A copper wire of diameter 1.6 mm carries a current of 20A . Find the maximum magnitude of the magnetic field $\vec{B}$ (in mT) due to this current

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answer is 5.

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Detailed Solution

$\begin{array}{l} d = 1.6 mm \\ So, r = 0 .8 mm = 0 .0008 m \\ i = 20 A \\ \vec{B} = \frac{μ_{0} I}{2 πr} = \frac{4 π \times 10^{- 7} \times 20}{2 \times π \times 8 \times 10^{- 4}} = 5 \times 10^{- 3} T = 5 mT \end{array}$

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