A square coil of side 10 cm has 20 turns and carries a current of 12 A. the coil is suspended vartically and the normal to the plane of the coil, makes an angle `theta` with the direction of a uniform horizontal magnetic field of 0.80 T. if the torque, experienced by the coil, equals 0.N-m , the value of ` theta ` is

Options

A. `0^(@)`
B. `pi/2 ` rad
C. `pi/3` rad
D. `pi/6` rad

Answer

d

Explanation

Area of coil A =`side^(2) =(0.1)^(2) =0.01 m^(2)`, Number of turns, N=20 current I=12 A
Normal to the coil make an angle `theta` with the direction of B, magnetic field , B=0.80 T. torque, experienced by the coil , `tau=0.96 N-m` since, total torque on the coil , `tau=(NIA) B sin theta` Substituting the values in above formula, we get 0.96 N-m =20x12 A x `0.01 m^(2) xx0.80 T xx sin theta`
`sin theta=0.96/1.92 =1/2`
`theta=(pi)/6` rad