Animated Solution for Physics - Current Electricity: When $5\text{V}$ potential difference is applied across a wire of length $0.1\text{m}$, the drift speed of electrons is $2.5\times 10^{-4}{\text{ms}}^{-1}$. If the electron density in the wire is $8\times 10^{28}{\text{m}}^{-3}$ the resistivity of the material is close to

Imagine you are standing inside a copper wire. All around you, a sea of free electrons is buzzing with chaotic thermal energy. But the moment a battery is connected—bam! An electric field sweeps through the wire, and this chaotic swarm begins a slow, collective march. This slow march is what we call the drift velocity ($v_d$).

In this problem, we are given a wire of length $L=0.1\text{ m}$ connected to a $5\text{ V}$ battery. The electrons are drifting at a speed of $v_d=2.5\times 10^{-4}\text{ m/s}$. We are also given the electron density, $n=8\times 10^{28}{\text{ m}}^{-3}$, which tells us how incredibly packed these charge carriers are. Our mission? To find the resistivity ($\rho$) of the material.

Before we jump into macroscopic formulas like Ohm's Law, let's look at what's happening at the atomic level. The current $I$ flowing through a wire is directly proportional to how many electrons are available, their charge, the cross-sectional area they flow through, and how fast they are drifting. This gives us our fundamental microscopic equation:

Here, $n$ is the electron density, $e$ is the elementary charge ($1.6\times 10^{-19}\text{ C}$), $A$ is the cross-sectional area, and $v_d$ is the drift velocity.

Now, let's bring in the macroscopic laws we all know and love. Ohm's Law states that the voltage $V$ across a conductor is proportional to the current $I$ and its resistance $R$:

We also know that the resistance $R$ depends on the material's resistivity $\rho$, its length $L$, and its cross-sectional area $A$:

Let's substitute this expression for $R$ into Ohm's Law:

Now comes the beautiful part. Let's substitute our microscopic expression for current ($I=neA{v}_d$) into this equation:

Notice what happens? The cross-sectional area $A$ is in the numerator of the current term and the denominator of the resistance term. It perfectly cancels out! This makes physical sense: a thicker wire allows more current to flow, but it also has less resistance. The two effects balance each other out completely.

Rearranging this to solve for our target variable, resistivity $\rho$:

We have our master equation. Now, it's just a matter of carefully substituting the given values and managing the powers of 10. Let's plug them in:

To avoid silly mistakes, let's group the base numbers and the powers of 10 in the denominator:

Let's simplify the base numbers: $8\times 2.5=20$. Then $20\times 0.1=2$. Finally, $2\times 1.6=3.2$.

Now for the powers of 10: $28-19-4=5$. So the denominator is $3.2\times 10^5$.

Rounding to two significant figures, we get our final answer:

This perfectly matches option (d).

As a final thought experiment, imagine what would happen if we heated this wire. The thermal agitation of the atoms would increase, causing the drifting electrons to collide more frequently. This would decrease their drift velocity $v_d$. Looking at our master equation $\rho =\frac{V}{ne{v}_{d}L}$, if $v_d$ goes down, the resistivity $\rho$ must go up. This is exactly why the resistance of most metals increases with temperature!