Magnitude of magnetic field (in $SI$ units) at the centre of a hexagonal shape coil of side $10\, cm$, $50$ turns and carrying current $I$ (Ampere) in units of $\frac{\mu_{0} I}{\pi}$ is
JEE MAIN 2020, Diffcult
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$B =\frac{6 \mu_{0} I }{4 \pi a \cos 30^{\circ}} \times 2 \sin 30^{\circ} \times 50$
$=\frac{\mu_{0} I}{\pi} \frac{150}{\sqrt{3} a }=\frac{50 \sqrt{3}}{0.1} \frac{\mu_{0} I}{\pi}$
$=500 \sqrt{3} \frac{\mu_{0} I}{\pi}$
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