Charge Velocity and Velocity of Field Propagation | Formula and Solved Example | 20 Objective Questions

When a switch is turned on, current begins to flow through a conductor very quickly. However, the individual electrons in the metal usually move along the conductor at a very small average speed. These two ideas are explained by electron drift velocity and velocity of electric-field propagation.

They describe different kinds of motion: one is the slow average movement of charge carriers, while the other describes how quickly the electrical influence travels through a circuit.

📑 Contents

1. Important Terms Explained
2. Charge Velocity (Electron Drift Velocity)
3. Velocity of Electric-Field Propagation
4. Why Current Starts Quickly
5. Drift Velocity vs Field Propagation Speed
6. Drift Velocity Formula and Derivation
7. Solved Numerical Example
8. Time Taken by an Electron to Travel 10 km
9. Important Points for Exams
10. 20 Important Objective Questions on Charge Velocity and Velocity of Field Propagation

1. Important Terms Explained

Conductor: A material, such as copper, that contains many mobile electrons and allows electric charge to flow easily.

Electron: A tiny particle carrying a negative electric charge. In a metal wire, some electrons can move through the material.

Electric current: The rate at which electric charge passes a point in a circuit. Its SI unit is the ampere (A).

Drift: A small average movement in one direction superimposed on the electrons’ irregular motion.

Electric field: The electrical influence that exerts a force on charged particles. Its symbol is \(E\), and its SI unit is volt per metre (V/m).

Propagation: The travel of a disturbance, signal, or influence from one place to another.

Free-electron number density: The number of mobile electrons present in each cubic metre of a conductor. It is represented by \(n\) and measured in \(\text{m}^{-3}\).

2. Charge Velocity (Electron Drift Velocity)

Electrons inside a metal are already moving rapidly and irregularly because of thermal energy. Their directions are continually changed by collisions with atoms and other imperfections in the material. Without an applied electric field, their random motions cancel on average, so there is no net drift in one direction.

When a voltage is applied across the conductor, an electric field is established inside it. Because electrons have negative charge, the force on them is opposite to the direction of the electric field. The electrons therefore acquire a small average velocity called electron drift velocity, represented by \(v_d\).

\[v_d=\frac{I}{neA}\]

Here, \(I\) is current in amperes, \(n\) is the number of free electrons per cubic metre, \(e\) is the magnitude of the electron charge (\(1.602\times10^{-19}\,\text{C}\)), and \(A\) is the conductor’s cross-sectional area in square metres.

Drift velocity is usually very small compared with the random speeds of electrons. Depending on the current and the conductor, it may be far below one millimetre per second.

3. Velocity of Electric-Field Propagation

Velocity of field propagation is the speed at which a change in the electromagnetic field or electrical signal travels along a circuit. When a switch is closed, the electrical influence spreads through the circuit; electrons throughout the wire begin responding to the local electric field rather than waiting for one electron to travel all the way from the switch to the load.

In many practical circuits, the signal propagation speed is a substantial fraction of the speed of light in vacuum, \(c\approx3.0\times10^8\,\text{m/s}\). The actual value depends on the geometry of the circuit and the electrical properties of the surrounding insulating material. It should not automatically be treated as exactly \(3.0\times10^8\,\text{m/s}\) in every wire.

Key idea: The field or signal can travel quickly, while the average drift of the electrons is slow. These are not contradictory statements because they describe different physical processes.

4. Why Does Current Start Quickly?

Imagine a long copper wire already containing mobile electrons. When a voltage is applied, the electromagnetic field spreads along the circuit. Electrons at many locations begin to respond to that field, so the circuit can begin carrying current long before an individual electron could drift from one end of the wire to the other.

A useful analogy is a long tube already filled with small balls. Pushing one ball into one end can cause a ball at the other end to move almost immediately, even though the first ball has not travelled the full length of the tube. The analogy is not exact, but it helps distinguish the propagation of an influence from the travel of an individual particle.

