State Faraday's law of electromagnetic induction.
Figure shows a rectangular conductor `PQRS` in which the conductor `PQ` is free to move in a uniform magnetic field `B` perpendicular to the plane of the paper.The field extends from `x=0` to `x=b` and is zero for `x gt b`.Assume that only the arm `PQ` posses resistance `r`.When the arm `PQ` is pulled outward from `x=0` with constant speed `v`, obtain the expressions for the flux and the induced emf. sketch the variations of these quantities with distance `0 le x le 2b`.

Explanation

Part I:Farady's law of induction:It states that the `emf` induced in a coil of `N` turns is directly related to the rate of charge of flux through it.
`therefore epsilon=-N(d phi_(B))/(dt)`
Where `phi_(B)` is the flux linked with one turn of the coil.If the circuit is closed, a current `I=epsilon/R` is setup in it.
Part II :
Refer to Following Fig.(a) the arm `PQ` of the rectangular conductor is moved form `x=0` outwards, the uniform magnetic field is perpendicular to the plane and extends from `x=0` at `x=b` and is zero these situation when the arm `PQ` passes substantial resistance `r`.Consider the situation when the arm `PQ` is pulled outwards from `x=0` to `x=2b`,and is then moved back to `x=0` with constant speed `v`.
Let us first consider the forward motion from `x=0` to `x=2b` The flux `Phi_(B)` linked with the circuit `SPQR` is `Phi_(B)=Blx , 0lex lt b`
=`Blb b le x lt 2b`
The induced `emf` is `epsilon=-(dPhi_(B))/(dt)`
=`-Blv 0le x lt b`
=`0 b le x lt 2b`
When the induced `emf` is non-zero.the current `I` is (in magnitude)
`l=(Blupsilon)/r`
The force required to keep the arm `PQ` in constant motion is `IiB`. its direction is to the left.In magnitude
`F=(B^(2)l^(2)upsilon)/r 0 ne x lt b `
`F=0 b le x lt 2b`
The joule heating loss is
`P=I^(2)r`
`=(B^(2)l^(2)upsilon^(2))/r 0 ne x lt b `
`=0 b le x lt 2`