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:

$A$ 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.

$A$ 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 $\text{ rad/s}$)?

$A$ coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{B}$ is in the plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it,the side $l$ of the triangle is

$A$ coil in the shape of an equilateral triangle of side $2 \,cm$ is suspended from a vertex such that it hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{-3} \,T$. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{-5} \,Nm$, the current to be passed through the coil is (in $A$)

$A$ current-carrying coil is subjected to a uniform magnetic field. The coil will orient itself so that its plane becomes