$A$ rectangular coil has $2n$ turns,an area $2a$ and carries a current $2I$. The plane of the coil is at $60^{\circ}$ to a horizontal uniform magnetic field of flux density $B$. The torque on the coil due to magnetic force 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.)

$A$ closely wound solenoid of $2000$ turns and area of cross-section $1.5 \times 10^{-4} \, m^2$ carries a current of $2.0 \, A$. It is suspended through its centre and perpendicular to its length,allowing it to turn in a horizontal plane in a uniform magnetic field of $5 \times 10^{-2} \, T$ making an angle of $30^o$ with the axis of the solenoid. The torque on the solenoid will be:

$A$ coil in the shape of an equilateral triangle of side $10 \, cm$ lies in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of $20 \, mT$. The torque acting on the coil when a current of $0.2 \, A$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x} \times 10^{-5} \, Nm$. The value of $x$ is ..... .

The $(\tau - \theta)$ graph for a current-carrying coil in a magnetic field is:

$A$ wire of length $L$ carries a current $i$. If the wire is turned into a circular coil and kept in a uniform magnetic field $B$,the maximum magnitude of torque in the given magnetic field will be