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:

  1. (b) Work done

$=MB(\cos \theta_1-\cos \theta_2)$

$=MB(\cos 0^{\circ}-\cos 60^{\circ})$

$=MB(1-\frac{1}{2})=\frac{2 \times 10^{4} \times 6 \times 10^{-4}}{2}=6 J$