Step 1 (Order of Magnitude & Governing Definition): [0.5 Mark]

The order of magnitude of the drift velocity ($v_d$) of electrons in a typical metallic conductor is approximately $10^{-4} \text{ m/s}$.

To deduce the relation, consider a cylindrical conductor of cross-sectional area $A$ and length $L$. Let $n$ be the number density of free electrons (number of electrons per unit volume).

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

• The total volume of the conductor is $V = A \times L$.
• The total number of free electrons ($N$) in this volume is: $$N = n \times V = nAL$$
• The total charge ($q$) contained in this volume is: $$q = N \times e = nALe$$
• If the electrons drift with a velocity $v_d$, the time ($t$) taken for the electrons to cross the length $L$ of the conductor is: $$t = \frac{L}{v_d}$$
• By definition, the current ($I$) is the rate of flow of charge: $$I = \frac{q}{t}$$
• Substituting the values of $q$ and $t$: $$I = \frac{nALe}{L / v_d}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Canceling $L$ from the numerator and denominator, we obtain the final relation:

Where:

• $I$ = Current (Amperes)
• $n$ = Number density of electrons ($\text{m}^{-3}$)
• $A$ = Cross-sectional area ($\text{m}^2$)
• $e$ = Charge of an electron ($\text{C}$)
• $v_d$ = Drift velocity ($\text{m/s}$)

💡 EL10 Concept Insight:

Think of the conductor as a pipe filled with marbles (electrons); even though each individual marble moves very slowly (like a snail), because there are so many of them packed together, they all push each other forward at the same time to create a steady flow of current.