$A$ current-carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon

$A$ uniform magnetic field of $2 \times 10^{-3} \ T$ acts along the positive $Y$-direction. $A$ rectangular loop of sides $20 \ cm$ and $10 \ cm$ carrying a current of $5 \ A$ is placed in the $Y-Z$ plane. The current flows in an anticlockwise sense with reference to the negative $X$-axis. Find the magnitude and direction of the torque.

$A$ circular coil has a moment of inertia of $0.8 \, kg \cdot m^2$ around any diameter and is carrying a current to produce a magnetic moment of $20 \, A \cdot m^2$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \, T$ is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{\circ}$ will be:

$A$ rectangular loop of sides $10 \ cm$ and $5 \ cm$ carrying a current $I$ of $12 \ A$ is placed in different orientations as shown in the figures below. If there is a uniform magnetic field of $0.3 \ T$ in the positive $z$ direction,in which orientations would the loop be in $(i)$ stable equilibrium and $(ii)$ unstable equilibrium?

$A$ circular coil having $N$ turns and radius $r$ carries a current $I$. It is held in the $XZ$ plane in a magnetic field $\vec{B} = B\hat{i}$. The torque on the coil due to the magnetic field is

$(a)$ $A$ 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.)