5. Drift Velocity vs Field Propagation Speed

Point of comparison	Electron drift velocity	Field/signal propagation speed
Meaning	Average net velocity of charge carriers in a conductor	Speed at which an electrical or electromagnetic change travels
Typical scale	Often very small; depends on current and conductor dimensions	Often a substantial fraction of the speed of light
Main factors	Current, carrier density, charge, and cross-sectional area	Circuit geometry and surrounding dielectric/material properties
Formula used in this example	\(v_d=I/(neA)\)	No single universal value; it depends on the transmission structure and medium
What moves?	The charge carriers have a small average net motion	The field disturbance or signal propagates through the circuit

6. Drift Velocity Formula and Derivation

Consider a conductor with cross-sectional area \(A\). Suppose the electrons have an average drift speed \(v_d\). In a time interval \(t\), the electrons drift a distance

\[l=v_dt\]

The volume of conductor swept through is \(Al\). If \(n\) is the number of free electrons per unit volume, the number of electrons in that volume is

\[N=nAl=nAv_dt\]

Each electron carries a charge with magnitude \(e\). Therefore, the total charge passing a cross-section in time \(t\) is

\[Q=Ne=nAv_dte\]

Electric current is charge transferred per unit time:

\[I=\frac{Q}{t}\]

Substitute \(Q=nAv_dte\):

\[I=\frac{nAv_dte}{t}=neAv_d\]

Rearranging to make \(v_d\) the subject gives the drift velocity formula:

\[\boxed{v_d=\frac{I}{neA}}\]

Unit check: \(I\) is measured in coulombs per second, \(n\) in \(\text{m}^{-3}\), \(e\) in coulombs, and \(A\) in \(\text{m}^2\). Thus the result has units of \(\text{m/s}\), which is a unit of velocity.

7. Solved Numerical Example

Problem: A current of \(1\,\text{A}\) flows through a copper conductor with cross-sectional area \(1\,\text{cm}^2\) and length \(10\,\text{km}\). The free-electron number density of copper is \(8.5\times10^{28}\,\text{m}^{-3}\). Find the electron drift velocity and estimate how long an electron would take to drift through the full length at this average speed. Take \(e=1.6\times10^{-19}\,\text{C}\).

Step 1: Write the given data

• Current: \(I=1\,\text{A}\)
• Area: \(A=1\,\text{cm}^2\)
• Length: \(L=10\,\text{km}\)
• Electron number density: \(n=8.5\times10^{28}\,\text{m}^{-3}\)
• Electron charge magnitude: \(e=1.6\times10^{-19}\,\text{C}\)

Step 2: Convert the units to SI

Since \(1\,\text{cm}=10^{-2}\,\text{m}\), the area conversion is

\[A=1\,\text{cm}^2=(10^{-2}\,\text{m})^2=10^{-4}\,\text{m}^2\]

Also, \(10\,\text{km}=10\times1000=10{,}000\,\text{m}=10^4\,\text{m}\).

Step 3: Calculate drift velocity

Use

\[v_d=\frac{I}{neA}\]

Substitute the values:

\[v_d=\frac{1}{(8.5\times10^{28})(1.6\times10^{-19})(1\times10^{-4})}\]

First, multiply the numerical values and powers of ten:

\[neA=(8.5\times1.6)\times10^{28-19-4}\]

\[neA=13.6\times10^5=1.36\times10^6\]

Therefore,

\[v_d=\frac{1}{1.36\times10^6}\approx7.35\times10^{-7}\,\text{m/s}\]

Since \(1\,\mu\text{m}=10^{-6}\,\text{m}\), this is

\[\boxed{v_d\approx0.735\,\mu\text{m/s}}\]

8. Time Taken by an Electron to Travel 10 km

For an estimate using the average drift speed, use the relation

\[\text{Time}=\frac{\text{Distance}}{\text{Speed}}\]

Here, the distance is \(L=10^4\,\text{m}\) and the drift speed is \(v_d=7.35\times10^{-7}\,\text{m/s}\).

\[t=\frac{10^4}{7.35\times10^{-7}}\]

\[t\approx1.36\times10^{10}\,\text{s}\]

Convert seconds to years using \(1\,\text{year}\approx365\times24\times3600=31{,}536{,}000\,\text{s}\):

\[\text{Years}=\frac{1.36\times10^{10}}{31{,}536{,}000}\approx431\,\text{years}\]

Answer: At the calculated average drift speed, the estimated travel time through 10 km is about 431 years.

