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 the 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 $l$ be moving in u shaped conductor in perpendicular magnetic field.

$d\varphi =B\left(vldt\right)=Bvldt$

$\frac{d\varphi }{dt}=Bvl$

Induced emf $\epsilon =\frac{d\varphi }{dt}=Bvl$

(b)Due to motion of the conductor,free electrons move from one end to other end.

Due to this,both ends generate positive and negative charge and electric force acts on it.

Now according to Lorentz law,

$F_{net}=F_e+F_m$

At equilibrium, $F_{net}=0$

$F_e+F_m=0$

$qE+q(\overrightarrow{v}\times \overrightarrow{B})=0$

$E=-(\overrightarrow{v}\times \overrightarrow{B})$

$\left|E\right|=Bv\sin \theta$,when velocity perpendicular when magnetic field $\theta =90$

$\left|E\right|=Bv$

Also,$\frac{d\xi }{dl}=\left|E\right|=Bv$

$d\xi =Bvdl$

Induced emf=$Bvl=\xi$