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:

A magnetic needle suspended parallel to a magnetic field requires \(\sqrt{3}~\text{J}\) of work to turn it through $\mathrm{}$\(60^\circ\). The torque needed to maintain the needle in this position will be:
1. \(3\) N-m
2. \(\sqrt{3} \) N-m
3. \(\frac32\) N-m
4. \(2\sqrt{3}\) N-m

A short bar magnet of magnet moment \(0.4\) ${\mathrm{JT}}^{-1}$ is placed in a uniform magnetic field of \(0.16\) $\mathrm{T}$. The magnet is in stable equilibrium when the potential energy is:
1. \(0.064\) J
2. \(-0.064\) J
3. zero
4.\(-0.082\) J