Animated Solution for Physics - Current Electricity: A cylindrical wire of radius $0.5\text{ mm}$ and conductivity $5\times 10^7\text{ S/m}$ is subjected to an electric field of $10\text{ mV/m}$. The expected value of current in the wire will be $x^3\pi \text{ mA}$. The value of $x$ is ......... .

The Microscopic View of Current

Have you ever wondered what happens inside a wire when a current flows through it? We often use the macroscopic version of Ohm's law, $V=IR$, but there is a deeper, microscopic reality at play. Let's dive into this problem and uncover the hidden mechanics of electrical conduction!

Imagine a cylindrical wire. We are given its physical dimensions and material properties. The radius of the wire is $r=0.5\text{ mm}$, which we must immediately convert to standard SI units as $0.5\times 10^{-3}\text{ m}$. The material's ability to conduct electricity is given by its conductivity, $\sigma =5\times 10^7\text{ S/m}$.

We are also told that an electric field is established inside the wire. This field is the driving force that pushes the electrons. The strength of this electric field is $E=10\text{ mV/m}$, which translates to $10\times 10^{-3}\text{ V/m}$. Our ultimate goal is to find the total current $I$ flowing through this wire.

Microscopic Ohm's Law

To connect these microscopic quantities to the macroscopic current, we invoke the microscopic form of Ohm's law. This powerful principle states that the current density $\vec{J}$ at any point in a conductor is directly proportional to the electric field $\vec{E}$ at that point. The constant of proportionality is the conductivity $\sigma$.

Mathematically, this is expressed as $J=\sigma E$.

But what is current density? It is simply the current flowing per unit cross-sectional area. Therefore, the total current $I$ is the product of the current density $J$ and the cross-sectional area $A$, giving us the relation $I=J\cdot A$. Combining these, we get our master equation: $I=\sigma EA$.

Now, let's carefully substitute our known values into this framework. First, we calculate the current density $J$.

Next, we need the cross-sectional area of our cylindrical wire. The area of a circle is $A=\pi r^2$.

To make our upcoming multiplication smoother, let's rewrite this area by shifting the decimal point:

With both the current density and the area in hand, we are ready to find the total current. We multiply $J$ and $A$ together:

Notice that $10^{-3}\text{ A}$ is exactly one milliampere ($\text{mA}$). Therefore, we can elegantly write our current as:

The problem states that the expected value of the current is $x^3\pi \text{ mA}$. By comparing our result with this expression, we can easily spot the value of $x$.

Taking the cube root of both sides, we arrive at our final answer:

This problem is a fantastic reminder of how macroscopic phenomena like current are deeply rooted in the microscopic properties of the material and the electric fields within it!