A magnetic needle lying parallel to a magnetic field requires $\mathrm{W}$ units of work to turn it through…

A magnetic needle lying parallel to a magnetic field requires $\mathrm{W}$ units of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be
  1. $\sqrt{3} \mathrm{~W}$
  2. W
  3. $\frac{\sqrt{3}}{2} \mathrm{~W}$
  4. $2 \mathrm{~W}$

Solution

$\tau=(\mathrm{H}) \tan 60^{\circ} \quad=\mathrm{W} \cdot \sqrt{3}$

Asked in: JEE Main 2003

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