Question: Q. 8. Two material bars $A$ and $B$ of equal area of crosssection, are connected in series to a DC supply. A is made of usual resistance wire and $B$ of an $n$-type semiconductor.
(i) In which bar is drift speed of free electrons greater?
(ii) If the same constant current continues to flow for a long time, how will the voltage drop across $A$ and $B$ be affected? Justify each error. A&E [CBSE Comptt. 2018]
Sol. (a) Drift speed in $B$ ( $n$-type semiconductor) is higher
Reason : $\mathrm{I}=n e A v_{d}$ is same for both $n$ is much lower in semiconductors.
(b) Voltage drop across $A$ will increase as the resistance of $A$ increases with increase in temperature. $1 / 2+1 / 2$ Voltage drop across $B$ will decrease as resistance of $B$ will decrease with increase in temperature.
$1 / 2+1 / 2$
[CBSE Marking Scheme 2018]
AI Q. 9. (a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field $\mathrm{E}$.
R&U [Delhi & O.D. 2018]
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Solution:
Ans. (a) Definition and SI unit of conductivity $1 / 2+1 / 2$
(b) Derivation of the expression for conductivity $1 \frac{1}{2}$ Relation between current density and electric field
(a) The conductivity of a material equals the reciprocal of the resistance of its wire of unit length and unit area of cross-section.
[Alternatively :
The conductivity $(\sigma)$ of a material is the reciprocalof its resistivity $(\rho)]$
(Also accept $\sigma=\frac{1}{\rho}$ )
Its SI unit is
$\left(\frac{1}{\text { ohm-metre }}\right) /$ ohm $^{-1} \mathrm{~m}^{-1} /$ (mhe $\mathrm{m}^{-1}$ ) / siemen $\mathrm{m}^{-1}$
(b) The acceleration, $\quad a=-\frac{e}{m} \vec{E}$
The average drift velocity, $v_{d}$, is given by
$$ v_{d}=-\frac{e E}{m} \tau $$
( $\tau=$ average time between collisions/relaxation time)
If $n$ is the number of free electrons per unit volume, the current $I$ is given by
$$ \begin{align*} I & =n e A\left|v_{d}\right| \ & =\frac{e^{2} A}{m} \tau n|E| \end{align*} $$
But
We, therefore, get
$$ I=|j| A \quad(j=\text { current density }) $$
$$ |j|=\frac{n e^{2}}{m} \tau|E| $$
The term $\frac{n e^{2}}{m} \tau$ is conductivity
$$ \begin{array}{ll} \therefore & \sigma=\frac{n e^{2} \tau}{m} \ \Rightarrow & J=\sigma E \end{array} $$
[CBSE Marking Scheme 2018]
