Two parallel rails with negligible resistance are $10.0 c m$ apart and are connected by a $5.00 Ω$ resistor. The circuit also contains two metal rods having resistance of $10.0 Ω$ and $15.0 Ω$ sliding along the rails. The rods are pulled away from the resistor at a constant speeds $8.00 m / s$ and $4.00 m / s$ respectively. A uniform magnetic field $0.010 T$ is applied perpendicular to the plane of the rails. Determine the current in the $5.00 Ω$ resistor.

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Solution

$E_{1} = B v_{1} l = 0.01 \times 8 \times 0.1$
$8 \times 10^{- 3} V$
$E_{2} = B v_{2} l = 4 \times 10^{- 3} V$
$r_{1} = 10 Ω$
$r_{2} = 15 Ω$
For external resistance of $5 Ω$ , the two batteries are in parallel.
Equivalent emf
$E == \frac{E_{1} / r_{1} - E_{2} / r_{2}}{1 / r_{1} + 1 / r_{2}} = [\frac{(8 / 10) - (4 / 15)}{(1 / 10) + (1 / 15)}] \times 10^{- 3} V$
$= 3.2 \times 10^{- 3} V$
$r = \frac{r_{1} r_{2}}{r_{1} {+ r}_{2}} = \frac{10 \times 15}{10 + 15} = 6 Ω$
$∴ i = \frac{E}{R + r} = (\frac{3.2 \times 10^{- 3}}{5 + 6}) = 0.3 \times 10^{- 3} A = 0.3 m A$

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