The figure below shows the north and south poles of a permanent magnet in which an $n$-turn coil of cross-sectional area $A$ is placed. When a current $i$ is passed through the coil,the plane of the coil makes an angle $\theta$ with respect to the direction of the magnetic field $B$. The torque on the coil 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.

$(a)$ $A$ circular coil of $30$ turns and radius $8.0 \; cm$ carrying a current of $6.0 \; A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \; T$. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
$(b)$ Would your answer change,if the circular coil in $(a)$ were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

$A$ rectangular coil (dimension $5\,cm \times 2\,cm$) with $100\,turns$,carrying a current of $3\,A$ in the clockwise direction,is kept centered at the origin and in the $X-Z$ plane. $A$ magnetic field of $1\,T$ is applied along the $X$-axis. If the coil is tilted through $45^{\circ}$ about the $Z$-axis,then the torque on the coil is.....$Nm$.

$A$ circular coil of $30$ turns and radius $8.0\, cm$ carrying a current of $6.0\, A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\, T$. The field lines make an angle of $60^o$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (in $, Nm$)

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