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A magnetic needle suspended parallel to a magnetic field requires $\sqrt3$ J of work to turn it through $60^\circ$. The torque needed to maintain the needle in this position will be
A
$\frac{3}{2}$ J
B
$2\sqrt3$ J
C
3 J
D
$\sqrt3$ J
Detailed Solution
Work done $= U_f - U_i = -MB\cos60^\circ - (-MB\cos0^\circ) = MB\left(1 - \frac{1}{2}\right) = \frac{MB}{2}$
$\sqrt3 = \frac{MB}{2} \Rightarrow MB = 2\sqrt3$ J
Torque needed: $\tau = MB\sin60^\circ = 2\sqrt3\times\frac{\sqrt3}{2}$
$\tau = 3$ J
$\sqrt3 = \frac{MB}{2} \Rightarrow MB = 2\sqrt3$ J
Torque needed: $\tau = MB\sin60^\circ = 2\sqrt3\times\frac{\sqrt3}{2}$
$\tau = 3$ J