$A$ closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{-4} \;m^{2}$ carries a current of $3.0 \;A$. If the solenoid is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25 \;T$ is applied,what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of the applied field?

$A$ circular coil having $N$ turns and radius $R$ carrying current $I$ is held in the $z-x$ plane in a magnetic field $B\hat{k}$. The torque on the coil due to the magnetic field in $N-m$ is:

$A$ $100$ turns coil shown in the figure carries a current of $2\, A$ in a magnetic field $B = 0.2\, Wb/m^2$. The torque acting on the coil is

$A$ circular coil of $30$ turns and radius $8 \,cm$ carrying a current of $6 \,A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \,T$. The field lines make an angle of $20^{\circ}$ with the normal of the coil. The magnitude of the counter torque that must be applied to prevent the coil from turning is (in $\,Nm$)

$(a)$ $A$ circular coil of $30$ turns and radius $8.0 \; cm$ carrying a current of $6.0 \; A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \; T$. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
$(b)$ Would your answer change,if the circular coil in $(a)$ were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

The radius of a circular ring of wire is $R$ and it carries a current of $I \, A$. At its centre,a smaller ring of radius $r$ with current $i$ and $N$ turns is placed. Assuming that the planes of the two rings are perpendicular to each other and the magnetic induction produced at the centre of the bigger ring is constant,then the torque acting on the smaller ring will be: