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

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$\frac{μ_{0} i}{2 πa}$

$\frac{μ_{0} i \sqrt{2}}{πa}$

$\frac{2 \sqrt{2} μ_{0} i}{πa}$

$\frac{μ_{0} i}{\sqrt{2} πa}$

answer is C.

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Detailed Solution

Magnetic induction field due to the side AB at the centre ‘O’

$B = \frac{μ_{0} i}{4 π (\frac{a}{2})} (2 \sin 45^{\circ})$

$B_{R} = 4 B = 4 \times \frac{μ_{0} i}{4 π (\frac{a}{2})} (2 \sin 45^{0})$

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