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Q.
A square conducting loop of side length $L$ carries a current $I$ . The magnetic field at the centre of the loop is
NTA AbhyasNTA Abhyas 2020Moving Charges and Magnetism
Solution:
Field due to wire $AB$ at the center,
$B_{A B}=\frac{\mu_0 I}{4 \pi\left(\frac{L}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)$
$B_{c e n t e r}=4B_{A B}$
$B_{\text {center }}=4 \frac{\mu_0 I}{4 \pi\left(\frac{L}{2}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$
$B_{c e n t e r}=2\sqrt{2}\frac{\mu _{0} I}{\pi L}$
$B_{c e n t e r} \propto \frac{1}{L}$
