Question

please answer this question

Answer 1

Let us arrange an apparatus as instructed.
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

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