Electric current

1. Overview

Electric current is the rate of flow of electric charge. In a conducting material, this flow is facilitated by charge carriers, which are particles possessing a net electric charge. The magnitude of the current is determined by the number of charge carriers passing through a cross-section of the conductor per unit time and the charge held by each carrier. On a microscopic level, current is not a simple instantaneous movement of all particles at high speed, but rather a slow net drift of charge carriers superimposed on their rapid, random thermal motion. This topic bridges the gap between macroscopic circuit measurements ($I$ and $Q$) and the microscopic properties of materials ($n$ and $v$).

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2. Key Definitions

• Electric Current ($I$): The rate of flow of electric charge through a given cross-section. It is a fundamental SI base quantity. (Unit: Ampere, A)
• Charge ($Q$): A fundamental physical property of matter that causes it to experience a force when placed in an electromagnetic field. (Unit: Coulomb, C)
• The Coulomb ($C$): The charge that passes a point in a circuit when a steady current of one ampere flows for one second. ($1\text{ C} = 1\text{ A s}$)
• Charge Carrier: Any particle (such as an electron or ion) that has an electric charge and is free to move through a medium, thereby creating an electric current.
• Number Density ($n$): The number of free charge carriers per unit volume of a material. This is a property of the material itself. (Unit: $\text{m}^{-3}$)
• Drift Velocity ($v$): The average displacement of charge carriers along the length of a conductor per unit time when a potential difference is applied. (Unit: $\text{m s}^{-1}$)
• Quantisation of Charge: The principle that electric charge is not continuous but exists in discrete packets that are integral multiples of the elementary charge $e$.

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3. Content

3.1 Electric Current as a Flow of Charge Carriers

Current is defined as the net movement of charged particles. The specific identity of the charge carrier depends entirely on the medium through which the current is flowing:

• Metals: The charge carriers are delocalised (free) electrons. In a metallic lattice, the outermost electrons of the atoms are not bound to specific nuclei and are free to move throughout the structure.
• Electrolytes (Liquids/Solutions): Current is carried by ions. Both positive ions (cations) and negative ions (anions) move in opposite directions toward the electrodes.
• Gases: Gases are typically insulators, but if ionised (e.g., by high voltage or radiation), the charge carriers are both electrons and positive ions.
• Semiconductors: Current is carried by electrons and "holes" (the absence of an electron which acts as a positive charge carrier).

Conventional Current vs. Electron Flow In the 19th century, before the discovery of the electron, current was defined as the flow of positive charge. We maintain this "Conventional Current" today:

• Conventional Current: Flows from the positive terminal to the negative terminal.
• Electron Flow: In metallic conductors, electrons actually flow from the negative terminal to the positive terminal.
• Exam Note: Always assume current refers to conventional current unless a question specifically asks about the direction of electron flow.

3.2 Quantisation of Charge

Charge is a quantised property. This means it cannot exist in any arbitrary amount; it must be a whole-number multiple of the smallest possible unit of free charge, known as the elementary charge ($e$).

The value of the elementary charge is: $e = 1.60 \times 10^{-19} \text{ C}$

The total charge $Q$ on any object or carrier is expressed as: $Q = \pm ne$

Where:

• $Q$ = Total charge (C)
• $n$ = An integer ($1, 2, 3, \dots$)
• $e$ = Elementary charge ($1.60 \times 10^{-19} \text{ C}$)

Implications for Exams:

1. If a calculation results in a charge that is not a multiple of $1.60 \times 10^{-19} \text{ C}$ (e.g., $2.4 \times 10^{-19} \text{ C}$), that value is physically impossible for a standalone particle.
2. An electron has a charge of $-1.60 \times 10^{-19} \text{ C}$.
3. A proton has a charge of $+1.60 \times 10^{-19} \text{ C}$.
4. Alpha particles ($\alpha$) have a charge of $+2e$ ($+3.20 \times 10^{-19} \text{ C}$).

