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Q.
The magnetic field intensity at the point $O$ of a loop with current $i$, whose shape is illustrated below is
Moving Charges and Magnetism
Solution:
$\vec{B}$ due to square part :-
$\vec{B}$ due to side $O A$ and $O C$ will be zero at point $O$
$\vec{B}$ due to side $A B$ and $B C$ will be equal so
$\vec{B_{1}}=2\left[\frac{\mu_{0} i}{4 \pi b}\left(\sin 45^{\circ}+0\right)\right]=\frac{\mu_{0} i}{2 \sqrt{2} \pi b} \otimes$
$\vec{B}$ due to circular part
$\vec{B_{2}}=\frac{\mu_{0} i}{2 a}\left[\frac{\left(\frac{3 \pi}{2}\right)}{2 \pi}\right]=\frac{3 \mu_{0} i}{8 a} \otimes$
$\vec{B_{\text {net }}}=\vec{B_{1}}+\vec{B_{2}}=\mu_{0}\left[\frac{3}{8 a}+\frac{1}{2 \sqrt{2} \pi b}\right]$
$=\frac{\mu_{0} i}{4 \pi}\left[\frac{3 \pi}{2 a}+\frac{\sqrt{2}}{b}\right]$
