The square loop $ABCD$,carrying a current $i$,is placed in a uniform magnetic field $B$,as shown. The loop can rotate about the axis $XX'$. The plane of the loop makes an angle $\theta$ $(\theta < 90^o)$ with the direction of $B$. Through what angle will the loop rotate by itself before the torque on it becomes zero?

$A$ loop is formed by a wire of constant length carrying a current and placed in an external magnetic field. The torque acting on it does not depend upon:

$A$ circular coil of radius $9 \, cm$ carrying a current of $2 \, A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field of $\pi \times 10^{-2} \, T$. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a time period of $\frac{1}{3} \, s$. If the moment of inertia of the coil about its axis of rotation is $9 \times 10^{-5} \, kg \cdot m^2$, the number of turns of 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.

$A$ rectangular coil of $400$ turns and $10^{-2} \ m^2$ area,carrying a current of $0.5 \ A$ is placed in a uniform magnetic field of $1 \ T$ such that the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The initial moment of force acting on the coil in $Nm$ is

$A$ circular coil of $20$ turns and radius $10\, cm$ is placed in a uniform magnetic field of $0.10\, T$ normal to the plane of the coil. If the current in the coil is $5\, A$,then the torque acting on the coil will be...... $Nm$.