Question

A wire in the form of a square of side a carries a current $i$. Then, the magnetic induction at the centre of the square wire is (magnetic permeability of free space $={\mu }_0$ ).

• A. $\frac{{\mu }_{0}i}{2\pi a}$
• B. $\frac{{\mu }_{0}i\sqrt{2}}{\pi a}$
• C. $\frac{2\sqrt{2}{\mu }_{0}i}{\pi a}$
• D. $\frac{{\mu }_{0}i}{\sqrt{2}\pi a}$

Answer 1

Answer: (C)

Magnetic field due to one side of the square at centre $O$.

$B_1=\frac{{\mu }_0}{4\pi }\cdot \frac{2i\sin 45^{\circ }}{a/2}$
$\Rightarrow B_1=\frac{{\mu }_0}{4\pi }\cdot \frac{2\sqrt{2}i}{a}$
Hence, magnetic field at centre due to all sides
$B_{\text{net }}=4B_1=\frac{{\mu }_0(2\sqrt{2}i)}{\pi a}$