Two parallel rails with negligible resistance are `10.0 cm` apart. The are connected by a `5.0Omega` resistor. The circuit also contains two metal rods having resistances of `10.0Omega` and `15.0Omega` 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 the rails. Determine the current in the `5.0Omega` resistor.

Explanation

In the Figure `R=5.0Omega, r_1=10Omega, r_2=15Omega, e_1=4xx10^-3 V` and `e_2=2xx10^-3V`

Refer Figure b. Net resistance of the circuit `=r_2+(Rr_1)/(R+r_1)=15+(10xx5)/(10+5)=55/3Omega`
`:.` Current `i=e_2/("Net resistance") =(2xx10^-3)/(55//3)6/55xx10^-3A`
`:.` Current through `R`,
`i_1=(r_1/(R+r_1))i=(10/(10+5))(6/55xx10^-3)A`
`=4/55xx10^-3A=4/55mA`
Refer Figure c. Net resistance Net resistance of the circuit `=r_1+(Rr_2)/(R+r_2)`
`=10+(5xx15)/(5+15)=55/4Omega`
`:.` Current `i'=e_1/("Net resistance")`
`=(4xx10^-3)/(55/4)=16/55xx10^-3A`
`:.` Current through `R, i_1^'=(r_2/(R+r_2))i'=(15/(15+5))(16/55)xx10^-3A`
`=12/55mA`
From superposition principle net current through `5.0Omega` resistor is
`i_1^'=i_1=8/55mA` from `d` to `c`