Question

The torque experienced by a loop of magnetic moment $7.5A{m}^2$ is $1.5\times {10}^{-4}Nm$. If the area vector makes an angle of ${30}^{\circ }$ with the magnetic field, find the magnitude of the magnetic field.

Answer 1

We know that,
$\tau =MBsin\theta \Rightarrow 1.5\times {10}^{-4}=7.5\times B\times sin{30}^{\circ }\Rightarrow B=\frac{1.5\times 2\times {10}^{-4}}{7.5}=4\times {10}^{-5}T$