Question

A coil of 10 turns and area 2 * 10^-2 m^2 is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 5 A . When the coil is held with its plane in North-South direction, it experiences a couple of 0.3 N-m and when its plane is East-West the corresponding couple is 0.4 N-m. The magnetic induction is
a) 0.2 T
b) 0.3 T
c) 0.4 T
d) 0.5 T

Answer 1

Couple due to magnetic field acting on magnetic moment of the current coil= M X B in vectors
Couple = M * B * Sin theta
Theta. = angle between M and B.
M = I * N*A = 5*10* 0.02 = 1 units
Couple = 0.3 = 1*B*sin theta

Further, wrt East West:
Couple = 0.4 = 1 *B*Cos theta

Squaring and adding we get
B = 0.5 T