Question

The arm `PQ` of the rectangular conductor is moved from `x=0`, outwards in the uniform magnetic field which extends from `x=0` to `x=b` and is zero for `x gt b` as shown. Only the arm `PQ` possess 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`. Obtain expression 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.

Answer 1

Consider the forward motion from x = 0 to x = 2b. When arem PQ is at x, magnetic flux linked with the circuit PQRS is
`phi = B l x` for `0 le x lt b`
`phi = B l b` for `b le x lt 2 b`
The induced e..m.f. is
`E = (-d phi)/(dt) = - B l v` for `0 le x lt b`
`E = (-d phi)/(dt) = 0` for `b le x lt 2 b`
When the induced e.m.f. in non-zero, the magnitude of current is ` I = (E)/(r) = (B l upsilon)/(r)`
Forcee required to keep the arm PQ in constant motion `= BI l = B((B l upsilon)/(r)) t`
`= (B^(2) l^(2) upsilon)/(r)`, for `0 le x lt b`,
`= zero`, For `b le x lt 2 b`
The direction of the force on PQ is the left.
The joule heating loss is
`P = I^(2) r = (B^(2) l^(2) upsilon)/(r)`, for `0 le x lt b`
`= 0`, for `b le x lt b`
Similar expressions are obtained fro inward motion of PQ from `x = 2 b ti x = 0`. The various quantites have been plotted in fig.