Physics - EMI/AC Question with Solution | TestHub

Magnetic field due to finite length of a wire, $$B=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}\left(\sin {\varphi }_1+\sin {\varphi }_2\right)$$
Here, ${\varphi }_1=0^{\circ },\varphi =45^{\circ }$ $$\begin{array}{l}B=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}\left(\sin 0^{\circ }+\sin 45^{\circ }\right)=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{\sqrt{2l}}\\ \Rightarrow B=\frac{\sqrt{2}{\mu }_{0}i}{8\pi l}\end{array}$$