Question

A bar magnet for dipole moment ${10}^{4}J{T}^{-1}$ is free to rotate around a horizontal plane. A horizontal magnetic field $4\times {10}^{-5}T$ exists on the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction ${60}^0$ from the field.

• A. $0.1J$
• B. $0.2J$
• C. $0.4J$
• D. $0.5J$

Answer 1

The correct option is B $0.2J$

$W=MB(\cos 0-\cos 60)$
$W=\frac{1}{2}MB$
$W=\frac{1}{2}\times 4\times {10}^{-5}\times {10}^4$
$W=0.2J$