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

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$\frac{μ_{0} i}{2 π a}$

$\frac{μ_{0} i}{\sqrt{2} π a}$

$\frac{2 \sqrt{2} μ_{0} i}{π a}$

$\frac{μ_{0} i \sqrt{2}}{π a}$

answer is B.

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Detailed Solution

Magnetic field due to one side of the square at centre O

$B_{1} = \frac{μ_{0}}{4 π} \cdot \frac{2 i \sin 45^{\circ}}{a / 2}$

$\Rightarrow B_{1} = \frac{μ_{0}}{4 π} \cdot \frac{2 \sqrt{2} i}{a}$

Hence magnetic field at centre due to all side $B_{n e t} = 4 B_{1} = \frac{μ_{0} (2 \sqrt{2} i)}{π a}$

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