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(C) When the loop moves downwards with velocity $V$ in a magnetic field $B$,an induced electromotive force $(EMF)$ is generated across the horizontal

Vedclass · Accepted Answer

$\frac{mg R}{B^2 \ell^2}$

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(C) When the loop moves downwards with velocity $V$ in a magnetic field $B$,an induced electromotive force $(EMF)$ is generated across the horizontal side of length $\ell$ inside the magnetic field.The induced $EMF$ is given by $\varepsilon = B \ell V$.The induced current in the loop is $I = \frac{\varepsilon}{R} = \frac{B \ell V}{R}$.The magnetic force acting on the horizontal side of the loop is $F_m = I \ell B = \left( \frac{B \ell V}{R} \right) \ell B = \frac{B^2 \ell^2 V}{R}$.For the loop to fall freely with a constant velocity $V$,the downward gravitational force must be balanced by the upward magnetic force:$mg = F_m$$mg = \frac{B^2 \ell^2 V}{R}$Solving for $V$,we get:$V = \frac{mg R}{B^2 \ell^2}$.

$A$ long rectangular conducting loop of width ' $\ell$ ',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

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