Question

A wire of length $l$ is bent in the form of a square and carries an electric current $i$. Find the magnetic field $B$ at the centre :

• A. $\frac{8\sqrt{2}{\mu }_{0}i}{\pi l}$
• B. $\frac{4\sqrt{2}{\mu }_{0}i}{\pi l}$
• C. $\frac{2\sqrt{2}{\mu }_{0}i}{\pi l}$
• D. $\frac{\sqrt{2}{\mu }_{0}i}{\pi l}$

Answer 1

Answer: (A)

$\frac{8\sqrt{2}{\mu }_{0}i}{\pi l}$

Length of given wire=L, so the length of each side of square is l/4.

Also, we know that magnetic field by one side of square at the center is given by $B=\frac{\mu I}{4\pi a}[Sin{\theta }_1+Sin{\theta }_2]$.

But in this case both ${\theta }_1={\theta }_2$ which is 45 degree.

So magnetic strength due to one side is $B=\frac{\mu }{4\pi }\frac{I}{l}8\sqrt{2}$. And due to all four sides is 4B Which equals $\frac{8\sqrt{2}\mu I}{\pi l}$.