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 xz-plane and the other half lies in the yz-plane. The origin O is center of the frame also. The loop carries current i. The magnetic field at the center is

Answer 1

Magnetic field due to AB:

${\overrightarrow{B}}_{AB}=\frac{{\mu }_{0}i}{4\pi \left(\frac{l}{2}\right)}(sin{45}^0+sin{45}^0)\hat{j}$

$\Rightarrow {\overrightarrow{B}}_{AB}=\sqrt{2}\frac{{\mu }_{0}i}{2\pi l}\hat{j}$

Magnitude field due to BC:

${\overrightarrow{B}}_{BC}=\frac{{\mu }_{0}i}{4\pi \left(\frac{l}{2}\right)}(sin{45}^0+sin{45}^0)\hat{j}=\frac{{\mu }_{0}i}{\sqrt{2}2\pi l}\hat{j}$

Similarly, magnetic field due to FA

${\overrightarrow{B}}_{FA}=\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}\hat{j}$
Hence,

${\overrightarrow{B}}_{FABC}={\overrightarrow{B}}_{FA}=+{\overrightarrow{B}}_{AB}+{\overrightarrow{B}}_{BC}$

$\Rightarrow {\overrightarrow{B}}_{FABC}=\frac{{\mu }_{0}i}{\pi l}[\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{2}]\hat{j}$

$\Rightarrow {\overrightarrow{B}}_{FABC}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\left(\hat{j}\right)$

Similarly,
We can find magnetic field due to CDEF

$\Rightarrow {\overrightarrow{B}}_{CDEF}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\left(\hat{j}\right)$

Resultant magnetic field,

$\Rightarrow {\overrightarrow{B}}_{net}={\overrightarrow{B}}_{FABC}+{\overrightarrow{B}}_{CDEF}$

$\Rightarrow {\overrightarrow{B}}_{net}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}(\hat{i}+\hat{j})$

Hence, option (C) is correct.