A coil of `100` turns and area `2 xx 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 `0.3 N//m`. When the plane is in East-West direction the torque is `0.4 Nm`. The value of magnetic induction is (Neglect earth's magnetic field)

Options

A. 0.2 T
B. 0. 3T
C. 0.4 T
D. 0.05 T

Answer

D

Explanation

N = 100 , `A = pi r^(2) = 2 xx 10^(-2), i= 5A`
`r_(1) = 0.3` Nm (North-south)
`r_(2)= 0.4` Nm (East-west)
M = NIA = `100 xx 5 xx 2 xx 10^(-2) = 10 Am^(2)`
Net torque -
`tau = sqrt(r_(1)^(2) + r_(2)^(2))` = MB
`implies sqrt((0.3)^(2) + (0.4)^(2)) = MB`
`implies B = (0.5)/(10) = 0.05` T