Current flows through uniform, square frames | vedclass

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Question

$\mathrm{i}_{1}=\frac{3}{4} \mathrm{i} ; \quad \mathrm{i}_{2}=\frac{\mathrm{i}}{4}$

Let magnetic field due to sides of square be $B_{5}$ $B_{s}=\fr

Vedclass · Accepted Answer

Vedclass · Answer

$\mathrm{i}_{1}=\frac{3}{4} \mathrm{i} ; \quad \mathrm{i}_{2}=\frac{\mathrm{i}}{4}$

Let magnetic field due to sides of square be $B_{5}$ $B_{s}=\frac{-\mu_{0} \frac{3}{4} i_{1}}{4 \pi \frac{L}{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \hat{k}+\frac{3 \mu_{0} \frac{i}{4}}{4 \pi \frac{L}{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$

$\mathrm{B}_{3}=0$

But magnetic field due to $2$ infinitely long wires is not zero so net magnetic Field is zero.

Current flows through uniform, square frames as shown. In which case is the magnetic field at the centre of the frame not zero?

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