Question

Determine the motional emf induced in a straight conductor moving in a uniform magnetic field with constant velocity on the basis of Lorentz force.

Answer 1

A frame of wire PQRS in the magnetic field `vec"B"` and wire BC is moving with velocity `vec"v"` along the x-axis

1. Consider a rectangular frame of wires PQRS of area (lx) situated in a constant magnetic field `(vec"B").`
2. As the wire QR of length l is moved out with velocity `vec"v"` to increase x, the area of the loop PQRS increases. Thus the flux of `vec"B"` through the loop increases with time.
3. According to the flux rule, the induced emf will be equal to the rate at which the magnetic flux through a conducting circuit changes.
4. The induced emf will cause a current in the loop. It is assumed that there is enough resistance in the wire so that the induced currents are very small producing a negligible magnetic field.
5. As the flux `phi` through the frame PQRS is Blx, the magnitude of the induced emf can be written as 
   |e| = `("d"phi)/"dt" = "d"/"dt"("Blx") = "Bl""dx"/"dt"` = Blv ….(1)
   where v is the velocity of wire QR increasing the length x of wires PQ and SR.
6. Now, a charge q which is carried along by the moving wire QR, experiences Lorentz force `vec"F"` = q (`vec"v" xx vec"B"`); which is perpendicular to both `vec"v"` and `vec" B"` and hence is parallel to wire QR.
7. The force `vec" F"` is constant along the length l of the wire QR (as v and B are constant) and zero elsewhere (∵ v = 0 for stationary part RSPQ of wireframe).
8. When the charge q moves a distance l along the wire, the work done by the Lorentz force is W = F.l = qvBsinθ.l
   where θ = angle between `vec" B"` and `vec"v".`
9. The emf generated is, e = `"Work"/"charge" = "W"/"q"` = vBsinθ.l
10. For maximum induced emf, sinθ = 1
    e_max = Blv ….(2)
11. Thus, from equations (1) and (2), for any circuit whose parts move in a fixed magnetic field, the induced emf is the time derivative of flux (`phi`) regardless of the shape of the circuit.