First slide

Magnetic dipole in a uniform magnetic field

Question

The work done in turning a magnet of magnetic moment M by angle of 90^o from the meridian, is n times the corresponding work done to turn in through an angle of 60^o. The value of n is given by

Moderate

A

2
B

1
C

1/2
D

1/4

Solution

$\begin{array}{l} W_{1} = MB (\cos 0^{\circ} - \cos 90^{\circ}) = MB (1 - 0) = MB \\ W_{2} = MB (\cos 0^{\circ} - \cos 60^{\circ}) \\ = MB (1 - 1 / 2) = MB / 2 \\ ∴ W_{1} = 2 W_{2} or n = 2 \end{array}$

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