Interpretation: This is an idealized estimate of the drift time for a charge carrier moving at the calculated average drift speed. It does not mean that a lamp or other electrical load waits hundreds of years to respond when the circuit is switched on. The electrical field/signal propagates much faster.

9. Important Points for Exams

⚡ Remember Point 1: Electron drift velocity is the small average net velocity acquired by electrons under an applied electric field.

⚡ Remember Point 2: In a metal, electrons drift opposite to the direction of the electric field and conventional current.

⚡ Remember Point 3: The drift velocity formula is \(v_d=I/(neA)\).

⚡ Remember Point 4: Field or signal propagation speed is not the same as electron drift velocity.

⚡ Remember Point 5: Convert areas carefully: \(1\,\text{cm}^2=10^{-4}\,\text{m}^2\), not \(10^{-2}\,\text{m}^2\).

⚡ Remember Point 6: The exact signal propagation speed depends on the circuit and its surrounding material; \(3\times10^8\,\text{m/s}\) is the speed of light in vacuum and is a useful approximate upper reference.

20 Important Objective Questions on Charge Velocity and Velocity of Field Propagation

📑 Contents

1. MCQ 1: What is electron drift velocity?
2. MCQ 2: What is the SI unit of drift velocity?
3. MCQ 3: In a metal conductor, electrons drift in which direction relative to the electric field?
4. MCQ 4: Which formula gives the magnitude of electron drift velocity in a conductor?
5. MCQ 5: In the drift-velocity formula, what does n represent?
6. MCQ 6: What is the magnitude of the charge of one electron used in the example?
7. MCQ 7: What does A represent in the formula \(v_d=I/(neA)\)?
8. MCQ 8: How should an area of \(1\,\mathrm{cm^2}\) be converted to SI units?
9. MCQ 9: What is the SI unit of electric field strength?
10. MCQ 10: What is meant by velocity of field propagation?
11. MCQ 11: Which statement best explains why a lamp can respond quickly even though electron drift is slow?
12. MCQ 12: Which speed is approximately the speed of light in vacuum?
13. MCQ 13: What is the relationship between current and drift velocity magnitude?
14. MCQ 14: In the derivation, what is the distance drifted in time t at average drift speed \(v_d\)?
15. MCQ 15: If current increases while n, e, and A remain constant, what happens to drift velocity?
16. MCQ 16: If cross-sectional area increases while current and electron density remain constant, what happens to drift velocity?
17. MCQ 17: In the worked example, what are the given current and cross-sectional area?
18. MCQ 18: Using \(n=8.5\times10^{28}\,\mathrm{m^{-3}}\), \(e=1.6\times10^{-19}\,\mathrm{C}\), \(I=1\,\mathrm{A}\), and \(A=10^{-4}\,\mathrm{m^2}\), what is the approximate drift velocity?
19. MCQ 19: For a 10 km conductor and drift velocity \(7.35\times10^{-7}\,\mathrm{m/s}\), approximately how long is the drift-time estimate?
20. MCQ 20: Which statement about the 431-year result is correct?

📝 MCQ 1. What is electron drift velocity?

• A) The speed of light in a vacuum
• B) The small average net velocity acquired by electrons in an applied electric field
• C) The random speed of an electron only
• D) The speed at which atoms vibrate

✅ Correct Answer: B) The small average net velocity acquired by electrons in an applied electric field

Explanation: Drift velocity is the average net motion of charge carriers caused by an applied electric field. Electrons still move irregularly, but the field adds a small average motion in one direction.

\[v_d\]

💡 Remember Point: Drift velocity describes average net motion, not the full random motion of each electron.

📝 MCQ 2. What is the SI unit of drift velocity?

• A) Ampere (A)
• B) Coulomb (C)
• C) Metre per second (m/s)
• D) Volt per metre (V/m)

✅ Correct Answer: C) Metre per second (m/s)

Explanation: Velocity is distance travelled per unit time, so its SI unit is metre per second. Ampere measures current, coulomb measures charge, and volt per metre measures electric field.

\[\mathrm{m/s}\]

💡 Remember Point: A velocity must have units of length divided by time.

📝 MCQ 3. In a metal conductor, electrons drift in which direction relative to the electric field?

