A very long wire $ABDMNDC$ is shown in figure carrying current $I. AB$ and $BC$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $DMND$ of radius $R. AB.$ $\mathrm{BC}$ parts are tangential to circular turn at $\mathrm{N}$ and $D$. Magnetic field at the centre of circle is
JEE MAIN 2020, Diffcult
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We say we have $3$ parts $(A, B, C)$
$\mathrm{B}=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}}$
$=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\sin 90^{\circ}-\sin 45^{\circ}\right) \otimes+\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}} \odot+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\sin 45^{\circ}+\sin 90^{\circ}\right) \odot$
$=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\sin 45^{\circ}+\pi\right)$
$=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)$
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