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A long, rectangular conducting loop of width $l$ mass $m$ and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $v$ so that it may continue to fall freely. The velocity $v$ is $(g=$ acceleration due to gravity)
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The correct answer is:
$\frac{m g R}{B^2 l^2}$
The motional emf is given by: $V=B v l$
Current in the loop,
$i=\frac{V}{R}=\frac{B v l}{R}$
During free fall the electromagnetic force felt by the loop balances its weight, and since the loop is long so we assume its falling with a terminal velocity.
$\begin{aligned}
& \therefore i l B=m g \\
& \Rightarrow B\left(\frac{B v l}{R}\right) l=m g \\
& \Rightarrow v=\frac{m g R}{B^2 l^2}
\end{aligned}$
Current in the loop,
$i=\frac{V}{R}=\frac{B v l}{R}$
During free fall the electromagnetic force felt by the loop balances its weight, and since the loop is long so we assume its falling with a terminal velocity.
$\begin{aligned}
& \therefore i l B=m g \\
& \Rightarrow B\left(\frac{B v l}{R}\right) l=m g \\
& \Rightarrow v=\frac{m g R}{B^2 l^2}
\end{aligned}$
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