$A$ 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

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?

$A$ current-carrying loop is placed in a uniform magnetic field $B$ in different orientations $I$, $II$, $III$, and $IV$ as shown in the figure. The correct order of decreasing potential energy is ($\hat{n}$ is the unit vector normal to the plane of the loop).

The torque acting on a coil carrying current '$I$',having '$n$' turns and area '$A$',situated parallel to a magnetic field of induction '$B$' is:

$A$ thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$,as shown in the figure. Initially,it hangs vertically downwards,because of acceleration due to gravity $g$,on two conducting supports at $P$ and $Q$. When a current $I$ is passed through the loop,the loop turns about the line $PQ$ by an angle $\theta$ given by

$A$ circular coil of radius $R$ and a current $I$,which can rotate about a fixed axis passing through its diameter,is initially placed such that its plane lies along the magnetic field $B$. The kinetic energy of the loop when it rotates through an angle $90^{\circ}$ is: (Assume that $I$ remains constant)