- Let us arrange an apparatus as shown in figure.
- It consists of a pair of parallel bare conductors which are spaced T meters a part in uniform magnetic field of ‘B’.
- We can hold another bare conductor in such a way that it is in contact with the two parallel wires.
- A galvanometer is connected to the ends of parallel conductors to complete an electric circuit.
- Now if the cross wire placed across parallel conductors is moved to the left, galvanometer needle will deflect in one direction.
- If the cross wire is moved to the right, its needle deflects in a direction oppo-site to the previous deflection.
- A current will set up in the circuit only when there is an EMF in the circuit. Let this EMF be e.
- According to principle of conservation of energy this electric energy must come from the work that we have done in moving the cross wire.
- If we ignore friction, the work done by this applied force = Fs (where s is the distance moved by cross conductor)
- The force applied on the cross wire by the field B is F = BIl ----> (1)
- The work done by us in moving the cross wire converts into electrical energy. So the work done is given by

W = FS

Substitute (1) W = $F _{ s }$ = BIls -----»(2)

$\triangle$$\phi$ = Bls ------ > (3)

From (2) and (3)

W = ($\triangle$$\phi$) I

Let us divide both sides by At

W/$\triangle$t = I $\triangle$$\phi$ /$\triangle$t ------> (4)

$\Sigma $ = $\triangle$$\phi$ /$\triangle$t

Electric power, P = $\Sigma $I ----->(5)

Electric power, P = I ($\triangle$$\phi$ /$\triangle$t)

Divide (2) by $\triangle$t

W / $\triangle$t = Fs / $\triangle$t = BIls / $\triangle$t -----> (6)

Here s / $\triangle$t gives the speed of the cross wire, let it be v.

Electric power

P = W / $\triangle$t = Fv = BIlv ----->(7)

Power is also given as force times velocity.

From (5) and (7), $\Sigma $I = BIlv