Consider the situation shown in . The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of `5 cm s^(-1)` . Find the current in the `10 Omega` resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.

Explanation

Here B=1T
`v=5xx10^(-2)m//s R+10 Omega`
`(a) When the switch is thrown to the middle rail `
`E=Bvl`
`=1xx5xx10^(-2)xx2xx10^(-2)V`
`=10xx10^(-4)=10^(-3) V.`
`Current in the 10 Omega resistor `
`=(E/R)`
`=10^(-3)/(10)=10^(-4)=0.1 mA.`
`(b) The switch is thrown to the lower rail`
`E=Blv`
`=1xx5xx10^(-2)xx4xx10^(-2)`
`=20xx10^(-4)V`
`Current =(20xx10^(-4))/(10)`
`=2xx10^(-4)V=0.2mA.`