A bar magnet of length `10cm` and having the pole strength equal to `10^(-3)` weber is kept in a magnetic field having magnetic induction (B) equal to `4pixx10^(-3)` Tesla. It makes an angle of `30^(@)` with the direction of mwgnetic induction. The value of the torque acting on the magnet is
`(mu_(0)=4pixx10^(-7) weber//ampxxm)`

Options

A. `2pi xx 10^(-7) N xx m`
B. `2pi xx 10^(-5) N xx m`
C. `0.5 N xx m`
D. `0.5 xx 10^(2) N xx m`

Answer

A

Explanation

`q_(m)= 10^(-3) wb, l= 0.1m, B= 4pi xx 10^(-3)T`
`theta= 30^(@)`
`tau= mB sin theta= q_(m)l B sin 30^(@)`
`= 10^(-3) xx 0.1 xx 4pi xx 10^(-3) xx (1)/(2) = 2pi xx 10^(-7) Nm`