment of the magnet?

   Hint. Rearrange the torque formula for a magnetic dipole to make m the subject.

   A magnetic dipole in a uniform field experiences a torque $\tau =mB\sin \theta$, exactly parallel to the electric dipole case.

   Rearranging for the magnetic moment:

   $m=\frac{\tau }{B\sin \theta }=\frac{4.5\times 10^{-2}}{0.25\times \sin 30^{\circ }}=\frac{4.5\times 10^{-2}}{0.125}$

   $m=0.36{\text{J T}}^{-1}$

   The unit J T${}^{-1}$ follows because torque has units of J and the field is in tesla.

   ✦ m = 0.36 J/T

2. 23 marksNCERT Exercises, Chapter 5

   A short bar magnet of magnetic moment $m=0.32$ J T${}^{-1}$ 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 corresponds to its (a) stable and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

   Hint. Use U = -mB cos(theta) and consider which angle minimises it.

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

   $U=-mB\cos \theta$

   (a) Stable equilibrium occurs where the energy is a minimum, which is at $\theta =0^{\circ }$, with the magnet aligned with the field:

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

   (b) Unstable equilibrium is at the energy maximum, $\theta =180^{\circ }$, with the magnet anti-aligned:

   $U=+mB=+4.8\times 10^{-2}\text{J}$

   In both positions the torque is zero, since $\sin \theta =0$. What distinguishes them is that a small displacement from $0^{\circ }$ produces a restoring torque, while at $180^{\circ }$ it produces a torque that drives the magnet further away.

   ✦ (a) Stable at 0 degrees, aligned with the field, U = -4.8 x 10^-2 J (b) Unstable at 180 degrees, U = +4.8 x 10^-2 J

3. 33 marksNCERT Exercises, Chapter 5

   A closely wound solenoid of 800 turns and area of cross-section $2.5\times 10^{-4}$ 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?

   Hint. A current loop has a magnetic moment NIA; a solenoid is a stack of such loops.

   Why a solenoid resembles a bar magnet. A current-carrying solenoid produces a magnetic field whose external pattern is essentially identical to that of a bar magnet: field lines emerge from one end and re-enter at the other, so the two ends behave as a north and a south pole.

   Suspended freely it aligns itself along the north-south direction, and it attracts or repels another magnet exactly as a bar magnet would.

   Magnetic moment. For a coil of $N$ turns carrying current $I$ and enclosing area $A$:

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

   $m=0.60{\text{J T}}^{-1}$

   The moment points along the axis, in the direction given by the right-hand rule applied to the current.

   ✦ The solenoid's external field pattern matches a bar magnet's, with its ends acting as poles; m = 0.60 J/T

4. 42 marksNCERT Exercises, Chapter 5

   If the solenoid of 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 the torque on the solenoid when its axis makes an angle of $30^{\circ }$ with the direction of the applied field?

   Hint. Use the magnetic moment computed in the previous question.

   A note on the printed question. The book refers to "the solenoid in Exercise 5.5", but Exercise 5.5 concerns a bar magnet, not a solenoid. The solenoid is the one defined in Exercise 5.3, and that is plainly what is intended — the cross-reference in the printed text is an uncorrected error.

   Using the moment found there, $m=0.60$ J T${}^{-1}$:

   $\tau =mB\sin \theta =0.60\times 0.25\times \sin 30^{\circ }$

   $\tau =0.60\times 0.25\times 0.5=7.5\times 10^{-2}\text{N m}$

   ✦ Torque = 7.5 x 10^-2 N m (using m = 0.60 J/T from Exercise 5.3, which the printed cross-reference misnumbers as 5.5)

5. 54 marksNCERT Exercises, Chapter 5

   A bar magnet of magnetic moment 1.5 J T${}^{-1}$ lies aligned with a uniform magnetic field of 0.22 T. (a) What work is required by an external torque to turn the magnet so that its moment is (i) normal to the field, (ii) opposite to the field? (b) What is the torque on the magnet in cases (i) and (ii)?

   Hint. Work equals the change in potential energy; torque is evaluated at the final orientation.

   The work done by an external torque equals the increase in potential energy, $W=U_f-U_i=-mB(\cos {\theta }_f-\cos {\theta }_i)$, starting from ${\theta }_i=0$.

   (a)(i) Turning to ${\theta }_f=90^{\circ }$:

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

   (a)(ii) Turning to ${\theta }_f=180^{\circ }$:

   $W=1.5\times 0.22\times (1-(-1))=0.66\text{J}$

   The second is exactly twice the first, since the cosine swings through twice the range.

   (b) The torque at each final orientation is $\tau =mB\sin \theta$:

   (i) At $90^{\circ }$: $\tau =1.5\times 0.22\times 1=0.33$ N m, which is the maximum possible.

   (ii) At $180^{\circ }$: $\tau =1.5\times 0.22\times 0=0$, because the magnet is anti-aligned and the torque vanishes even though the energy is greatest.

   ✦ (a)(i) 0.33 J (ii) 0.66 J (b)(i) 0.33 N m (ii) zero

6. 64 marksNCERT Exercises, Chapter 5

   A closely wound solenoid of 2000 turns and area of cross-section $1.6\times 10^{-4}$ 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 its magnetic moment? (b) What is the force and torque on it if a uniform horizontal field of $7.5\times 10^{-2}$ T is set up at $30^{\circ }$ with the solenoid's axis?

   Hint. Remember that a uniform field exerts no net force on a magnetic dipole, only a torque.

   (a) The magnetic moment of the solenoid is:

   $m=NIA=2000\times 4.0\times 1.6\times 10^{-4}=1.28{\text{J T}}^{-1}$

   (b) Force. In a uniform field the forces on the two poles are equal and opposite, so they cancel and the net force is zero. A net force would require a non-uniform field.

   Torque. The two forces form a couple:

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

   $\tau =1.28\times 7.5\times 10^{-2}\times 0.5=4.8\times 10^{-2}\text{N m}$

   So the solenoid rotates towards alignment with the field but does not translate.

   ✦ (a) m = 1.28 J/T (b) Force = zero in a uniform field; torque = 4.8 x 10^-2 N m

7. 74 marksNCERT Exercises, Chapter 5

   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 its centre on (a) the axis, (b) the equatorial line (normal bisector) of the magnet.

   Hint. The axial field is twice the equatorial field at the same distance, and the two point in opposite senses.

   For a short bar magnet the two standard results differ by a factor of two, which is the point of asking for both.

   (a) On the axis:

   $B=\frac{{\mu }_0}{4\pi }\cdot \frac{2m}{r^3}=10^{-7}\times \frac{2\times 0.48}{(0.10{)}^3}=10^{-7}\times \frac{0.96}{10^{-3}}$

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

   This points along the magnetic moment, that is from the south pole to the north pole inside the magnet, and away from the magnet at the point considered.

   (b) On the equatorial line:

   $B=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{r^3}=10^{-7}\times \frac{0.48}{10^{-3}}=4.8\times 10^{-5}\text{T}$

   This points opposite to the magnetic moment.

   So the axial field is exactly twice the equatorial field at the same distance, and the two are antiparallel — both facts mirror the electric dipole exactly.

   ✦ (a) 9.6 x 10^-5 T, along the magnetic moment (b) 4.8 x 10^-5 T, opposite to the magnetic moment

Solutions written by the tuition.in editorial team and checked against leph105.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.