Question

A short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800\, G$ experiences a torque of $0.016\, Nm$. The magnetic moment of the magnet is

• A. $4\, Am ^{2}$
• B. $0.5\, Am ^{2}$
• C. $2\, Am ^{2}$
• D. $0.40\, Am ^{2}$

Answer 1

Answer: (D)

The magnetic moment of the magnet
$\tau=m B \sin \theta, \theta=30^{\circ}$
hence, $\sin \theta=1 / 2 \left(\because 1\, G =10^{-4} T \right)$
Thus, $0.016=m \times\left(800 \times 10^{-4} T \right) \times(1 / 2)$
$\Rightarrow m=160 \times 2 / 800=0.40\, Am ^{2}$