Question

Find the magnetic field B at the center of a rectangular loop of length l and width b,carrying a current i.

Answer 1

B at P due to AB, $\frac{{\mu }_{0}i}{4\pi \left(\frac{b}{2}\right)}.\frac{2l}{\sqrt{l^2+4\left(\frac{b}{2}\right)}}=\frac{{\mu }_{0}i}{\pi b}.\frac{l}{\sqrt{l^2+b^2}}$

B at P due to BC, $\frac{{\mu }_{0}i}{4\pi \left(\frac{l}{2}\right)}.\frac{2b}{\sqrt{b^2+l^2}}=\frac{{\mu }_{0}i}{\pi l}.\frac{b}{\sqrt{l^2+b^2}}$

B at P due to CD, $\frac{{\mu }_{0}i}{4\pi \left(\frac{b}{2}\right)}.\frac{2l}{\sqrt{l^2+b^2}}=\frac{{\mu }_{0}i}{\pi b}.\frac{b}{\sqrt{l^2+b^2}}$

B at P due to DA, $\frac{{\mu }_{0}i}{4\pi \left(\frac{l}{2}\right)}.\frac{2b}{\sqrt{l^2+b^2}}=\frac{{\mu }_{0}i}{\pi b}.\frac{b}{\sqrt{l^2+b^2}}$

So, net B=$2[\frac{{\mu }_{0}i}{\pi b}.\frac{l}{\sqrt{l^2+b^2}}+\frac{{\mu }_{0}i}{\pi l}.\frac{b}{\sqrt{l^2+b^2}}]$

=$2\frac{{\mu }_{0}i}{\pi \sqrt{l^2+b^2}}[\frac{l}{b}+\frac{b}{l}]$

=$\frac{2{\mu }_{0}i}{\pi \sqrt{l^2+b^2}}\frac{b^2+l^2}{bl}$

=$\frac{2{\mu }_{0}i\sqrt{b^2+l^2}}{\pi bl}$