3.3 The Current Equation ($Q = It$)

Current is the rate of change of charge. For a constant (steady) current, the relationship is:

$I = \frac{\Delta Q}{\Delta t}$

Rearranging to find the total charge transferred over a period of time:

$Q = It$

Where:

• $I$ is current in Amperes (A)
• $Q$ is charge in Coulombs (C)
• $t$ is time in seconds (s)

Worked Example 1 — Charge in a Circuit

Question: A steady current of $4.5 \text{ mA}$ flows through a resistor for $15$ minutes. Calculate the total number of electrons that pass through the resistor in this time.

Step 1: Convert units to SI $I = 4.5 \text{ mA} = 4.5 \times 10^{-3} \text{ A}$ $t = 15 \text{ minutes} = 15 \times 60 = 900 \text{ s}$

Step 2: Calculate total charge $Q$ $Q = It$ $Q = (4.5 \times 10^{-3}) \times 900$ $Q = 4.05 \text{ C}$

Step 3: Calculate the number of electrons $n$ using $Q = ne$ $n = \frac{Q}{e}$ $n = \frac{4.05}{1.60 \times 10^{-19}}$ $n = 2.53125 \times 10^{19}$

Answer: $2.5 \times 10^{19}$ electrons (2 s.f.)

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3.4 Derivation of $I = Anvq$

You are required to derive this expression, which links the macroscopic current to the microscopic behavior of charge carriers.

1. Consider a section of a conductor Imagine a wire with cross-sectional area $A$. We focus on a segment of length $L$. The charge carriers in this segment move with an average drift velocity $v$.

2. Calculate the volume of the segment The volume $V$ of this cylindrical segment is: $V = \text{Area} \times \text{Length} = A \times L$

3. Calculate the total number of charge carriers ($N$) Let $n$ be the number density (number of carriers per unit volume). $N = n \times V = nAL$

4. Calculate the total charge ($Q$) in the segment If each individual carrier has a charge $q$, the total charge $Q$ contained in this volume is: $Q = N \times q = (nAL) \times q$

5. Calculate the time ($t$) for the charge to pass a point The time taken for all the charge carriers in this segment to pass through the end cross-section is the time it takes for the carriers at the very start of the segment to travel the distance $L$: $t = \frac{\text{distance}}{\text{speed}} = \frac{L}{v}$

6. Combine into the current definition $I = Q/t$ $I = \frac{nALq}{L/v}$ The $L$ terms cancel out: $I = Anvq$

Understanding the Variables:

• $I$ (Current): Measured in Amperes (A).
• $A$ (Cross-sectional Area): Measured in $\text{m}^2$. Be careful with $\text{mm}^2$ conversions.
• $n$ (Number Density): Measured in $\text{m}^{-3}$.
  • Conductors (Metals): $n \approx 10^{28} \text{ m}^{-3}$ (very high).
  • Semiconductors: $n \approx 10^{15} \text{ to } 10^{23} \text{ m}^{-3}$.
  • Insulators: $n \approx 0$.
• $v$ (Drift Velocity): Measured in $\text{m s}^{-1}$. This is the net velocity of the carriers.
• $q$ (Charge per carrier): Measured in Coulombs (C). For electrons, $q = e = 1.60 \times 10^{-19} \text{ C}$.

Worked Example 2 — Comparing Drift Velocities

Question: Two wires, P and Q, are connected in series. Wire P has a diameter $d$, and wire Q has a diameter $2d$. Both wires are made of the same material. Calculate the ratio of the drift velocity in wire P to the drift velocity in wire Q ($\frac{v_P}{v_Q}$).

Step 1: Identify constants Since they are in series, the current $I$ is the same in both. Since they are the same material, the number density $n$ and charge $q$ are the same.

Step 2: Relate $v$ to Area $A$ From $I = Anvq$, we see that $v = \frac{I}{Anq}$. Since $I, n, q$ are constant, $v \propto \frac{1}{A}$.

Step 3: Relate Area to Diameter $A = \pi r^2 = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$. Therefore, $A \propto d^2$.

Step 4: Combine proportionalities $v \propto \frac{1}{d^2}$. $\frac{v_P}{v_Q} = \frac{(d_Q)^2}{(d_P)^2}$ $\frac{v_P}{v_Q} = \frac{(2d)^2}{(d)^2} = \frac{4d^2}{d^2} = 4$

Answer: The drift velocity in wire P is 4 times greater than in wire Q. Ratio = $4:1$.

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3.5 The Microscopic View: Why is $v$ so small?

