Physics - Magnetism Question with Solution | TestHub

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.