Question

In the figure shown, ABCDEFA is a square loop of side l, but is folded in two equal parts so that half of it lies in the xz plane and the other half lies in the yz plane. The origin O is the centre of the frame also. The loop carries current i. The magnetic field at the centre is $\frac{\sqrt{x}{\mu }_{0}i}{\pi l}\left(\hat{i}+\hat{j}\right)$ where x is . (Answer upto 2 decimal places)

Answer 1

Due to FABC, the magnetic field at O is along y-axis and due to CDEF, the magnetic field is along x-axis.



Hence the field will be of the form A$\left(\hat{i}+\hat{j}\right]$Calculating the field due to FABC:

Due to AB:
$\stackrel{}{B_{AB}}=\frac{{\mu }_{0}i}{4\pi \frac{l}{2}}(\sin {45}^0+\sin {45}^0)\hat{j}=\sqrt{2}\frac{{\mu }_{0}i}{2\pi l}\hat{j}DuetoBC:\stackrel{}{B_{BC}}=\frac{{\mu }_{0}i}{4\pi l\frac{l}{2}}(\sin 0^0+\sin {45}^0)\hat{j}=\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}\hat{j}SimilarlyduetoFA:\stackrel{}{B_{FA}}=\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}\hat{j}Hence,\stackrel{}{B_{FABC}}=\frac{{\mu }_{0}i}{\pi l}(\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{2})\hat{j}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\hat{j}SimilarlyduetoCDEF:\stackrel{}{B_{CDEF}}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\hat{i}Hence,net\overrightarrow{B}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}(\hat{i}+\hat{j})$