Question

A current $I$ is flowing through the sides of the rectangle as shown below. The magnetic field at the crossing point of the diagonal is

• A. $\frac{4{\mu }_{0}I}{\pi a}$
• B. $\frac{4{\mu }_{0}I}{\sqrt{3}\pi a}$
• C. $\frac{{\mu }_{0}I}{\pi a}$
• D. $\frac{{\mu }_{0}I}{3\pi a}$

Answer 1

The correct option is B $\frac{4{\mu }_{0}I}{\sqrt{3}\pi a}$


$\text{PQRS}$ is a rectangle and $O$ is the intersection point of its diagonals.

From $\Delta \text{OTQ}$

$\tan \theta =\frac{TQ}{OT}=\frac{\sqrt{3}a/2}{a/2}=\sqrt{3}$

$\therefore \theta ={\tan }^{-1}\left(\sqrt{3}\right)={60}^{\circ }$

Now calculating magnetic field due to $PQ$ part at point $O$,

$B_{PQ}=\frac{{\mu }_0}{4\pi }\frac{I}{OT}[\sin {\theta }_1+\sin {\theta }_2]$

$B_{PQ}=\frac{{\mu }_0}{4\pi }\frac{I}{\left(a/2\right)}[\sin {60}^{\circ }+\sin {60}^{\circ }]$

$B_{PQ}=\frac{{\mu }_0}{4\pi }\frac{I}{a}2\sqrt{3}$



From $\Delta \text{ROU}$

$\tan \alpha =\frac{RU}{OU}=\frac{a/2}{\sqrt{3}a/2}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^{\circ }$

Magnetic field due to $QR$ part at point $O$ is given by

$B_{QR}=\frac{{\mu }_0}{4\pi }\frac{I}{\left(\sqrt{3}a/2\right)}[\sin {30}^{\circ }+\sin {30}^{\circ }]$

$B_{QR}=\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}a}$

From right- hand thumb rule, it is clear that direction of magnetic field $B$ at the point of intersection due to all the sides will be same either perpendicular inside or outside the plane of wire.

Also, $B$ at $\text{O}$ due to wire $\text{(PQ and SR)}$ is same and $\text{(QR and PS)}$ is also same.

$\therefore B_{\text{net}}=2(B_{PQ}+B_{SP})$

$B_{\text{net}}=2(\frac{{\mu }_0}{4\pi }\frac{I}{a}2\sqrt{3}+\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}a})$

$B_{net}=2\frac{{\mu }_0}{4\pi }\frac{I}{a}(2\sqrt{3}+\frac{2}{\sqrt{3}})$

$B_{net}=\frac{4{\mu }_{0}I}{\sqrt{3}\pi a}$

Hence, option $\left(b\right)$ is correct.