Question

A bar magnet of moment \(0.4 \times 10^{-3} \, {Am}^2\) is kept in a magnetic field of \(2\pi \times 10^{-3} \, {T}\). The magnet makes an angle of \(45^\circ\) with the direction of magnetic field. The torque acting on the magnet is:

• A. \(7.65 \pi \times 10^{-7} \, {Nm}\)
• B. \(6.55 \pi \times 10^{-4} \, {Nm}\)
• C. \(5.65 \times 10^{-2} \, {Nm}\)
• D. \(5.65 \times 10^{-7} \, {Nm}\)

Hint:

When calculating torque on a magnetic dipole, remember that \(\sin(45^\circ)\) is approximately \(0.707\).

Answer 1

The Correct Option is D

Apply the torque formula for a magnet in a magnetic field. The torque \( \tau \) on a magnetic dipole is given by: \[ \tau = \mu B \sin(\theta) \] where \( \mu \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta \) is the angle with the magnetic field. \[ \mu = 0.4 \times 10^{-3} \, {Am}^2 \] \[ B = 2\pi \times 10^{-3} \, {T} \] \[ \theta = 45^\circ \] \[ \tau = (0.4 \times 10^{-3}) \times (2\pi \times 10^{-3}) \times \sin(45^\circ) \] \[ \tau \approx 5.65 \times 10^{-7} \, {Nm} \]