$(a)$ $A$ 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) Number of turns on the circular coil,$n=30$.
Radius of the coil,$r=8.0 \; cm = 0.08 \; m$.
Area of the coil,$A = \pi r^2 = \pi(0.08)^2 \approx 0.0201 \; m^2$.
Current flowing in the coil,$I=6.0 \; A$.
Magnetic field strength,$B = 1.0 \; T$.
Angle between the field lines and the normal to the coil surface,$\theta = 60^{\circ}$.
The coil experiences a magnetic torque $\tau = n I A B \sin \theta$. To prevent the coil from turning,an equal and opposite counter-torque must be applied.
$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin 60^{\circ}$.
$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \frac{\sqrt{3}}{2} \approx 3.133 \; Nm$.
$(b)$ The torque on a current-carrying coil in a magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where the magnetic moment $\vec{m} = n I \vec{A}$. Since the magnitude of the torque depends only on the area $A$ and not on the specific shape of the coil,the answer would not change if the circular coil were replaced by a planar coil of any irregular shape that encloses the same area.