First slide

Magnetic dipole in a uniform magnetic field

Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60^o. The torque needed to maintain the needle in this position will be

Easy

A

2W
B

$\sqrt{3} W$
C

W
D

$(\frac{\sqrt{3}}{2}) W$

Solution

$\begin{array}{l} W = - MB (\cos 60^{\circ} - \cos 0^{\circ}) \\ = - MB (\frac{1}{2} - 1) = \frac{1}{2} MB ∴ MB = 2 W \end{array} Now τ = MBsin θ = 2 Wsin 60^{\circ} = 2 W (\sqrt{3} / 2) = \sqrt{3} W$

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