A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `0*25T` experiences a torque of magnitude equal to `4*5xx10^-2J`. What is the magnitude of magnetic moment of the magnet?

Explanation

Torque experienced by the bar magnet of magnetic moment M placed in an external magnetic field at an angle with it is given by
`tau = MB sin theta `
here , ` theta = 30^@ , B = 0.25T `
`tau = 4.5 xx 10^(-2) J, M = ?`
` therefore M = (tau)/(B sin theta) = (4.5 xx 10^(-2))/(0.25 xx sin 30^@) `
or ` M = (4.5 xx 2)/(25)`
` = 0.36 JT^(-1)`