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 $O$.

$B_{1} =\frac{\mu_{0}}{4 \pi} \cdot \frac{2 i \sin 45^{\circ}}{a / 2}$
$\Rightarrow B_{1} =\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \sqrt{2} i}{a}$
Hence, magnetic field at centre due to all sides
$B_{\text {net }}=4 B_{1}=\frac{\mu_{0}(2 \sqrt{2} i)}{\pi a}$