Answer

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**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 \[m\].

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