In a metal, free electrons move randomly due to thermal energy at speeds of approximately $10^5 \text{ to } 10^6 \text{ m s}^{-1}$. However, they constantly collide with the vibrating positive ions of the metallic lattice.

When an electric field is applied (by connecting a battery), the electrons experience an electric force that accelerates them. Because of the frequent collisions, they do not accelerate indefinitely but instead gain a very slow net drift in the direction opposite to the electric field.

• Typical Drift Velocity: $\approx 10^{-4} \text{ m s}^{-1}$ (less than $1 \text{ mm}$ per second).
• Why do lights turn on instantly?: Even though individual electrons move slowly, the electric field is established through the conductor at nearly the speed of light. This field exerts a force on all free electrons in the circuit simultaneously, so they all begin to drift at once.

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4. Key Equations

Equation	Symbols & Units	Data Sheet?
$Q = It$	$Q$: Charge (C), $I$: Current (A), $t$: Time (s)	No (Memorise)
$Q = \pm ne$	$n$: Number of carriers (integer), $e$: Elementary charge (C)	No (Memorise)
$I = Anvq$	$A$: Area ($\text{m}^2$), $n$: Number density ($\text{m}^{-3}$), $v$: Drift velocity ($\text{m s}^{-1}$), $q$: Charge (C)	Yes
$e = 1.60 \times 10^{-19} \text{ C}$	Elementary charge constant	Yes

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5. Common Mistakes to Avoid

• ❌ Area Unit Conversion Errors: This is the most common source of lost marks.
  • $1 \text{ mm}^2 = 1 \times 10^{-6} \text{ m}^2$ (since $(10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$).
  • $1 \text{ cm}^2 = 1 \times 10^{-4} \text{ m}^2$ (since $(10^{-2} \text{ m})^2 = 10^{-4} \text{ m}^2$).
  • ✓ Right: Always convert to $\text{m}^2$ before substituting into $I = Anvq$.
• ❌ Confusing $n$ and $N$:
  • $n$ is the number density (e.g., $8.5 \times 10^{28} \text{ m}^{-3}$).
  • $N$ is the total number of carriers (a dimensionless integer).
  • ✓ Right: Check the units. If it's "per unit volume" or "$\text{m}^{-3}$", it is $n$.
• ❌ Time Units: Using minutes or hours in $Q = It$.
  • ✓ Right: Always convert time to seconds.
• ❌ Misinterpreting $q$ in Electrolytes: In an electrolyte, if a divalent ion like $Cu^{2+}$ is the carrier, $q = 2e = 3.2 \times 10^{-19} \text{ C}$.
  • ✓ Right: Read the question carefully to see if the carrier is an electron ($e$) or an ion ($Ze$).
• ❌ Drift Velocity Magnitude: Thinking a result of $10^{-4} \text{ m s}^{-1}$ is "too small" and changing the power of ten.
  • ✓ Right: Drift velocities are naturally very small. Trust your calculation if your units are correct.

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6. Exam Tips

1. The "Show That" Derivation: You must be able to derive $I = Anvq$ from scratch. Examiners look for the intermediate steps: Volume $\rightarrow$ Number of carriers $\rightarrow$ Total charge $\rightarrow$ Time $\rightarrow$ Current.
2. Significant Figures: Cambridge 9702 generally requires answers to the same number of significant figures as the least precise data given in the question (usually 2 or 3 s.f.). Never give an answer to 1 s.f. unless specifically directed.
3. Base Units: Remember that the Ampere (A) is an SI base unit, but the Coulomb (C) is a derived unit ($1 \text{ C} = 1 \text{ A s}$). You may be asked to express the Coulomb in SI base units.
4. Proportionality Reasoning: Many multiple-choice questions (MCQs) involve a wire changing diameter or two different materials in series. Use the method shown in Worked Example 2:
   • Identify what is constant ($I$ is constant in series).
   • Write the formula in terms of the variable you need ($v = I / Anq$).
   • See how the variable changes with the input (e.g., $v \propto 1/A$).
5. Defining Quantisation: If asked to define the quantisation of charge, use the phrase: "Charge only exists in discrete amounts which are integral multiples of the elementary charge." This hits all the marking keywords.