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 = 15Omega`
` 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 :
`10i + 5(i_1 - i_2) = 4.10^(-3)`
`rArr 15 i_1 + (5(i_1 - i_2) = 4.10^(-3)`
` rArr 15i_1 - 5i_2 = 4 xx 10^(-3) to (1) `
Right loop `: 15i_2 - 5(i_1 - i_2) = 2.10^(-3)`
`rArr 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`
`rArr` current through `5Omega = i_1 - i_2`
` = 8/55 xx 10^(-3) A = 8/55 mA`