Question

In $N,e,\tau$ and $m$ are representing electron density, charge, relaxation time, and mass of an electron respectively, then the resistance of a wire of length $l$ and cross-sectional area $A$ is given by
A) $\dfrac{{2ml}}{{N{e^2}A\tau }}$
B) $\dfrac{{2m\tau A}}{{N{e^2}l}}$
C) $\dfrac{{N{e^2}\tau A}}{{2ml}}$
D) $\dfrac{{N{e^2}A}}{{2m\tau l}}$

Answer 1

First we have to use the formula for current passing through a wire
Then, we use the relation between the drift velocity(${v_d}$ ) and Electric field intensity ($E$) formula.
Also we have to use the expression on ohm's law.
Doing some simplification we get the required answer

Formula used:
-$I = NeA{v_d}$
Where $N$= electron density,
$e$= charge of an electron,
$A$= cross-sectional area of the wire,
${v_d}$=average drift velocity.
-${v_d} = \dfrac{1}{2}\dfrac{{eE}}{m}\tau = \dfrac{1}{2}\dfrac{{eV}}{{ml}}\tau$
Where $m$= mass of the electron,
$\tau$= relaxation time.
$V = potential - difference = \dfrac{E}{l}$
-$V = IR$
where $R$ =The resistance of the wire.

Complete step by step answer:
In a wire of length $l$ and cross-sectional area, $A$ if a current passes, the current can be represented by,
$I = NeA{v_d}....\left( 1 \right)$
Where $N$ = electron density,
$e$ = charge of an electron,
$A$ = cross-sectional area of the wire,
${v_d}$ =average drift velocity.
The drift velocity in terms of Electric field intensity that are created due to the passing current in the wire,${v_d} = \dfrac{1}{2}\dfrac{{eE}}{m}\tau = \dfrac{1}{2}\dfrac{{eV}}{{ml}}\tau ...\left( 2 \right)$
Since, we know that $V =$ potential $-$ difference $= \dfrac{E}{l}$
Here $m$ = mass of the electron,
$\tau$ = relaxation time.
$E =$ electric field intensity.
Now from ohm’s law, we get,
$V = IR...\left( 3 \right)$
Putting the value of equation $\left( 2 \right)$ in $\left( 1 \right)$, we get
$I = NeA \times \dfrac{1}{2}\dfrac{{eV}}{{ml}} \times \tau$
Let us multiply we get,
$\Rightarrow I = \dfrac{{N{e^2}AV}}{{2ml}}\tau$
Put the value of $I$in equation $\left( 3 \right)$ we get,
$V = \dfrac{{N{e^2}AV}}{{2ml}}\tau \times R$
Taking $R$ as LHS and remaining as RHS on divided we get,
$\Rightarrow R = \dfrac{{2mlV}}{{N{e^2}AV\tau }}$
Cancel the same term we get,
$\Rightarrow R = \dfrac{{2ml}}{{N{e^2}A\tau }}$
Therefore the resistance of the wire is $R = \dfrac{{2ml}}{{N{e^2}A\tau }}$

Hence, the right answer is in option (A).

Note: The drift velocity is calculated by the following method,
Let the electron cover the distance $l$ i.e the length of the wire with a velocity $v$ at a time $t$ with an acceleration $a$,
The equation of motion should be,
$l = \dfrac{1}{2}a{t^2}$
On dividing $t$ on both sides we get,
$\dfrac{l}{t} = \dfrac{1}{2}at$
$v = \dfrac{1}{2}at$
Now we rewrite the acceleration in terms of force i.e the electric field intensity,
i.e. $a = \dfrac{{eE}}{m}$
putting the value of $a = \dfrac{{eE}}{m}$ on $v = \dfrac{1}{2}at$ we get,
$\Rightarrow v = \dfrac{1}{2}\dfrac{{eE}}{m}t$
Now, if at relaxation time $\tau$ the velocity is ${v_d}$ ,
The above equation is modified with,
${v_d} = \dfrac{1}{2}\dfrac{{eE}}{m}\tau$.
This is the equation of the drift velocity of an electron of charge $e$ and mass $m$.