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=0.25T`
Torque on the bar magnet T=`4.5xx10^(-2)J`
Angine between the bar magnet and the external magnetic field `tehta=30^(@)`
torque is related to magnetic moment (M) as:
`T=MB sin theta`
`therefore M=(T)/(B sin theta)`
`=(4.5xx10^(-2))/(0.25xxsin30^(@))=0.36J T^(-1)`
Hence, the magnetic moment of the magnet is `0.36J T^(-1)`