Question

A wire is bent into three successive quadrants. The quadrant $ab$ lies in the $xy$ plane. $b$ in $xy$ plane and $ca$ in the $zx$ plane. What is the magnetic moment of this system if a current $I$ flows through it? Give: $R=$radius each quadrant.

• A. $\frac{\sqrt{3}\pi r^{2}I}{4}$
• B. $\frac{\sqrt{2}\pi r^{2}I}{4}$
• C. $\frac{\pi r^{2}I}{2}$
• D. $\frac{\pi r^{2}I}{4}$

Answer 1

The correct option is A $\frac{\sqrt{3}\pi r^{2}I}{4}$

Given: current=I ,radius=R

Solution: Let us suppose that the wire in each quadrant forms a loop whose

magnetic moment we have to find :

$\overrightarrow{M}=I\times \frac{\pi r^2}{4}\left(\hat{k}\right)+I\times \frac{\pi r^2}{4}\widehat{\left(j\right)}+I\times \frac{\pi r^2}{4}\left(\hat{i}\right)$

Remember direction is given by right hand thumb rule.

$|\overrightarrow{M}|=\sqrt{(I\times \frac{\pi r^2}{4}{)}^2+(I\times \frac{\pi r^2}{4}{)}^2+(I\times \frac{\pi r^2}{4}{)}^2}$

$M=\sqrt{3}\times I\times \frac{\pi r^2}{4}$

Hence the correct option is A