$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

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:

$A$ metallic loop is placed in a magnetic field. If a current is passed through it,then

$A$ 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?

The figure below shows the north and south poles of a permanent magnet in which an $n$-turn coil of cross-sectional area $A$ is placed. When a current $i$ is passed through the coil,the plane of the coil makes an angle $\theta$ with respect to the direction of the magnetic field $B$. The torque on the coil is:

Write an equation for the torque acting on a current-carrying loop which subtends an angle $\theta$ with a uniform magnetic field.