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A rectangular loop of length $l$ and breadth $b$ is placed at distance of $x$ from infinitely long wire carrying current $i$ such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity $v$, the magnitude of the emf in the loop is : ( $\mu_0=$ permeability of free space $)$
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Verified Answer
The correct answer is:
$\frac{\mu_0 i l b v}{2 \pi x(l+x)}$
We can show the situation as :
Since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure.
Net magnetic field on the loop due to wire
$B=\frac{\mu_0 i}{2 \pi}\left(\frac{1}{x}-\frac{1}{l+x}\right)$
$=\frac{\mu_0 i l}{2 \pi x(l+x)}$
So, the magnitude of the emf in the loop
$e=v B b=\frac{\mu_0 \text { ilvb }}{2 \pi x(l+x)}$
Since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure.
Net magnetic field on the loop due to wire
$B=\frac{\mu_0 i}{2 \pi}\left(\frac{1}{x}-\frac{1}{l+x}\right)$
$=\frac{\mu_0 i l}{2 \pi x(l+x)}$
So, the magnitude of the emf in the loop
$e=v B b=\frac{\mu_0 \text { ilvb }}{2 \pi x(l+x)}$
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