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A long cylindrical wire carrying current of 10 A. has radius of $5 \mathrm{~mm}$, then find its magnetic field induction at a point $2 \mathrm{~mm}$ from the centre of the wire. Current is uniformly distributed.
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The correct answer is:
$1.6 \times 10^{-4} \mathrm{~T}$
$\begin{aligned}B & =\frac{\mu_0 I r}{2 \pi R^2} \\ B & =\frac{4 \pi \times 10^{-7} \times 10 \times 2 \times 10^{-3}}{2 \pi \times\left(5 \times 10^{-3}\right)^2}=\frac{40 \times 10^{-10}}{25 \times 10^{-6}} \\ & =1.6 \times 10^{-4} \mathrm{~T}\end{aligned}$
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