A small bar magnet experiences a torque of 0.016 Nm when placed with its axis at 30° with an external field of 0.04 T. If the bar magnet is replaced by a solenoid of cross-sectional area of `1 cm^2` and 1000 turns but having the same magnetic moment as that of bar magnet, then the current flowing through the solenoid is

Options

A. 2A
B. 4A
C. 6A
D. 8A

Answer

D

Explanation

Given
torque on a bar magnet , `tau = 0.016 Nm , `
` theta = 30^@ ` and magnetic field, B= 0.04 T
Since , `tau = mB sin theta`
`rArr m = (2 xx 0.016)/(0.04)`
` m = 0.8 Am^2` ….(i)
For solenoid , area of cross section ,
`A = 1 cm^2 = 10^(-4) m^2`
Number of turns in solenoid ,
N = 1000
`therefore ` magnetic moment of solenoid
`m = NIA`
`rArr I = (m)/(NA)`
` = (0.8)/(1000 xx 10^(-4) )` [ from eq. (i) ]
` = 8A`