First slide

Magnetism

Question

A bar magnet having a magnetic moment of ${4×10}^{4} {JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=3 \times 10^{- 4} T$ exits in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{0}$ from the field is

Moderate

A

12 J
B

6 J
C

2 J
D

0.6 J

Solution

Work done in turning the magnet of moment ${M=4×10}^{4} {JT}^{-1}$ in a field ${B=3×10}^{-4} T$ through an angle $60^{0}$ is
$W = U_{f} - U_{i} = [- M B \cos 60^{o}] - [- M B \cos 0^{o}] = \frac{M B}{2}$
$⇒W= \frac{4 \times 10^{4} \times 3 \times 10^{- 4}}{2}$
$⇒W=12 \times \frac{1}{2}$
$⇒W=6 J$ .

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