• A) In the same direction as the electric field
• B) Opposite to the electric field
• C) Perpendicular to the electric field in all cases
• D) They never respond to an electric field

✅ Correct Answer: B) Opposite to the electric field

Explanation: An electron carries negative charge, so the electric force on it points opposite to the electric field. Its average drift direction is therefore opposite to the field direction.

\[\vec F=q\vec E,\quad q=-e\]

💡 Remember Point: Negative charges experience force opposite to the electric-field direction.

📝 MCQ 4. Which formula gives the magnitude of electron drift velocity in a conductor?

• A) \(v_d=neA/I\)
• B) \(v_d=I/(neA)\)
• C) \(v_d=IneA\)
• D) \(v_d=ne/(IA)\)

✅ Correct Answer: B) \(v_d=I/(neA)\)

Explanation: The current magnitude is related to drift speed by \(I=neAv_d\). Rearranging this expression gives \(v_d=I/(neA)\).

\[v_d=\frac{I}{neA}\]

💡 Remember Point: Current increases with drift speed when the other quantities stay constant.

📝 MCQ 5. In the drift-velocity formula, what does n represent?

• A) Current in amperes
• B) Conductor length in metres
• C) Number of mobile electrons per unit volume
• D) Electron charge in coulombs

✅ Correct Answer: C) Number of mobile electrons per unit volume

Explanation: The symbol \(n\) is the free-electron number density: the number of mobile electrons in each cubic metre of conductor. Its SI unit is \(\mathrm{m^{-3}}\).

\[[n]=\mathrm{m^{-3}}\]

💡 Remember Point: Number density is a count per unit volume.

📝 MCQ 6. What is the magnitude of the charge of one electron used in the example?

• A) \(1.6\times10^{-19}\,\mathrm{C}\)
• B) \(1.6\times10^{19}\,\mathrm{C}\)
• C) \(8.5\times10^{28}\,\mathrm{C}\)
• D) \(3.0\times10^8\,\mathrm{C}\)

✅ Correct Answer: A) \(1.6\times10^{-19}\,\mathrm{C}\)

Explanation: The magnitude of the elementary charge is approximately \(1.6\times10^{-19}\,\mathrm{C}\). An electron's charge is negative, but the drift-speed formula uses the positive magnitude \(e\).

\[e\approx1.6\times10^{-19}\,\mathrm{C}\]

💡 Remember Point: Use the magnitude of electron charge when calculating drift-speed magnitude.

📝 MCQ 7. What does A represent in the formula \(v_d=I/(neA)\)?

• A) Electric-field strength
• B) Cross-sectional area of the conductor
• C) Electron acceleration
• D) Time in seconds

✅ Correct Answer: B) Cross-sectional area of the conductor

Explanation: The symbol \(A\) represents the area of the conductor's cross-section, measured in square metres. A larger area means more electrons can pass through each second at the same drift speed.

\[v_d=\frac{I}{neA}\]

💡 Remember Point: Convert area to square metres before substituting into the formula.

📝 MCQ 8. How should an area of \(1\,\mathrm{cm^2}\) be converted to SI units?

• A) \(10^{-2}\,\mathrm{m^2}\)
• B) \(10^{-3}\,\mathrm{m^2}\)
• C) \(10^{-4}\,\mathrm{m^2}\)
• D) \(10^4\,\mathrm{m^2}\)

✅ Correct Answer: C) \(10^{-4}\,\mathrm{m^2}\)

Explanation: Since \(1\,\mathrm{cm}=10^{-2}\,\mathrm{m}\), square both sides: \(1\,\mathrm{cm^2}=(10^{-2})^2\,\mathrm{m^2}=10^{-4}\,\mathrm{m^2}\).

\[1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}\]

💡 Remember Point: When converting area, square the length-conversion factor.

📝 MCQ 9. What is the SI unit of electric field strength?

• A) \(\mathrm{m/s}\)
• B) \(\mathrm{A}\)
• C) \(\mathrm{V/m}\)
• D) \(\mathrm{m^{-3}}\)

✅ Correct Answer: C) \(\mathrm{V/m}\)

Explanation: Electric field strength is measured in volts per metre (V/m). It describes the electric potential change per unit distance in the relevant field direction.

\[E\text{ is measured in }\mathrm{V/m}\]

💡 Remember Point: Do not confuse electric field units with velocity units.

