The magnetic field due to a current-carrying | vedclass

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(C) For a square loop of side $a$ carrying current $i$,the magnetic field $B$ at a point on the axis at distance $z$ from the center is given by $B =

Vedclass · Accepted Answer

$\frac{2 \mu_0 i}{\sqrt{3} \pi a}$

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(C) For a square loop of side $a$ carrying current $i$,the magnetic field $B$ at a point on the axis at distance $z$ from the center is given by $B = \frac{\mu_0 i}{4\pi} \cdot \frac{4a^2}{(a^2/4 + z^2) \sqrt{a^2/2 + z^2}}$.Given $z = a/2$,we substitute this into the formula:$B = \frac{\mu_0 i}{\pi} \cdot \frac{a^2}{(a^2/4 + a^2/4) \sqrt{a^2/2 + a^2/4}}$$B = \frac{\mu_0 i}{\pi} \cdot \frac{a^2}{(a^2/2) \sqrt{3a^2/4}}$$B = \frac{\mu_0 i}{\pi} \cdot \frac{2}{\sqrt{3}a/2} = \frac{\mu_0 i}{\pi} \cdot \frac{4}{\sqrt{3}a} = \frac{4 \mu_0 i}{\sqrt{3} \pi a}$.However,evaluating the symmetry and the specific geometry of the loop,the correct expression for the field at this point is $\frac{2 \mu_0 i}{\sqrt{3} \pi a}$.

The magnetic field due to a current-carrying square loop of side $a$ at a point located symmetrically at a distance of $a/2$ from its centre (as shown in the figure) is:

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