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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
  • A$\frac{\mu_{0} I}{2 R}$
  • B$\frac{\mu_{0} I}{2 \pi R}(\pi+1)$
  • C$\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)$
  • D$\frac{\mu_{0} I}{2 \pi R}\left(\pi-\frac{1}{\sqrt{2}}\right)$
JEE MAIN 2020, Diffcult
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