$A$ 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?

$A$ 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

$A$ wire of length $10 \ m$ carrying a current of $1 \ A$ is bent into a circular loop. If a magnetic field of $2 \pi \times 10^{-4} \ T$ is applied on the loop,then the maximum torque acting on it 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.)

Four wires,each of length $2.0\,m$,are bent into four loops $P, Q, R$ and $S$ and then suspended in a uniform magnetic field. If the same current is passed in each,then the torque will be maximum on the loop:

$A$ circular coil of $10$ turns and radius $10 \,cm$ is placed in a uniform magnetic field of $0.1 \,T$ normal to the plane of the coil. If the current in the coil is $5 \,A$, then the magnitude of the torque on the coil is