Question

A bar magnet has a magnetic moment of $200\, A \,m^2$. The magnet is suspended in a magnetic field of $0.30\, N\, A^{-1}\, m^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$, will be

• A. $30 \,N\,m$
• B. $30 \sqrt{3} \,N \,m$
• C. $60\, N\, m$
• D. $60 \sqrt{3} \,N \,m$

Answer 1

Answer: (A)

$30 \,N\,m$

Answer 2

Given, $M=200 A-m^{2} B=0.30 NA ^{-1} M^{-1}$
and $\theta=30^{\circ}$
We know that the Torque,
$\tau = M \times B$
$\Rightarrow |\tau| =M B \sin \theta=200 \times 0.3 \times \frac{1}{2}$
$=100 \times 0.3=30 N - m$