Question

A wire loop carrying current is placed in the x-y plane as shown in figure. An external uniform magnetic induction field $\overrightarrow{B}=B\hat{i}$ is applied. If torque on wire is $\overrightarrow{\tau }=X\ast I{a}^{2}B[\frac{\pi }{3}-\frac{\sqrt{3}}{4}]\hat{j}Nm$. Find X?

Answer 1

Area of section$=A=$area of sector$-$ area of $△PMN$
$=\frac{1}{2}{\left(radius\right)}^2\times$ (angle subtended in radian) $-\frac{1}{2}MN\times PS$
$=\frac{1}{2}\times a^2\times \frac{2\pi }{3}-\frac{1}{2}\times 2a\sin 60\times a\cos 60$
$=a^2\frac{\pi }{3}-a^2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}$
$A=a^2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$
Magnetic moment due to loop$=NIA$
$=1\times I\times a^2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})\hat{k}$
Direction by right hand screw rule
Torque$=\overrightarrow{M}\times \overrightarrow{B}$
$=I{a}^2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})B(\hat{k}\times \hat{i})$
$\overrightarrow{\tau }=I{a}^{2}B(\frac{\pi }{3}-\frac{\sqrt{3}}{4})\hat{j}$
But given $\overrightarrow{\tau }=X\times I{a}^{2}B(\frac{\pi }{3}-\frac{\sqrt{3}}{4})\hat{j}$
$\Rightarrow X=1$