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

Here, two conductors are moving in uniform magnetic field. So we will use the motional approach. The rod ab will act as a source of emf,
`e_(1)=Bvl=(0.01)(4)(0.1)=4xx10^(-3)V`
and internal resistance, `r_(1)=10Omega` ltBrgt Similarly, rod ef will also act as a source of emf,
`e_(2)=(0.01)(2)(0.1)=2xx10^(-3)V`
and internal resistance, `r_(2)=15Omega`
From right hand rule we can see that
`V_(b)gtV_(a)and V_(e)gtV_(f)`
Now, either by applying Kirchhoff's laws or applying principle of superposition (discussed in the chapter of current electricity) we can find current through `5Omega` resistor. We will here use the superposition principle. You solve it by using Kirchhoff's laws.
Now, in the figures, `R=5Omega,r_(1)=10Omega,r_(2)=15Omega`,
`e_(1)=4xx10^(-3)Vand e_(2)=2xx10^(-3)V`.

(b) Net resistance of the circuit
`=r_(2)+(Rr_(1))/(R+r_(1))=15+(10xx5)/(10+5)=(55)/(3)Omega`
`therefore"Current, i"=(e_(2))/("Net resistance")`
`=(2xx10^(-3))/(55//3)=(6)/(55)xx10^(-3)A`
`therefore` Current through R,
`i_(1)=((r_(1))/(R+r_(1)))i=((10)/(10+5))((6)/(55)xx10^(-3))A`
`=(4)/(55)xx10^(-3)A=(4)/(55)mA`
(c) Net resistance of circuit `r_(1)+(Rr_(2))/(R+r_(2))=10+(5xx15)/(5+15)=(55)/(4)Omega`
`therefore"Current,i'"=(e_(1))/("Net resistance")=(4xx10^(-3))/(55//4)=(16)/(35)xx10^(-3)A`
`therefore` Current through R, `i_(1)=((r_(2))/(R+r_(2)))i'`
`=((15)/(15+5))((16)/(55))xx10^(-3)A=(12)/(55)mA`
From superposition principle net current through `5Omega` resistor is
`i_(1)'-i_(1)=(8)/(55)"mA from d to c"`