$A$ circular coil carrying a current of $2.5 \ A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field. When the coil is made to oscillate, the time period of oscillation is $T$. If the current through the coil is $10 \ A$, the time period of oscillation is

$A$ circular coil has a moment of inertia of $0.8 \, kg \cdot m^2$ around any diameter and is carrying a current to produce a magnetic moment of $20 \, A \cdot m^2$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \, T$ is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{\circ}$ will be:

$A$ 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.

$A$ current-carrying loop is placed in a uniform magnetic field in four different orientations,$I$,$II$,$III$,and $IV$. Arrange them in the decreasing order of potential energy.

$A$ uniform magnetic field of $3000\,G$ is established in the positive $z-$ direction. $A$ rectangular loop of sides $10\,cm$ and $5\,cm$ carries a current of $12\,A$. The loop is placed in the $xy-$ plane as shown in the figure. What is the magnitude of the torque on the loop?

$A$ circular coil of $30$ turns and radius $8 \,cm$ carrying a current of $6 \,A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \,T$. The field lines make an angle of $20^{\circ}$ with the normal of the coil. The magnitude of the counter torque that must be applied to prevent the coil from turning is (in $\,Nm$)