Answer –
We are given the magnetic field strength, B = 0.25 T
Torque on the bar magnet is given by, T = 4.5×10-2J
Θ = 30° is the angle between the external magnetic field and the bar magnet
Torque is related to magnetic moment (M) by the relation-
\[T=MB\sin \theta \]
\[\therefore M=\frac{T}{B\sin \theta }=\frac{4.5\times {{10}^{-2}}}{0.25\times \sin \theta }\]
Hence, the magnetic field is given by –
\[M=0.36J{{T}^{-1}}\]