A tightly wound $100$ turns coil of radius $10 \mathrm{~cm}$ carries a current of $7 \mathrm{~A}$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \mathrm{SI}$ units):
NEET 2024, Medium
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The magnitude of magnetic field due to circular coil of $N$ turns is given by
$B_c=\frac{\mu_0 i N}{2 R} $
$=\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1} $
$=4.4 \times 10^{-3} \mathrm{~T} $
$=4.4 \mathrm{mT}$
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