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

Hint: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 \[m\].