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A wire in the form of a square of side $a$ carries a current $i$. Then, the magnetic induction at the centre of the square is
(Magnetic permeability of free space $=\mu_0$ )
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(Magnetic permeability of free space $=\mu_0$ )
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Verified Answer
The correct answer is:
$\frac{2 \sqrt{2} \mu_d}{\pi a}$
Magnetic field produced by side ' $A B^{\prime}$ at the centre.
$\begin{aligned} B_1 & =\frac{\mu_0 j^j}{4 \pi r}\left(\sin \phi_1+\sin \phi_2\right) \\ & =\frac{\mu_{\omega^j}}{4 \pi \frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right) \\ & =\frac{2 \mu_0 i}{4 \pi a}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \Rightarrow=\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\end{aligned}$
$\therefore$ Total magnetic field at the centre
$\begin{aligned} B & =4 B_1 \\ & =4\left(\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\right) \\ & =\frac{2 \sqrt{2} \mu_0 i}{\pi a}\end{aligned}$
$\begin{aligned} B_1 & =\frac{\mu_0 j^j}{4 \pi r}\left(\sin \phi_1+\sin \phi_2\right) \\ & =\frac{\mu_{\omega^j}}{4 \pi \frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right) \\ & =\frac{2 \mu_0 i}{4 \pi a}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \Rightarrow=\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\end{aligned}$
$\therefore$ Total magnetic field at the centre
$\begin{aligned} B & =4 B_1 \\ & =4\left(\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\right) \\ & =\frac{2 \sqrt{2} \mu_0 i}{\pi a}\end{aligned}$
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