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 ......... .

Imagine a cylindrical wire acting as a conduit for electric charges. We are given its physical and electrical properties: the radius $r=0.5\text{ mm}$, the conductivity $\sigma =5\times 10^7\text{ S/m}$, and the applied electric field $E=10\text{ mV/m}$. Our ultimate goal is to determine the total current flowing through this wire.

Before we dive into the calculations, it is crucial to ensure all our units are in the standard SI format to avoid any catastrophic errors later on.

Let's convert the radius and the electric field: $r=0.5\times 10^{-3}\text{ m}$ $E=10\times 10^{-3}\text{ V/m}$

To find the current, we first need to understand how intensely the charges are flowing at a microscopic level. This is where the microscopic form of Ohm's law comes into play. It elegantly relates the current density ($J$) directly to the conductivity ($\sigma$) and the electric field ($E$).

The formula is:

Let's substitute our known values into this equation:

Upon multiplying these values, we find the current density:

This tells us that $5\times 10^5$ Amperes of current are flowing through every square meter of the wire's cross-section.

Now, how do we transition from current density to the total current ($I$)? We simply multiply the current density by the total cross-sectional area ($A$) of the wire. Since the wire is cylindrical, its cross-section is a circle.

The relation is:

Let's plug in the value of $J$ and the expression for the area. Be extremely careful to square the radius correctly while keeping it in meters:

Squaring the radius gives us $0.25\times 10^{-6}$. Multiplying everything together:

To match the format requested in the problem, we need to convert this current into milliamperes (mA) by multiplying by $10^3$:

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

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

$$x=5$$

What an elegant and satisfying result! Always remember that mastering the microscopic perspective of current flow unlocks a deeper understanding of electromagnetism.