$A$ uniform magnetic field of $0.3 \; T$ is established along the positive $Z$-direction. $A$ rectangular loop in the $XY$-plane with sides $10 \; cm$ and $5 \; cm$ carries a current of $I = 12 \; A$ as shown. The torque on the loop is

$A$ coil having $10 \text{ A m}^2$ magnetic moment is placed in a vertical plane and is free to rotate about its horizontal axis, which coincides with its diameter. $A$ uniform magnetic field of $2 \text{ T}$ in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of $90^{\circ}$ under the influence of the magnetic field. The moment of inertia of the coil is $0.1 \text{ kg m}^2$. What will be its angular speed (in $\text{ rad/s}$)?

$A$ uniform magnetic field $B$ of $0.3\, T$ is along the positive $Z-$ direction. $A$ rectangular loop $(abcd)$ of sides $10\, cm \times 5\, cm$ carries a current $I$ of $12\, A$. Out of the following different orientations,which one corresponds to stable equilibrium?

$A$ circular coil of $10$ turns and radius $10 \,cm$ is placed in a uniform magnetic field of $0.1 \,T$ normal to the plane of the coil. If the current in the coil is $5 \,A$, then the magnitude of the torque on the coil is

$A$ small coil $C$ with $N = 200$ turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in the figure. The cross-sectional area of the coil is $A = 1.0 \, cm^2$,and the length of the arm $OA$ of the balance beam is $l = 30 \, cm$. When there is no current in the coil,the balance is in equilibrium. On passing a current $I = 22 \, mA$ through the coil,the equilibrium is restored by putting an additional counterweight of mass $\Delta m = 60 \, mg$ on the balance pan. Find the magnetic induction $B$ at the spot where the coil is located. (in $, T$)

$A$ 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)