Question

(i) Define the term drift velocity.
(ii) On the basis of electron drift derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?
(iii) Why alloys like constantan and maganin are used for making standard resistors?
OR
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1m and resistance 10 $\Omega$. Calculate the potential gradient along the wire and balance length AO(=1).

Answer 1

(i) Drift Velocity: The average velocity with which the free electron drift under the influence of an electric field.

$\overrightarrow{V_d}=-\frac{e\overrightarrow{E}}{m}\tau$

Or, $V\propto I$

(ii) As$\overrightarrow{V_d}=-\frac{e\overrightarrow{E}}{m}\tau ....\left(i\right)$

Current flowing through the conductor ---

$I=nA{V}_{d}e....\left(ii\right)$

from equation (i) and (ii).

$I=nA\left(\frac{eE}{m}\tau \right)e⟹I=\frac{nA{e}^{2}E\tau }{m}....\left(iii\right)$

If $V$ is potential difference applied across the two ends of the conductor, then

$E=\frac{V}{l}$

Putting this $E$ in equation (iii),

$I=\frac{nA{e}^{2}V\tau }{ml}⟹\frac{V}{I}=\frac{m}{n{e}^2\tau }\frac{l}{A}$

According to ohm's law,

$\frac{V}{I}=R$ (resistance of the conductor)

$R=\frac{m}{n{e}^2\tau }\frac{l}{A}....\left(iv\right)$

But, $R=\rho \frac{l}{A}.....\left(v\right)$

Comparing (iv) and (v)

$\rho =\frac{m}{n{e}^2\tau }$

Resistivity of a conductor depends on the following factors:

(1) It is inversely proportional to the number of free electrons per unit volume $\left(n\right)$ of the conductor.

(2) It is inversely proportional to the average relaxation time $\left(\tau \right)$ of the free electrons in the conductor.

(iii) Being an alloy, its resistivity is very high and temperature coefficient of resistance is very low.

(i) Principle of potentiometer: The potential difference across any two points of current carrying wire, having uniform cross-sectional area and material, of the potentiometer is directly proportional to the length between the two points.

That is, $V\propto l$

Proof:

$V=IR=I\left(\frac{\rho l}{A}\right)$

i.e., $V=\left(\frac{I\rho }{A}\right)l$

For uniform current and cross-sectional area, we have

$\frac{\rho l}{A}=\text{constant}=\text{potential drop per unit length}$

i.e., $V\propto l$

(ii) Given: potentiometer circuit AB is a uniform wire of length 1m and resistance $10\Omega$

To find the potential gradient along the wire and balance length AO$(=1).$

Solution: According to the given criteria,

$E=2V,R=15\Omega ,R_{AB}=10\Omega$

Potential difference across the wire, $=\frac{2}{25}\times 10=0.8V/m$.

Therefore, potential gradient $=\frac{0.8}{1}=0.8V/m$

Potential difference across AO $=\frac{1.5}{1.5}\times 0.3=0.3V$

Therefore, Length, AO $=\frac{0.3}{0.8}\times 100=37.5cm$

which is the reqired balance length of the wire