Physics - Magnetism Question with Solution | TestHub

$\because \;$Magnetic dipole moment of current loop is$\overrightarrow{\text{M}}=\text{I}\overrightarrow{\text{A}}$

here area of loop A = Area of sector of circle - area of$∆$

$\frac{1}{3}\left(\pi a^2\right)-\frac{1}{2}\left(\mathrm{a\; cos}60°\right)\times \left(\mathrm{a\; sin}60°\right)\times 2$



$\text{A}=\pi {\text{a}}^2\times \frac{1}{3}-2\times \frac{1}{2}\times \frac{\text{a}}{2}\frac{\sqrt{3}}{2}\text{a}\text{.}=\frac{\pi {\text{a}}^2}{3}-\frac{\sqrt{3}}{4}{\text{a}}^2\text{,}\overrightarrow{\text{A}}=\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right){\text{a}}^2\widehat{\text{k}}$

$\because \;$torque on loop placed in uniform magnetic field is

$\overrightarrow{\tau }=\overrightarrow{\text{M}}\times \overrightarrow{\text{B}}=\text{I}\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}=\text{I}\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right){\text{a}}^2\widehat{\text{k}}\times \text{B}\widehat{\text{i}}=\text{IB}{\text{a}}^2\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right)\widehat{\text{j}}$