The current flowing through resistance $R$ if the rod $M N$ moves toward left and the rod $M^{'} N^{'}$ moves toward the right is

\frac{B l v}{R + (r / 2)}

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\frac{2 B l v}{R + r}

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zero

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\frac{3 B l v}{2 (R + r)}

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Solution

The correct option is C zero
he EMF induced by moving one slider
$| E | = B v L$
If both of them in the opposite direction
$E_{e q} = E - E = 0$
$i = 0$

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The current flowing through resistance R if the rod MN moves toward left and the rod M′N′ moves toward the right is

The correct option is C zerohe EMF induced by moving one slider|E|=BvLIf both of them in the opposite directionEeq=E−E=0i=0

The current flowing through resistance $R$ if the rod $M N$ moves toward left and the rod $M^{'} N^{'}$ moves toward the right is

The correct option is C zero
he EMF induced by moving one slider
$| E | = B v L$
If both of them in the opposite direction
$E_{e q} = E - E = 0$
$i = 0$