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A wire in the form of a square of side ‘2m’ carries a current 2A. Then the magnetic induction at the centre of the square wire is (magnetic permeability of free space $= μ_{0})$

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$\frac{μ_{0} \sqrt{2}}{π}$

$\frac{μ_{0}}{\sqrt{2 π}}$

$\frac{2 \sqrt{2} μ_{0}}{π}$

$\frac{μ_{0}}{2 π}$

answer is C.

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Detailed Solution

$B = 4 [\frac{μ_{0}}{4 π} \frac{I}{d} (\sin 45^{\circ} + \sin 45^{\circ})]$

$B = 4 \times \frac{μ_{0}}{4 π} \times \frac{2}{1} \times 2 \times \frac{1}{\sqrt{2}} \Rightarrow B = \frac{μ_{0}}{π} (2 \sqrt{2})$

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