First slide

Magnetism

Question

A bar magnet having a magnetic moment of ${2×10}^{4} {JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=4 \times 10^{- 4} T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{0}$ from the field is

Moderate

A

12 J
B

4 J
C

2 J
D

0.6 J

Solution

Work done in turning the magnet of moment $M = 2 \times 10^{4} {JT}^{- 1}$ in a field $B = 4 \times 10^{- 4} T$
(‘H’ is not present in the space) through an angle $60^{0}$ is
$W = U_{f} - U_{i} = [- M B \cos 60^{o}] - [- M B \cos 0^{o}] = \frac{M B}{2}$
$\Rightarrow W = \frac{2 \times 10^{4} \times 4 \times 10^{- 4}}{2}$
$⇒W=4 J$ .

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