Magnetism and Matter - Result Question 3
6. A bar magnet having a magnetic moment 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
[2009]
(a) $12 J$
(b) $6 J$
(c) $2 J$
(d) $0.6 J$
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Answer:
Correct Answer: 6. (b)
Solution:
- (b) Work done
$=MB(\cos \theta_1-\cos \theta_2)$
$=MB(\cos 0^{\circ}-\cos 60^{\circ})$
$=MB(1-\frac{1}{2})=\frac{2 \times 10^{4} \times 6 \times 10^{-4}}{2}=6 J$
