Two identical current carrying coils with same centre are placed with their planes perpendicular to each other. If current $I=\sqrt{2} \mathrm{~A}$ and radius of the coil is $\mathrm{R}=1 \mathrm{~m}$, then magnetic field at centre is equal to ( $\mu_0=$ permeability of free space)

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Aicharya · Accepted Answer

$\begin{aligned} i & =\sqrt{2} A \ B_{n e t} & =\sqrt{B_1^2+B_2^2} \ & =\sqrt{\left(\frac{\mu_0 i}{2 R}ight)^2+\left(\frac{\mu_0 i}{2 R}ight)^2}=\sqrt{2\left(\frac{\mu_0 \mathrm{i}}{2 R}ight)^2} \ & =\sqrt{2 \frac{\mu_0^2 \mathrm{i}^2}{4 \mathrm{R}^2}}=\sqrt{\frac{1}{2}\left(\frac{\mu_0 \mathrm{i}}{R}ight)^2} \ & =\frac{1}{\sqrt{2}} \frac{\mu_0 \mathrm{i}}{\mathrm{R}}=\frac{1}{\sqrt{2}} 	imes \frac{\mu_0 \sqrt{2}}{1}=\mu_0\end{aligned}$

Two identical current carrying coils with same centre are placed with their planes perpendicular to each…

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