A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60º is W. Now the torque required to keep the magnet in this new position is
$\dfrac{2W}{\sqrt{3}}$
$\dfrac{W}{\sqrt{3}}$
$\sqrt{3}W$
$\dfrac{\sqrt{3}W}{2}$
Explanation:
Wext = Uf – Vi
= – MB cos 60° – (–MB)
= MB(1 – cos 60°) = MB/2 = W
r = MB sin 60° = MB $\dfrac{\sqrt{3}}{2}$ =$ \sqrt{3}$W