📝 MCQ 10. What is meant by velocity of field propagation?

• A) The average drift speed of electrons
• B) The speed at which an electrical or electromagnetic change travels along a circuit
• C) The speed of atoms in copper
• D) The rate of charge stored in a battery

✅ Correct Answer: B) The speed at which an electrical or electromagnetic change travels along a circuit

Explanation: Field or signal propagation speed describes how quickly an electrical influence travels through the circuit. It is different from the average velocity of individual charge carriers.

\[\text{Field propagation speed}\ne v_d\]

💡 Remember Point: A signal can propagate quickly even when electron drift is slow.

📝 MCQ 11. Which statement best explains why a lamp can respond quickly even though electron drift is slow?

• A) Electrons travel from the switch to the lamp at light speed
• B) The electric-field influence propagates through the circuit much faster than electron drift
• C) The conductor contains no electrons until the switch is closed
• D) Drift velocity becomes infinite when a switch is closed

✅ Correct Answer: B) The electric-field influence propagates through the circuit much faster than electron drift

Explanation: The conductor already contains mobile electrons. Closing the switch establishes a changing electromagnetic field that propagates through the circuit, causing electrons throughout the circuit to respond locally.

\[\text{Signal speed}\gg v_d\quad\text{(typically)}\]

💡 Remember Point: The electrical signal does not require one electron to travel the whole circuit first.

📝 MCQ 12. Which speed is approximately the speed of light in vacuum?

• A) \(3.0\times10^{-8}\,\mathrm{m/s}\)
• B) \(3.0\times10^8\,\mathrm{m/s}\)
• C) \(7.35\times10^{-7}\,\mathrm{m/s}\)
• D) \(1.6\times10^{-19}\,\mathrm{m/s}\)

✅ Correct Answer: B) \(3.0\times10^8\,\mathrm{m/s}\)

Explanation: The speed of light in vacuum is approximately \(3.0\times10^8\,\mathrm{m/s}\). Signal propagation in a real circuit is often a substantial fraction of this value, depending on the circuit and surrounding material.

\[c\approx3.0\times10^8\,\mathrm{m/s}\]

💡 Remember Point: Do not assume every circuit signal travels at exactly the vacuum speed of light.

📝 MCQ 13. What is the relationship between current and drift velocity magnitude?

• A) \(I=neAv_d\)
• B) \(I=v_d/(neA)\)
• C) \(I=n/(eAv_d)\)
• D) \(I=neA/v_d\)

✅ Correct Answer: A) \(I=neAv_d\)

Explanation: The number of electrons passing a cross-section each second depends on the number density, area, and drift speed. Multiplying by the charge magnitude gives \(I=neAv_d\).

\[I=neAv_d\]

💡 Remember Point: This relation can be rearranged to find current or drift speed.

📝 MCQ 14. In the derivation, what is the distance drifted in time t at average drift speed \(v_d\)?

• A) \(l=t/v_d\)
• B) \(l=v_dt\)
• C) \(l=v_d/t\)
• D) \(l=v_d+t\)

✅ Correct Answer: B) \(l=v_dt\)

Explanation: For uniform average motion, distance equals speed multiplied by time. Therefore the drift distance is \(l=v_dt\).

\[l=v_dt\]

💡 Remember Point: Distance = speed × time.

📝 MCQ 15. If current increases while n, e, and A remain constant, what happens to drift velocity?

• A) It decreases
• B) It remains zero
• C) It increases in direct proportion to current
• D) It becomes equal to the speed of light

✅ Correct Answer: C) It increases in direct proportion to current

Explanation: From \(v_d=I/(neA)\), drift velocity is directly proportional to current when number density, charge magnitude, and area are fixed.

\[v_d\propto I\]

💡 Remember Point: Check which variables are held constant before judging proportionality.

📝 MCQ 16. If cross-sectional area increases while current and electron density remain constant, what happens to drift velocity?

• A) It increases in direct proportion to area
• B) It decreases
• C) It does not change
• D) It becomes negative

✅ Correct Answer: B) It decreases

Explanation: The area is in the denominator of \(v_d=I/(neA)\). Thus, for fixed current and electron density, a larger cross-sectional area means a smaller drift-speed magnitude.

\[v_d\propto\frac{1}{A}\]

💡 Remember Point: Drift speed is inversely proportional to cross-sectional area when other quantities are fixed.

