Question

A square loop of edge $b$ made of a uniform wire. Current $I$ enters the loop at point $A$ and leaves at $C$. Find the net magnetic field at the point $P$ which is on the perpendicular bisector of $AD$ at a distance of $b/3$ from it.

• A. $\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}-\frac{1}{10})⨂$
• B. $\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}-\frac{1}{10})⨀$
• C. $\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}+\frac{1}{10})⨂$
• D. $\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}+\frac{1}{10})⨀$

Answer 1

The correct option is B $\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}-\frac{1}{10})⨀$


Magnetic fields due to part $AB$ and $DC$ at $P$ will cancel each other.

Let $B_1$ be the field due to side $\text{AD}$.


The magnetic field due to a finite wire is,

$B_1=\frac{{\mu }_{0}i}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

Here, $\sin {\theta }_1=\sin {\theta }_2=\sin \alpha$ and $i=I/2$

$\therefore B_1=\frac{{\mu }_{0}I}{4\pi d}\left(\sin \alpha \right)...\left(1\right)$

Also, $\sin \alpha =\frac{b/2}{\left(\sqrt{13}b\right)/6}$

$\Rightarrow \sin \alpha =\frac{3}{\sqrt{13}}$

From $\left(1\right)$, we get,

$B_1=\frac{{\mu }_{0}I}{4\pi d}\frac{3}{\sqrt{13}}=\frac{{\mu }_{0}I}{4\pi \left(b/3\right)}\frac{3}{\sqrt{13}}$

$\therefore B_1=\frac{9{\mu }_{0}I}{4\sqrt{13}\pi b}⨀$

Similarly, magnetic field due to side $\text{BC}$,

As, $\sin {\theta }_1=\sin {\theta }_2=\sin \theta$ and $i=I/2$

$\therefore B_2=\frac{{\mu }_{0}I}{4\pi d}\left(\sin \theta \right)...\left(2\right)$

Also, $\sin \theta =\frac{b/2}{\sqrt{(2b/3{)}^2+(b/2{)}^2}}=\frac{3}{5}$

From $\left(1\right)$, we get,

$B_2=\frac{{\mu }_{0}I}{4\pi d}\frac{3}{\sqrt{13}}=\frac{{\mu }_{0}I}{4\pi \left(b/3\right)}\frac{3}{5}$

$\therefore B_2=\frac{9{\mu }_{0}I}{40\pi b}⨂$

So, net magnetic field,

$B_{net}=B_1-B_2=\frac{9{\mu }_{0}I}{4\pi b}(\frac{1}{\sqrt{13}}-\frac{1}{10})⨀$

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