The magnetic field due to a square loop of side a carrying a current $I$ at its centre is

\frac{μ_{0} i}{2 a}

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\frac{μ_{0} i}{\sqrt{2} π a}

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\frac{μ_{0} i}{2 π a}

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\sqrt{2} \frac{μ_{0} i}{π a}

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Solution

The correct option is D $\sqrt{2} \frac{μ_{0} i}{π a}$
$B = [\frac{m u_{0} i}{4 π a} (cos 45^{o} - cos 135^{o})] \times 4$

$= \frac{μ_{0} i}{π a} \frac{2}{\sqrt{2}}$

$= \sqrt{2} \frac{μ_{0} i}{π a}$

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The magnetic field due to a square loop of side a carrying a current I at its centre is

The correct option is D √2μ0iπaB=[mu0i4πa(cos45o−cos135o)]×4=μ0iπa2√2=√2μ0iπa

The magnetic field due to a square loop of side a carrying a current $I$ at its centre is

The correct option is D $\sqrt{2} \frac{μ_{0} i}{π a}$
$B = [\frac{m u_{0} i}{4 π a} (cos 45^{o} - cos 135^{o})] \times 4$

$= \frac{μ_{0} i}{π a} \frac{2}{\sqrt{2}}$

$= \sqrt{2} \frac{μ_{0} i}{π a}$