📝 MCQ 17. In the worked example, what are the given current and cross-sectional area?

• A) \(I=10\,\mathrm{A}, A=1\,\mathrm{m^2}\)
• B) \(I=1\,\mathrm{A}, A=10^{-4}\,\mathrm{m^2}\)
• C) \(I=1.6\,\mathrm{A}, A=10^{-2}\,\mathrm{m^2}\)
• D) \(I=8.5\,\mathrm{A}, A=10^4\,\mathrm{m^2}\)

✅ Correct Answer: B) \(I=1\,\mathrm{A}, A=10^{-4}\,\mathrm{m^2}\)

Explanation: The example uses a current of 1 A and a cross-sectional area of 1 cm². Converting the area gives \(10^{-4}\,\mathrm{m^2}\).

\[I=1\,\mathrm{A},\quad A=10^{-4}\,\mathrm{m^2}\]

💡 Remember Point: Convert all quantities into SI units before calculating.

📝 MCQ 18. Using \(n=8.5\times10^{28}\,\mathrm{m^{-3}}\), \(e=1.6\times10^{-19}\,\mathrm{C}\), \(I=1\,\mathrm{A}\), and \(A=10^{-4}\,\mathrm{m^2}\), what is the approximate drift velocity?

• A) \(7.35\times10^{-7}\,\mathrm{m/s}\)
• B) \(7.35\times10^7\,\mathrm{m/s}\)
• C) \(1.36\times10^{10}\,\mathrm{m/s}\)
• D) \(0.735\,\mathrm{m/s}\)

✅ Correct Answer: A) \(7.35\times10^{-7}\,\mathrm{m/s}\)

Explanation: Substitute into \(v_d=I/(neA)\). The denominator is \((8.5)(1.6)\times10^{28-19-4}=1.36\times10^6\), so \(v_d=1/(1.36\times10^6)\approx7.35\times10^{-7}\,\mathrm{m/s}\).

\[v_d\approx7.35\times10^{-7}\,\mathrm{m/s}\]

💡 Remember Point: Track powers of ten carefully during multiplication.

📝 MCQ 19. For a 10 km conductor and drift velocity \(7.35\times10^{-7}\,\mathrm{m/s}\), approximately how long is the drift-time estimate?

• A) 431 seconds
• B) 431 days
• C) 431 years
• D) 3.0 years

✅ Correct Answer: C) 431 years

Explanation: Convert 10 km to \(10^4\,\mathrm{m}\), then calculate \(t=L/v_d=10^4/(7.35\times10^{-7})\approx1.36\times10^{10}\,\mathrm{s}\). Dividing by about 31,536,000 seconds per year gives approximately 431 years.

\[t=\frac{L}{v_d}\approx1.36\times10^{10}\,\mathrm{s}\approx431\,\text{years}\]

💡 Remember Point: This idealized drift-time estimate is not the electrical signal travel time.

📝 MCQ 20. Which statement about the 431-year result is correct?

• A) A lamp must wait 431 years after switching on
• B) It estimates carrier drift over 10 km at the calculated average drift speed, not signal propagation time
• C) It is the speed of the electric field
• D) It means copper has no mobile electrons

✅ Correct Answer: B) It estimates carrier drift over 10 km at the calculated average drift speed, not signal propagation time

Explanation: The long time is obtained by dividing conductor length by the small average drift speed. It does not represent how quickly the electric field or signal travels, so a circuit can respond much sooner.

\[t_{\text{drift}}=\frac{L}{v_d}\]

💡 Remember Point: Always distinguish particle drift time from electrical signal propagation time.

📌 Quick Formula Revision

Electron drift velocity:

\[v_d=\frac{I}{neA}\]

Current and drift velocity:

\[I=neAv_d\]

Drift distance:

\[l=v_dt\]

Time for drift over length L:

\[t=\frac{L}{v_d}\]

Final reminder: Electron drift velocity is the slow average movement of charge carriers. Field or signal propagation is much faster and explains why a circuit can respond quickly after switching.