Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-section 5 mm^2 is v.

Q: Why do we multiply the area by $10^{-6}$?

The area is given in square millimeters ($\text{mm}^2$). Since $1\text{ mm} = 10^{-3}\text{ m}$, squaring both sides gives $1\text{ mm}^2 = (10^{-3}\text{ m})^2 = 10^{-6}\text{ m}^2$. Standard SI units are required for the formula to work correctly.

Q: Why is the drift velocity so incredibly small?

While electrons move randomly at very high speeds (around $10^6\text{ m/s}$) due to thermal energy, they constantly collide with the metal atoms. The electric field only provides a tiny, slow net forward 'drift' superimposed on this chaotic motion.

Q: Does the length of the wire affect the drift velocity?

No, as long as the current $I$ and cross-sectional area $A$ remain constant, the drift velocity is independent of the wire's length. This is clear from the formula $v_d = \frac{I}{neA}$.

Q: What exactly does '$n$' represent in the formula?

The variable '$n$' is the electron number density. It represents the number of free, conduction electrons available per unit volume (cubic meter) of the material.

Q: Why do the electrons drift in the opposite direction of the current?

By historical convention, the direction of electric current is defined as the flow of positive charge. Since electrons carry a negative charge, their physical motion is opposite to the direction of conventional current.

Question

Drift speed of electrons, when $1.5	ext{ A}$ of current flows in a copper wire of cross-section $5	ext{ mm}^2$ is $v$. If the electron density in copper is $9 	imes 10^{28} 	ext{ /m}^3$, the value of $v$ (in mm/s) is close to (Take, charge of electron to be $= 1.6 	imes 10^{-19}	ext{ C}$)

Options:

(A) 0.02

(B) 0.2

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Visualizing the Flow of Charge

Imagine you are looking inside a copper wire. When a voltage is applied, it creates an electric field that pushes the free electrons. However, these electrons don't just fly straight through; they constantly collide with the vibrating copper atoms. Because of these collisions, their net forward motion is surprisingly slow. This slow, average forward speed is what we call the **drift velocity** ($v_d$).

In this problem, we are given a macroscopic quantity—the current $I = 1.5	ext{ A}$—and we need to find the microscopic drift speed of the electrons.

### The Master Equation

The bridge between the macroscopic world (current) and the microscopic world (drift velocity) is given by the fundamental equation:

$$I = neAv_d$$

Here, $n$ is the electron number density (how many free electrons are packed into a cubic meter), $e$ is the elementary charge, $A$ is the cross-sectional area of the wire, and $v_d$ is the drift velocity.

Since we want to find the drift speed, let's rearrange the formula:

$$v_d = \frac{I}{neA}$$

### The Unit Conversion Trap

Before we plug in the numbers, we must ensure all units are in the standard SI system. This is where many students make a fatal error. The area is given as $5	ext{ mm}^2$.

Since $1	ext{ mm} = 10^{-3}	ext{ m}$, squaring both sides gives $1	ext{ mm}^2 = 10^{-6}	ext{ m}^2$. Therefore, our area is:

$$A = 5 	imes 10^{-6}	ext{ m}^2$$

### Executing the Calculation

Now, let's substitute all our values into the rearranged equation:

$$v_d = \frac{1.5}{9 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 5 	imes 10^{-6}}$$

To make the math easier, let's group the regular numbers and the powers of 10 separately:

$$v_d = \frac{1.5}{(9 	imes 1.6 	imes 5) 	imes 10^{28 - 19 - 6}}$$

Calculating the denominator:
$9 	imes 1.6 	imes 5 = 72$
$10^{28 - 19 - 6} = 10^3$

So, the expression simplifies beautifully to:

$$v_d = \frac{1.5}{72 	imes 10^3}	ext{ m/s}$$

$$v_d = \frac{1.5}{72} 	imes 10^{-3}	ext{ m/s}$$

Dividing $1.5$ by $72$ gives approximately $0.0208$. And since $10^{-3}	ext{ m}$ is exactly $1	ext{ mm}$, we can write:

$$v_d \approx 0.0208	ext{ mm/s}$$

Looking at our options, the value is closest to **$0.02	ext{ mm/s}$**. Notice how incredibly slow this is! It would take an electron nearly a minute to travel just one millimeter.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{The Master Formula}

\text{Substituting Values}

\text{Simplifying the Denominator}

\text{Final Calculation}

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Visualizing the Flow of Charge

The Master Equation

The Unit Conversion Trap

Executing the Calculation

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