Question

A wire is bent into three segments, each of radius $r$. Each segment is a quadrant of a circle. The first segment lies in $xy-$ plane, the second in $yz-$plane and the third in $xz-$plane. The centre of each quadrant is at origin. If a current $I$ flows through each quadrant, the net magnetic field at origin will be

• A. $\frac{{\mu }_{o}I}{4r}$
• B. $\frac{{\mu }_{o}I}{8r}$
• C. $\frac{\sqrt{2}{\mu }_{o}I}{8r}$
• D. $\frac{\sqrt{3}{\mu }_{o}I}{8r}$

Answer 1

The correct option is D $\frac{\sqrt{3}{\mu }_{o}I}{8r}$

The magnitude of the magnetic field at the center of one qudrant is,

$B=\frac{{\mu }_{o}I}{4\pi r^2}\frac{\pi r}{2}=\frac{{\mu }_{o}I}{8r}$

Since, the magnitude of magnetic field due to each arc is the same at origin,

$\therefore B_{xy}=B_{yz}=B_{xz}=\frac{{\mu }_{o}I}{8r}$

Using the "Fleming right-hand thumb rule", the direction of magnetic fields are,

${\overrightarrow{B}}_{xy}=B_{xy}(-\hat{k})$,

${\overrightarrow{B}}_{yz}=B_{yz}(-\hat{i})$, and

${\overrightarrow{B}}_{xz}=B_{xz}(-\hat{j})$

So, the net magnetic field at origin,

$B_{net}=\sqrt{B_{xy}^2+B_{yz}^2+B_{xz}^2}=\frac{{\mu }_{o}I}{8r}\sqrt{3}$

Hence, option $\left(d\right)$ is the correct answer.