Question

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

Answer 1

$E_1=B{v}_{1}l=0.01\times 8\times 0.1$
$8\times {10}^{-3}V$
$E_2=B{v}_{2}l=4\times {10}^{-3}V$
$r_1=10\Omega$
$r_2=15\Omega$
For external resistance of $5\Omega$, the two batteries are in parallel.
Equivalent emf
$E==\frac{E_1/r_1-E_2/r_2}{1/r_1+1/r_2}=\left[\frac{\left(8/10\right)-\left(4/15\right)}{\left(1/10\right)+\left(1/15\right)}\right]\times {10}^{-3}V$
$=3.2\times {10}^{-3}V$
$r=\frac{r_1{r}_2}{r_1{+r}_2}=\frac{10\times 15}{10+15}=6\Omega$
$\therefore i=\frac{E}{R+r}=\left(\frac{3.2\times {10}^{-3}}{5+6}\right)=0.3\times {10}^{-3}A=0.3mA$