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) Consider a rod PQ of length l, moving in a magnetic field $\overrightarrow{B}$ with a constant velocity $\overrightarrow{v}$. The length of the rod is perpendicular to the magnetic field and also the velocity is perpendicular to both the rod and the field. The free electrons of the rod also move at this velocity $\overrightarrow{v}$ because of which it experiences a magnetic force.


$\overrightarrow{F_b}=q\overrightarrow{v}\times \overrightarrow{B}$

This force is towards P from Q.

Thus, the free electrons will move towards P and positive charge will appear at Q. An electrostatic field E is developed within the wire from Q to P. This field exerts a force.

${\overrightarrow{F}}_e=q\overrightarrow{E}$

on each free electron. The charge keeps on gathering until

${\overrightarrow{F}}_b={\overrightarrow{F}}_e$

$\Rightarrow |q\overrightarrow{v}\times \overrightarrow{B}|=\left|q\overrightarrow{E}\right|$

$vB=E$

After this, resultant force on the free electrons of the wire PQ becomes zero. The potential difference between the ends Q and P is given by.

$V=El=vBl$

Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf, $e=Bvl$

(b) Lorentz force acting on a charge q which is moving with a speed v in a (normal) uniform magnetic field B, is Bqv.

All the charges present will experience the same force.

Work done to move the charge from P to Q,

$W=Bqv\times l$

$e=\frac{W}{q}=\frac{Bqvl}{q}=Blv$