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A closely packed coil having 1000 turns has an average radius of $62.8 \mathrm{~cm}$. If current carried by the wire of the coil is $1 \mathrm{~A}$, the value of magnetic field produced at the centre of the coil will be (permeability of free space $=4 \pi \times 10^7 \mathrm{H} / \mathrm{m}$ ) nearly:
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Verified Answer
The correct answer is:
$10^{-3} \mathrm{~T}$
As, magnetic field at centre of coil,
$\begin{aligned}
B & =\frac{\mu_0 \mathrm{~N} i}{2 \mathrm{R}} \\
\mathrm{B} & =\frac{4 \pi \times 10^{-7} \times 1000 \times 1}{2 \times 62.8 \times 10^{-2}} \\
& =\frac{4 \times 3.14 \times 10^{-7} \times 10^3}{2 \times 62.8 \times 10^{-2}} \\
& =10^{-3} \mathrm{~T}
\end{aligned}$
$\begin{aligned}
B & =\frac{\mu_0 \mathrm{~N} i}{2 \mathrm{R}} \\
\mathrm{B} & =\frac{4 \pi \times 10^{-7} \times 1000 \times 1}{2 \times 62.8 \times 10^{-2}} \\
& =\frac{4 \times 3.14 \times 10^{-7} \times 10^3}{2 \times 62.8 \times 10^{-2}} \\
& =10^{-3} \mathrm{~T}
\end{aligned}$
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