A coil of 100 turns and area `2 times 10^-2 m^2` pivoted about a vertical diameter in a uniform magnetic field carries a current of 5A. When the coil is held with its plane in North-South direction. It experiences a torque of 0.3N/m. When the plane is in East-west direction the torque is 0.4N/m. the value of magnetic induction is

Options

A. 0.2T
B. 0.3T
C. 0.4T
D. 0.05T

Answer

D

Explanation

Given `A=2 times 10^-2 m^2`
N=100
I=5A
When the coil is held with its plane in North- South direction then its torque
`t_1=0.3 Nm`
`t_2=MB sin theta`
When the plane is in East- West direction then its torque
`t_2=0.4 Nm`
`t_2=MB sin (90^@- theta)`
`t_2=Mb cos theta`
Ratio= `(MB sin theta)/(MB cos theta)=t_1/t_2`
`tan theta=3/4`
or `theta= tan^-1 (3/4)`
then `sin theta=0.6`
`therefore B=t_1/(M sin theta) `
`implies B=0.3/(NIA sin theta) `
`implies B= 0.3/(100 times 5 times 2 times 10^-2 times 0.6)`
`implies B=0.05 T`