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