Consider a loop ABCD placed in a uniform magnetic field of strength B. Suppose the arm CD can slide in the direction shown. On sliding, there will be an induced e.m.f. and thus a current along ABCD. Consider an electron in CD, which has drift velocity, say \(\overrightarrow{v_d}\) along DC in addition to v. The velocity v is perpendicular to vd. The magnetic Lorentz force on the electron has two components therefore: e vdB along DC and e vd B in direction opposite to v. The electron acceleration along DC on account of the force evB is opposed by the average force due to collisions with atoms, resulting in a net zero force along DC but a constant, drift velocity vd. This is just like the steady current which results when forces act on electrons in a conductor due to electric fields. The force e vd B will of course cause the electrons to strike the left edge of the conductor exerting a force, say of magnitude F on it. The conductor reacts with an opposite force F on the electron. Since the electron finally is unaccelerated, F must equal evd B. The agent pulling CD is obviously having to work against the force F exerted on CD. In a small time interval ∆t, the work done by the agent is
dW = Force x distance travelled in ∆t
= e vd B dl ...........(1)
The electron moves through a distance dl = vd dt in time dt and hence
dW = e. vd B dl ..................(2)
This is the work done by the agent in pulling CD during which a charge has moved through the distance dl. The work done by the agent during the time over which a unit charge moves along the entire length of CD is then:
W = vd.B.l = B vd l = B \(\frac{dl}{dt}.l\)
W = B\(\frac{dl^2}{dt}\)
= B\(\frac{dA}{dt}\) (A = l2 = area of loop)
or W = Φ .............(3)
W is clearly the e.m.f. e across the entire loop since CD is the only moving part of the loop. Eq. (3) is thus the same as Faraday’s flux rule.
