A square loop of side length $a$ carries a current $I$. What is the magnetic field at the center of the square loop?

Solution

1. The magnetic field due to a finite wire of length $a$ at a perpendicular distance $a/2$ is $B_{segment}=\frac{{\mu }_{0}I}{4\pi (a/2)}(\sin 45^{\circ }+\sin 45^{\circ })$.
2. Simplifying, $$\begin{array}{rl}{B}_{segment}& =\frac{{\mu }_{0}I}{2\pi a}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})\\ & =\frac{{\mu }_{0}I}{2\pi a}\sqrt{2}\\ & =\frac{\sqrt{2}{\mu }_{0}I}{2\pi a}\end{array}$$.
3. Since there are four sides, the total field is $$\begin{array}{rl}B& =4\times \frac{\sqrt{2}{\mu }_{0}I}{2\pi a}\\ & =\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}\end{array}$$.
4. Hence the answer is (B).