(a) A circular coil of 30 turns and radius `8.0cm`. Carrying a current of `6.0A` is suspended vertically in a uniform horizontal magnetic field of magnitude `1.0T`. The field lines make an angle of `60^@` with the normal to 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).

Explanation

(a) The orientation of coil with respect to the magnetic field B is shown below:


The magnitude of magnetic torque acting on the circular coil of radius r and number of turns, to current I flowing through it is given by:
` tau = n I AB sin theta`
Here, `n = 30 , r = 8 cm`
` :. " Area of coil " , A = pi (r)^(2) = pi (8)^(2) xx 10^(-4) m^(2)`
` I = 6 A, B = 1 T, theta = 60^(@)`
` rArr " " tau = n BIA sin theta`
` = 30 xx 1 xx 6 xx (pi (8)^(2))/(100 xx 100) xx sin 60^(@)`
` = 3.133 ` Nm
(b) No, the answer is unchanged because the formula `tau = n IAB sin theta` is true for a planar loop of any shape. Only, it depends on the size of the planar loop.