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The magnetic field at the centre of the square of side $a$ is
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$\frac{\sqrt{2} μ_{0}}{πa}$

$\frac{\sqrt{2} μ_{0} I}{3 πa}$

$\frac{2}{3} \frac{μ_{0} I}{a}$

Zero

answer is B.

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Detailed Solution

$B = 2 [\frac{μ_{0}}{4 π} \frac{\frac{2 I}{3}}{\frac{a}{2}} (\sin 45 ° + \sin 45 °) - \frac{μ_{0}}{4 π} \cdot \frac{\frac{I}{3}}{\frac{a}{2}}]$

$= \frac{\sqrt{2} μ_{0} I}{3 πa}$

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