(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° 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 were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).

Given, a current carrying circular coil. 

Number of turns in the coil, N = 30
Current, I = 6.0A
Uniform horizontal magnetic field, B = 1.0 T
Angle made by coil with the magnetic field lines, $\mathrm{\alpha } = 60°$ 
Radius of the circular coil, r = 8.0 cm = 8 x 10^-2m.

Therefore,
Area of the coil, $\mathrm{A} = {\mathrm{\pi r}}^2$
                         $= \frac{22}{7}\times {\left(8\times {10}^{-2}\right)}^2$
                       $\mathrm{A} = 2.01 \times {10}^{-2}{\mathrm{m}}^2$ 

(a)  Now,
Torque that must be applied to prevent the coil from turning
                      $\boxed{\tau = \mathrm{NBIA} \sin \mathrm{\alpha }}$ 

        $=30 \times 6.0 \times 1.0 \times (2 \times {10}^{-2}) \times \sin 60°$ 
 
      $$\begin{gathered} \tau = 30 \times 6 \times 1 \times 2 \times \frac{\sqrt{3}}{2}\times {10}^{-2} \\ = 3.12 \mathrm{Nm}. \end{gathered}$$ 

(b)  If the area of the loop is the same, the torque will remain unchanged as the torque on the planar loop does not depend upon the shape.