$A$ uniform current-carrying ring of mass $m$ and radius $R$ is suspended by a massless string as shown. $A$ uniform magnetic field $B_0$ exists in the region to keep the ring in a horizontal position. Determine the current $I$ in the ring.

$A$ circular coil of radius $10 \,cm$ with $100$ turns carrying a current of $0.5 \,A$ lies in a magnetic field of $2 \,T$ such that the normal drawn to the plane of the coil makes an angle $\theta$ with the direction of the field. Work done in rotating the coil to change the angle $\theta$ from $0^{\circ}$ to $180^{\circ}$ is

$A$ 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

$A$ solenoid of length $0.4 \ m$ and having $500$ turns of wire carries a current $3 \ A$. $A$ thin coil having $10$ turns of wire and radius $0.1 \ m$ carries current $0.4 \ A$. The torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid is $(\mu_0 = 4\pi \times 10^{-7} \ SI \ units, \pi^2 = 10, \sin 90^{\circ} = 1)$.

$A$ small circular loop of conducting wire has radius $a$ and carries current $I$. It is placed in a uniform magnetic field $B$ perpendicular to its plane such that when rotated slightly about its diameter and released,it starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$,then find the expression for $T$.

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