Refer to fig. The arm PQ of the rectangular conductor is moved from x=0 outwards. The uniform magnetic field is perpendicular to the plane and extends from x=0 to x=b and is zero for `x gt b`. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x=0 to x=2b, and is then moved back x=0 with constant speed `upsilon`. Obtain expressions for the flux the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Explanation

Considering the outward motion from x = 0 to x = 2b, the magnetic flux linked with the loop PQRS is
`phi=Blx` for `0lex lt b`
`phi=Blb` for `blex lt b`
The induced emf will be
`epsilon=(-dphi)/(dt)`
= - Blv for `0lex lt b`
`epsilon=(-dphi)/(dt)=0` for `blex lt 2b`
When the induced emf is non-zero, the current is given by
`I=(epsilon)/(r)=(Blv)/(r)`

The force required to keep the arm PQ in constant motion is given by:
`F=IlB=(B^(2)l^(2)v)/(r)` for `0lex lt b`
F=0 for `blex lt 2b`
The direction of force on PQ is to the left.
The power dissipated as joule heat is
`P=I^(2)r=(B^(2)l^(2)v^(2))/(r)` for `0lex lt b`
P=0 for `blex lt 2b`
Similar expression can be obtained for the inward motion from x= 2b to x = 0. This whole process can be understood with the help of given graph.