$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

$A$ bar magnet of magnetic moment $1.5 \, J \, T^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$.
$(a)$ What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: $(i)$ normal to the field direction,$(ii)$ opposite to the field direction?
$(b)$ What is the torque on the magnet in cases $(i)$ and $(ii)?$

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......$cm$.

If the dipole moment of a short bar magnet is $1.25 \ A-m^2$,find the magnetic field on its axis at a distance of $0.5 \ m$ from the centre of the magnet.

The pole strength of a bar magnet is $48 \, A \cdot m$ and the distance between its poles is $25 \, cm$. The torque required to hold the magnet at an angle of $30^{\circ}$ with a uniform magnetic field of flux density $0.15 \, N \cdot A^{-1} \cdot m^{-1}$ is . . . . . . $N \cdot m$.

There are two current-carrying planar coils made from identical wires of length $L$. $C_1$ is circular (radius $R$) and $C_2$ is square (side $a$). They are constructed such that they have the same frequency of oscillation when placed in the same uniform magnetic field $B$ and carry the same current $I$. Find $a$ in terms of $R$.