Two short bar magnets $A$ and $B$ (having magnetic moments $M_{1}$ and $M_{2}$ respectively) are kept one above the other with their magnetic axes perpendicular to each other. If their resultant magnetic field at a point on the axis of magnet $A$ is inclined at $45^{\circ}$ with the axis of magnet $A$,then the ratio of magnetic moments $\frac{M_{2}}{M_{1}}$ is $[\tan 45^{\circ} = 1]$.

The work done in rotating a bar magnet,which is initially in the direction of a uniform magnetic field,through $45^{\circ}$ is $W$. The additional work to be done to rotate the magnet further through $15^{\circ}$ is

$A$ short bar magnet of magnetic moment $5.25 \times 10^{-2} \;J\, T^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet,the resultant field is inclined at $45^{\circ}$ with the earth's field on
$(a)$ its normal bisector and
$(b)$ its axis.
Magnitude of the earth's field at the place is given to be $0.42 \;G$. Ignore the length of the magnet in comparison to the distances involved.

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?

Two short bar magnets of magnetic moments '$M$' each are arranged at the opposite corners of a square of side '$d$',such that their centres coincide with the corners and their axes are parallel to one side of the square. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is: