This force must oppose the applied force. The direction of applied force determines the direction of current through the cross wire. Here we are doing positive work. The work done by us in moving the cross wire converts into electric energy. So the work done is given by,

W = Fs =BIls .............................(9) (using equation-8)

When we put the cross wire across parallel conductors it makes a complete electrical circuit which encloses a certain amount of magnetic flux. Now as we move the cross wire to the left, the area of the loop (formed by the parallel conductors and cross wire) decreases and the flux through the loop also decreases. The decrease in flux is given by,

ΔΦ = Bls .............................(10)

Here B is perpendicular to the area (ls). From equations 9 and 10 W = (ΔΦ) Ι

Let us divide both sides of this equation by Δt

W/Δ t = I (ΔΦ/ Δt) .............................(11)

Electric power, P = IΔΦ/ Δt We know that electric power is the product of current and emf or voltage. ε=ΔΦ/Δt is obviously equal to induced EMF. Electric power, P = εΙ .............................(12)

Thus the electric power generated in the circuit is equal to product of induced EMF and the current. Thus the mechanical energy utilised to move the cross wire in one second is converted into electric power (ΔΦ/Δt) I. This is nothing but conservation of energy.

Dividing equation (9) by Δt, we have

W/Δt = Fs/Δt =BIls/Δt ..................(13)

Here s/Δt gives the speed of the cross wire, let it be taken as v. Then we get, Electric power P = W/Δ t = Fv = BIlv ......(14)

Power is also given as force times velocity. From equations (12) and (14), we get W/Δ t = εI ⇒ εI = BIlv
We get,
ε = Blv     This is called motional EMF.