Electric Current and Drift Velocity

Historical Note: Benjamin Franklin's "Mistake"

Fun fact: The conventional current direction was established by Benjamin Franklin in the 1700s, before electrons were discovered. Franklin assumed that electricity flowed from positive to negative terminals, which turned out to be backwards! When electrons were discovered in the late 1800s, scientists realized that the actual charge carriers (electrons) flow in the opposite direction.

However, by then the conventional current direction was so well-established in electrical engineering and physics that it was kept as the standard. This is why we still use conventional current direction in circuit analysis today, even though we know electrons actually flow in the opposite direction.

Memory Trick: NEVADA

Remember the drift velocity formula using NEVADA:

• NE = \(ne\) (number density × elementary charge)
• V = drift velocity (\(v_d\))
• A = cross-sectional area (\(A\))
• D = diameter (related to area)
• A = area again (\(A\))

So: \(ne \times v_d \times A = I\)

Rearranging: \(v_d = \frac{I}{neA}\)

This helps you remember that \(nev_dA = I\), which rearranges to give the drift velocity formula!

Example 1: Calculating Current from Charge Flow

Problem: A wire carries 2.5 × 10¹⁹ electrons past a point in 3.0 seconds. What is the current in the wire?

Solution:

1. Total charge: \(q = (2.5 \times 10^{19} \text{ electrons}) \times (1.602 \times 10^{-19} \text{ C/electron})\)
2. Calculation: \(q = 2.5 \times 10^{19} \times 1.602 \times 10^{-19} = 4.005 \text{ C}\)
3. Current: \(I = \frac{q}{t} = \frac{4.005 \text{ C}}{3.0 \text{ s}} = 1.335 \text{ A}\)

Answer: The current is 1.34 A.

Example 2: Drift Velocity Calculation

Problem: A copper wire with diameter 2.0 mm carries a current of 5.0 A. If copper has 8.5 × 10²⁸ free electrons per m³, what is the drift velocity of the electrons?

Solution:

1. Cross-sectional area: \(A = \pi r^2 = \pi(1.0 \times 10^{-3} \text{ m})^2 = 3.14 \times 10^{-6} \text{ m}^2\)
2. Drift velocity formula: \(v_d = \frac{I}{nqA}\)
3. Substitution: \(v_d = \frac{5.0 \text{ A}}{(8.5 \times 10^{28} \text{ m}^{-3})(1.602 \times 10^{-19} \text{ C})(3.14 \times 10^{-6} \text{ m}^2)}\)
4. Calculation: \(v_d = \frac{5.0}{8.5 \times 10^{28} \times 1.602 \times 10^{-19} \times 3.14 \times 10^{-6}}\)
5. Result: \(v_d = 1.17 \times 10^{-4} \text{ m/s} = 0.117 \text{ mm/s}\)

Answer: The drift velocity is 0.117 mm/s.

Example 3: Charge Carrier Concentration

Problem: A wire with cross-sectional area 1.0 × 10⁻⁶ m² carries a current of 3.0 A. If the drift velocity is 2.0 × 10⁻⁴ m/s, what is the concentration of charge carriers?

Solution:

1. Rearranging drift velocity formula: \(n = \frac{I}{qv_dA}\)
2. Substitution: \(n = \frac{3.0 \text{ A}}{(1.602 \times 10^{-19} \text{ C})(2.0 \times 10^{-4} \text{ m/s})(1.0 \times 10^{-6} \text{ m}^2)}\)
3. Calculation: \(n = \frac{3.0}{1.602 \times 10^{-19} \times 2.0 \times 10^{-4} \times 1.0 \times 10^{-6}}\)
4. Result: \(n = 9.36 \times 10^{28} \text{ m}^{-3}\)

Answer: The charge carrier concentration is 9.36 × 10²⁸ m⁻³.

Example 4: Time for Charge to Travel

Problem: A copper wire is 10.0 m long and carries a current of 2.0 A. If the drift velocity is 1.0 × 10⁻⁴ m/s, how long does it take for an electron to travel the length of the wire?

Solution:

1. Time formula: \(t = \frac{d}{v}\)
2. Substitution: \(t = \frac{10.0 \text{ m}}{1.0 \times 10^{-4} \text{ m/s}}\)
3. Calculation: \(t = 1.0 \times 10^5 \text{ s}\)
4. Convert to hours: \(t = \frac{1.0 \times 10^5 \text{ s}}{3600 \text{ s/hour}} = 27.8 \text{ hours}\)

Answer: It takes 27.8 hours for an electron to travel 10.0 m.