The torque required to hold a small circular coil of $10$ turns,$2 \times 10^{-4} \ m^2$ area,and carrying $0.5 \ A$ current in the middle of a long solenoid of $10^3$ turns per meter carrying $3 \ A$ current,with its axis perpendicular to the axis of the solenoid,is:

$(a)$ $A$ 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.)

$A$ uniform magnetic field $B$ of $0.3\, T$ is along the positive $Z-$ direction. $A$ rectangular loop $(abcd)$ of sides $10\, cm \times 5\, cm$ carries a current $I$ of $12\, A$. Out of the following different orientations,which one corresponds to stable equilibrium?

$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

Write an expression for the torque acting on a current-carrying loop suspended in a uniform magnetic field.

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: