$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

Two identical bar magnets,each of magnetic moment $M$,are kept perpendicular to each other at a certain distance. The magnetic induction at a point that is at the same distance $d$ from the center of both magnets is: (where $\mu_{0}$ is the permeability of free space)

Figure shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
$(a)$ In which configuration the system is not in equilibrium?
$(b)$ In which configuration is the system in $(i)$ stable,and $(ii)$ unstable equilibrium?
$(c)$ Which configuration corresponds to the lowest potential energy among all the configurations shown?

The distance of two points on the axis of a magnet from its centre is $10 \, cm$ and $20 \, cm$ respectively. The ratio of magnetic intensity at these points is $12.5 : 1$. The length of the magnet will be......$cm$.

Two short bar magnets of magnetic moments $M$ and $\sqrt{3} M$ are joined like a cross (+). This cross is suspended from its centre with its plane horizontal in the earth's magnetic field. When the cross comes to equilibrium,the angle made by the weaker magnet with the earth's magnetic field $B_H$ is (in $^{\circ}$)