Question

A bar magnet having a magnetic movement of $2 \times 10^{4} JT ^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B =6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :

• A. 12 J
• B. 6 J
• C. 2 J
• D. 0.6 J

Answer 1

Answer: (B)

6 J

Answer 2

$W =MB\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$=2 \times 10^{4} \times 6 \times 10^{-4}\left(\cos \theta-\cos 60^{\circ}\right)$
$=12 \times \frac{1}{2}=6\, J$