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=5.0 Omega, r_(1) = 10 Omega, r_(2) = 15 Omega`,
`e_(1) =Blv_(1) = 0.01 xx 0.1 xx 4 = 4 xx 10^(-3)V`
`e_(2) =Blv_(2) = 0.01 xx 0.1 xx 2 = 2 xx 10^(-3)V`
Applying Kirchoff.s law to the left loop :
`10 i_(1) +5(i_(1) - i_(2)) = 4 xx 10^(-3) implies 15i_(1) -5i_(2) = 4 xx 10^(-3) to (1)`
Right loop : `15i_(1) - 5(i_(1) - i_(2)) = 2 xx 10^(-3)`
`implies 20i_(2) -5i_(1) = 2 xx 10^(-3) to (2)`
Solving (1) and (2) gives : `i_(1)=(18)/(55) xx 10^(-3) A` and `i_(2) = (10)/(55) xx 10^(-3)` A
`implies` Current through `5 Omega=i_(1)-i_(2)=(8)/(55) xx 10^(-3)A=(8)/(55) mA`.