Two parallel rails with negligible resistance are 10.0 cm apart. They are connected by a $5\Omega$ resistor. The circuit also contains two metal rods having resistances of $10.0\Omega$ and $15.0\Omega$ along the rails. The rods are pulled away from the resistor at constant speeds $4.00 m/s$ and $2.00 m/s$ respectively. A uniform magnetic field of magnitude $0.01 T$ is applied perpendicular to the, plane of rails. Determine the current in the $5.0 \Omega$ resistor.

In the figures $R=5.0\Omega , r_1=10\Omega , r_2=\Omega , e_1=4\times {10}^{-3}V and e_2=2\times {10}^{-3}V$

Refer Figure (b) Net resistance of the circuit $=r_2+\frac{R{r}_1}{R+r_1}=15+\frac{10\times 15}{10+15}=\frac{55}{3}\Omega$

current, $i=\frac{e_2}{Net resis\tan ce}=\frac{2\times {10}^{-3}}{{\displaystyle \raisebox{1ex}{$55$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{6}{55}\times {10}^{-3}A$

Current through R,

$i_1=\left(\frac{r_1}{R+r_1}\right)i=\left(\frac{10}{10+5}\right)\left(\frac{6}{55}\times {10}^{-3}\right)A$

$=\frac{4}{55}\times {10}^{-3}A=\frac{4}{55}mA$

Refer figure (b) Net resistance of the circuit $=r_1+\frac{R{r}_2}{R+r_2}=10+\frac{5\times 15}{5+15}=\frac{55}{4}\Omega$

Current,

 $$\begin{gathered} i\text{'}=\frac{e_2}{Net resis\tan ce} \\ =\frac{4\times {10}^{-3}}{{\displaystyle \raisebox{1ex}{$55$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{16}{55}\times {10}^{-3}A \end{gathered}$$

Current through R,

$$\begin{gathered} i{\text{'}}_1=\left(\frac{r_2}{R+r_2}\right)i=\left(\frac{15}{15+5}\right)\left(\frac{16}{55}\times {10}^{-3}\right)A \\ =\frac{12}{55}mA \end{gathered}$$

From superposition principle net current through $5.0 \Omega$ resistor is

$i{\text{'}}_1-i_1=\frac{8}{55}mA$ from $d$ to $c$.