Question

In the figure shown ABCDEFA was 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 centre of the frame also. The loop carries current. The magnetic field at the centre is :

• A. $\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}(\hat{i}-\hat{j})$
• B. $\frac{{\mu }_{0}i}{4\pi l}(\hat{i}-\hat{j})$
• C. $\frac{\sqrt{2}{\mu }_{0}i}{\pi l}(\hat{i}+\hat{j})$
• D. $\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}(\hat{i}+\hat{j})$

Answer 1

The correct option is C $\frac{\sqrt{2}{\mu }_{0}i}{\pi l}(\hat{i}+\hat{j})$

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 $A[\hat{i}+\hat{j}]$
Calculating field due to FABC :
due to AB :
${\overrightarrow{B}}_{AB}=\frac{{\mu }_{0}i}{4\pi \left(\frac{l}{2}\right)}(sin{45}^{\circ }+sin{45}^{\circ })\hat{j}=\sqrt{2}\frac{{\mu }_{0}i}{2\pi l}\hat{j}$
Due to BC:
${\overrightarrow{B}}_{BC}=\frac{{\mu }_{0}i}{4\pi l\left(\frac{l}{2}\right)}(sin{0}^{\circ }+sin{45}^{\circ })\hat{j}=\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}\hat{j}$
Similarly due to FA: ${\overrightarrow{B}}_{FA}=\frac{{\mu }_{0}i}{2\sqrt{2}\pi l}\hat{j})$
${\overrightarrow{B}}_{PABC}=\frac{{\mu }_{0}i}{\pi l}[\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{2}]\hat{i}\Rightarrow {\overrightarrow{B}}_{FABC}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\left(\hat{j}\right)$
Hence
similarly due ot CDEF:
${\overrightarrow{B}}_{CDEF}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}\left(\hat{i}\right)\Rightarrow {\overrightarrow{B}}_{net}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}(\hat{i}+\hat{j})$