$A$ current-carrying coil experiences a torque due to a magnetic field. The value of the torque is $80 \%$ of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is

$A$ coil is placed in $y-z$ plane making an angle of $30^{\circ}$ with the $x$-axis. The current through the coil is $I$,and the number of turns is $N$. If a magnetic field of strength $B$ is applied in the positive $x$-direction,find the torque experienced by the coil (Radius of the coil is $R$).
$(N=100, I=1 \text{ A}, R=2 \text{ m}, B=\frac{1}{\pi} \text{ T})$

$(a)$ $A$ 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.)

$A$ wire of length $L$ carries a current $I$. If the wire is turned into a square coil of single turn,the maximum magnitude of the torque in a given magnetic field $\overrightarrow{B}$ 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?

$A$ circular coil having $N$ turns and radius $R$ carrying current $I$ is held in the $z-x$ plane in a magnetic field $B\hat{k}$. The torque on the coil due to the magnetic field in $N-m$ is: