A conducting rod of length `l` slides at constant velocity `v` on two parallel conducting rails, placed in a uniform and constant magnetic field `B` perpendicular to the plane of the rails as shown in Fig.3.49. A resistance `R` is connected between the two ends of the rails.

(a) Idenify the cause which produces change in magnetic flux.
(b) Identify the direction of current in the loop.
( c) Determine the emf induced in the loop.
(d) Compute the electric power dissipated in the resistor.
(e) Calcualte the mechenical power required to pull the rod at a cinstant velovity.

Explanation

(a) The change in area produces the magnetic flux.
(b) The direction of current in the loop is aniclockwise. As the rod moves toward right, the number of crosses in the loop increases with time and to oppose the increasing number of crosses in the loop, the current in the loop must be anticlockwise.

( c) According to Faraday's law
`|epsilon| = (dphi_(B))/(dt) = (d)/(dt)(Blx) = Bl(dx)/(dt) = Blv`.
(d) The magnitude of current is `I = (epsilon)/(R) = (Blv)/R`.
The electric power dissipated in the resistor is
`P_(ele) = I^(2)R = (B^(2)l^(2)v^(2))/(R)`
(e) The mechenical power is `P_(mech) = F_(ext)v`.
The external force is equal and opposite to Ampere's force:
`F_(ext) = -F_(Ampere)`
Ampere's force is given by
`F_(Ampere) = intIdvec(l) xx B` or `F_(Ampere) = BIl`
Now, `P_(mech) = (BIl)v`.
Substituting the value of `I`, we get `P_(mech) = (B^(2)l^(2)v^(2))/(R)`.
We see that machanicale power is same as electric power. it means mechanical power supplied to the rod is dissipated in the form of heat through resistor.