NCERT Solutions · Class 12 Physics Magnetism and Matter

Torque on a magnetic dipole: \[\tau = mB\sin\theta \] Given \(\displaystyle \theta = 30^\circ\), \(\displaystyle B = 0.25\ \mathrm{T}\), \(\displaystyle \tau = 4.5\times10^{-2}\ \mathrm{J}\). \[m = \frac{\tau}{B\sin\theta} = \frac{4.5\times10^{-2}}{0.25\times\sin 30^\circ} = \frac{4.5\times10^{-2}}{0.25\times0.5} \] \[m = \frac{4.5\times10^{-2}}{0.125} = 0.36\ \mathrm{J\,T^{-1}} \] Magnetic moment \(\displaystyle m = 0.36\ \mathrm{J\,T^{-1}}\).

A current-carrying solenoid behaves like a bar magnet because each turn is a current loop with its own magnetic dipole moment along the solenoid's axis; these add up, so the whole solenoid produces a dipole-like field with one end acting as a N-pole and the other as a S-pole (by the right-hand rule), just as field lines emerge from a bar magnet's N-pole and curve back into its S-pole.Magnetic moment of a solenoid: \[m = NIA \] Given \(\displaystyle N = 800\), \(\displaystyle I = 3.0\ \mathrm{A}\), \(\displaystyle A = 2.5\times10^{-4}\ \mathrm{m^2}\). \[m = 800\times3.0\times2.5\times10^{-4} = 0.6\ \mathrm{J\,T^{-1}} \] Associated magnetic moment \(\displaystyle m = 0.6\ \mathrm{J\,T^{-1}}\) (i.e. \(\displaystyle 0.6\ \mathrm{A\,m^2}\)), directed along the solenoid's axis.

The referenced solenoid ($\displaystyle 800$ turns, area \(\displaystyle 2.5\times10^{-4}\ \mathrm{m^2}\), current $\displaystyle 3.0$ A) has magnetic moment \(\displaystyle m = NIA = 0.6\ \mathrm{J\,T^{-1}}\).Torque on the dipole: \[\tau = mB\sin\theta \] With \(\displaystyle B = 0.25\ \mathrm{T}\), \(\displaystyle \theta = 30^\circ\): \[\tau = 0.6\times0.25\times\sin 30^\circ = 0.6\times0.25\times0.5 \] \[\tau = 7.5\times10^{-2}\ \mathrm{N\,m} \] Torque on the solenoid \(\displaystyle \tau = 7.5\times10^{-2}\ \mathrm{N\,m}\).

(a) Magnetic moment of the solenoid: \[m = NIA \] Given \(\displaystyle N = 2000\), \(\displaystyle I = 4.0\ \mathrm{A}\), \(\displaystyle A = 1.6\times10^{-4}\ \mathrm{m^2}\). \[m = 2000\times4.0\times1.6\times10^{-4} = 1.28\ \mathrm{J\,T^{-1}} \](b) In a uniform magnetic field the net force on any current loop (and hence on the solenoid, a stack of loops) is zero — the forces on opposite sides cancel — so: \[F = 0 \] The torque, with \(\displaystyle B = 7.5\times10^{-2}\ \mathrm{T}\) and \(\displaystyle \theta = 30^\circ\): \[\tau = mB\sin\theta = 1.28\times7.5\times10^{-2}\times\sin 30^\circ = 1.28\times0.075\times0.5 \] \[\tau = 4.8\times10^{-2}\ \mathrm{N\,m} \] \(\displaystyle m = 1.28\ \mathrm{J\,T^{-1}}\); Force \(\displaystyle =0\); Torque \(\displaystyle =4.8\times10^{-2}\ \mathrm{N\,m}\), tending to align the solenoid's axis with \(\displaystyle \vec{B}\).

Field of a short bar magnet, magnetic moment \(\displaystyle m = 0.48\ \mathrm{J\,T^{-1}}\), at distance \(\displaystyle r = 10\ \mathrm{cm} = 0.10\ \mathrm{m}\).(a) On the axis: \[B_{\text{axial}} = \frac{\mu_0}{4\pi}\frac{2m}{r^3} \] \[B_{\text{axial}} = 10^{-7}\times\frac{2\times0.48}{(0.10)^3} = 10^{-7}\times\frac{0.96}{1.0\times10^{-3}} \] \[B_{\text{axial}} = 10^{-7}\times 960 = 9.6\times10^{-5}\ \mathrm{T} \] directed along the axis, parallel to \(\displaystyle \vec{m}\) (from S to N pole, i.e. pointing away from the magnet on the N-pole side).(b) On the equatorial line (normal bisector): \[B_{\text{eq}} = \frac{\mu_0}{4\pi}\frac{m}{r^3} \] \[B_{\text{eq}} = 10^{-7}\times\frac{0.48}{1.0\times10^{-3}} = 10^{-7}\times 480 = 4.8\times10^{-5}\ \mathrm{T} \] directed antiparallel to \(\displaystyle \vec{m}\) (i.e. from N to S, opposite to the magnet's moment).\(\displaystyle B_{\text{axial}} = 9.6\times10^{-5}\ \mathrm{T}\) (parallel to \(\displaystyle \vec{m}\)); \(\displaystyle B_{\text{eq}} = 4.8\times10^{-5}\ \mathrm{T}\) (antiparallel to \(\displaystyle \vec{m}\)), i.e. exactly half the axial value.