Given, uniform magnetic field

                        B = 0.25 T

 

The magnitude of torque $\tau =4\text{.}5\times 10^{-2}\text{J}$

Angle between magnetic moment and magnetic field $\theta =30^{\text{o}}$

Torque experienced on a magnet placed in external magnetic field

                       $\tau =\text{M}\times \text{B}$

                        $\tau =\text{MB }\text{sin }\theta$     $\left(\because \text{A}\times \text{B}=\text{AB }\text{sin }\theta \right)$

             4.5 x 10^-2 = M x 0.25 x sin 30^o

                       $\text{M}=\frac{4\text{.}5\times 10^{-2}}{0\text{.}25\times \text{sin }30^{\text{o}}}$

                          $=\frac{4\text{.}5\times 10^{-2}\times 2}{0\text{.}25\times 1}$         $\left(\because \text{sin }30^{\text{o}}=\frac{1}{2}\right)$

                            = 0.36 J/T.