A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.35 T experiences a torque of magnitude equal to `4.5xx10^(-2)Nm`. The magnitude of magnetic moment of the given magnet is

Options

A. `0.26JT^(-1)`
B. `2.6JT^(-1)`
C. `0.26JT^(-1)`
D. `0.026JT^(-1)`

Answer

C

Explanation

Here, `theta=30^(@), B=0.35T`
`tau=4.5xx10^(-2)Nm`
then, `m=(tau)/(B sin theta)=(4.5xx10^(-2))/(0.35xxsin 30^(@))`
`m=(4.5xx10^(-2))/(0.35xx(1)/(2))=(2xx4.5xx10^(-2))/(0.35)`
`m=0.26JT^(-1)`