The current flowing through resistance $R$ if both the rods move with the same speed $v$ toward 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{B l v}{2 (R + r)}

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Solution

The correct option is D $\frac{B l v}{R + (r / 2)}$
The EMF induced by moving one slider: $| E | = B v L$
If both of them in the same direction: $E_{e q} = E = B v L$
$R_{e q} = R + \frac{r . r}{r + r} = R + r / 2$
$i = \frac{V}{R_{e q}} = \frac{B v l}{R + r / 2}$

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The current flowing through resistance R if both the rods move with the same speed v toward right is

The correct option is D BlvR+(r/2)The EMF induced by moving one slider: |E|=BvLIf both of them in the same direction: Eeq=E=BvLReq=R+r.rr+r=R+r/2i=VReq=BvlR+r/2

The current flowing through resistance $R$ if both the rods move with the same speed $v$ toward right is

The correct option is D $\frac{B l v}{R + (r / 2)}$
The EMF induced by moving one slider: $| E | = B v L$
If both of them in the same direction: $E_{e q} = E = B v L$
$R_{e q} = R + \frac{r . r}{r + r} = R + r / 2$
$i = \frac{V}{R_{e q}} = \frac{B v l}{R + r / 2}$