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

Magnetic field strength `B=0.25 T`
Torque on the bar magnet `tau=4.5xx10^(-2)J`
Angle between the bar magnet and the external magnetic field, `theta=30^(@)`
Torque is related to magnetic moment (M) as :
`tau=Mbsin theta :. M=(tau)/(B sin theta) , =(4.4xx10^(-2))/(0.25 xxsin30^(@))=0.36JT^(-1)`
Hence, the magnetic moment of the magnet is `0.36JT^(-1)`