Question

(a) a rod of length l is moved horizontally with a uniform velocity 'v' in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. derive an expression for the emf induced across the ends of the rod. (b)how does one understand this motional emf by invoking the lorentz force acting on the free charge carriers of the conductor? explain.

Answer 1

(a) Let a straight conductor of length l is moving in U shaped conductor in perpendicular magnetic field B as shown in figure.
We know,
dФ = B(change in area from XY to X'Y') [conductor moves from XY to X'Y']
here dФ is the change in flux .
dФ = B(lvdt) = Bvl.dt [ ∵ length = l and Breadth = v.dt {speed × time =distance}]
so, dФ/dt = Bvl
Hence, |induced emf | = dФ/dt = Bvl

(b) due to motion of conductor , free electrons moves from one end to other ends . Due to this, both ends generate positive and negative charge and electric force act on it.
Now, according to Lorentz law,
Fnet = Fe + Fm
At equilibrium Fnet = 0 ,
∴Fe + Fm = 0
qE + q(v × B) = 0
⇒E = -(v × B)
⇒|E| = Bvsinθ when velocity perpendicular upon magnetic field θ = 90°
|E| = Bv
Also we know, dξ/dl = |E| = Bv
dξ = Bvdl ∴ |induced emf | = Bvl