A bar magnet has a magnetic moment of `200 A m`. The magnet is suspended in a magnetic field of `0.30 N A^(-1) m^(-1)`. The torque required to rotate the magnet from its equilibrium position through an angle of `30^(@)`, will be

Options

A. `30 N m`
B. `30sqrt(30) N m`
C. `60 N m`
D. `60sqrt(3) N m`

Answer

A

Explanation

Torque experienced by a magnet suspended in a uniform magnetic field `B` is given by
`tau=MBsintheta`
Here, `M=200"A m"^(2)`, B= 0.30 N `A^(-1) m^(-1)` and `theta= 30^(@)`
`thereforetau=200xx0.30xxsin30^(@)`
`tau=30"N m"`