A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium

Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by $60^\circ$ is $W$. Now the torque required to keep the magnet in this new position is:

Accepted Answer

1. **Work Done (Energy):** The work done ($W$) in rotating a magnetic dipole with magnetic moment $m$ in a uniform magnetic field $B$ from an initial angle $	heta_1$ to a final angle $	heta_2$ is equal to the change in potential energy:
   $$ W = U_f - U_i = -mB \cos 	heta_2 - (-mB \cos 	heta_1) = mB(\cos 	heta_1 - \cos 	heta_2) $$
   Given that the magnet starts from the equilibrium state, $	heta_1 = 0^\circ$. It is rotated by $60^\circ$, so $	heta_2 = 60^\circ$.
   $$ W = mB(\cos 0^\circ - \cos 60^\circ) = mB\left(1 - \frac{1}{2}ight) = \frac{mB}{2} $$
   From this, we deduce: $mB = 2W$.
2. **Torque Calculation:** The torque ($	au$) magnitude required to hold the magnet at an angle $	heta$ is given by:
   $$ 	au = mB \sin 	heta $$
   At the new position $	heta = 60^\circ$:
   $$ 	au = mB \sin 60^\circ $$
3. **Substitution:** Substitute the value of $mB$ derived from the work equation:
   $$ 	au = (2W) \left( \frac{\sqrt{3}}{2} ight) = \sqrt{3}W $$

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