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)=15 Omega`
`e_(1)=Blv_(1)=0.01xx0.1xx4=4xx10^(-3)V`
`e_(2)=Blv_(2)=0.01xx0.1xx2=2xx10^(-3)B`
Applying Kircoff's law to the left loop:
`10i_(1)+5(i_(1)-i_(2))=4xx10^(-3)`
`rArr15i_(1)-5i_(2)=4xx10^(-3) rarr(1)`
Right loop: `15i_(2)-5(i_(1)-i_(2))=2xx10^(-3)`
`rArr20i_(2)-5i_(1)=2xx10^(-3) rarr(2)`
Solving (1) and (2) gives:
`i_(1)=(18)/(55)xx10^(-3)A` and `i_(2)=(10)/(55)xx10^(-3)A`
`rArr` current through `5 Omega=i_(1)-i_(2)`
`=(8)/(55)xx10^(-3)A=(8)/(55)mA`