Question

4. A conductor of length 𝐿 moves at a steady speed vat right angles to a uniform magnetic field 

of flux density 𝐵. Show that the e.m.f. 𝐸 across the ends of the conductor is given by the 

equation 𝐸 = 𝐵𝑙𝑣. 

Answer 1

Consider the area enclosed by the moving rod, rails and resistor. B is perpendicular to this area, and the area is increasing as the rod moves. Thus the magnetic flux enclosed by the rails, rod and resistor is increasing. When flux changes, an EMF is induced according to Faraday’s law of induction.

To find the magnitude of EMF induced along the moving rod, we use Faraday’s law of induction without the sign:

$$EMF=N\frac{\Delta \Phi}{\Delta t}$$

In this equation, N=1 and the flux Φ=BAcosθ. We have θ=0º and cosθ=1, since B is perpendicular to A. Now ΔΦ=Δ(BA)=BΔA, since B is uniform. Note that the area swept out by the rod is ΔA=ℓx. Entering these quantities into the expression for EMF yields:

$$EMF=\frac{B\Delta A}{\Delta t}=\frac{Bl\Delta x}{\Delta t}=Blv$$