When a bar magnet is rotated from its position parallel to the external magnetic field \(B=10^{-3}\) T to a direction opposite to the field (anti-parallel), the work done is \(3\) J.
Then, the maximum torque experienced by this magnet in this field is:
1. \(3\times10^{-3}\) N-m
2. \(3\times10^{3}\) N-m
3. \(6\) N-m
4. \(1.5\) N-m

Two magnetic dipoles, \(X\) and \(Y,\) are separated by a distance \(d,\) with their axes oriented perpendicular to each other. The dipole moment of \(Y\) is twice that of \(X.\) A charged particle with charge \(q\) moves with velocity \(v\) through their midpoint \(P,\) which makes an angle \(\theta=45^\circ\) with the horizontal axis, as shown in the diagram. Assuming \(d\) is much larger than the dimensions of the dipoles, the magnitude of the force acting on the charged particle at this instant is:

Three identical bar magnets, each having a dipole moment \(M,\) are placed at the origin—oriented along the \(x\text-\)axis, the \(y\text-\)axis, and the \(z\text-\)axis respectively. The net magnetic moment of the dipoles has the magnitude:
1. \(3M\)
2. \(\sqrt2M\)
3. \(\sqrt3M\)
4. zero

A small bar magnet placed with its axis at \(30^\circ\) with an external field of \(0.06\) T experiences a torque of \(0.018\) Nm. the minimum work required to rotate it from its stable to unstable equilibrium position is:
1. \(7.2\times 10^{-2}~\text{J}\)
2. \(11.7\times 10^{-3}~\text{J}\)
3. \(9.2\times 10^{-3}~\text{J}\)
4. \(6.4\times 10^{-2}~\text{J}\)

Two short bar magnets, each with a magnetic moment \(M,\) are arranged as shown in the figure. What will the magnetic field at point \(P\) be?