Electricity and Magnetism

Electric Charge and Fields

Electric Charge

Electric charge is a fundamental property of matter. There are two types of charge: positive and negative. Like charges repel and unlike charges attract.

The SI unit of charge is the coulomb (C). The smallest unit of free charge is the elementary charge:

$e = 1.6 \times 10^{-19} \mathrm{ C}$

A proton carries charge $+e$ and an electron carries charge $-e$. The charge on any object is an integer multiple of $e$ (quantisation of charge):

$Q = ne$

Where $n$ is an integer.

Conductors, Insulators, and Semiconductors

Material Type	Description	Examples
Conductor	Charges move freely; low resistivity	Copper, aluminium, gold, silver
Insulator	Charges are tightly bound; very high resistivity	Rubber, glass, plastic, wood
Semiconductor	Intermediate properties; conductivity can be controlled	Silicon, germanium

Methods of Charging

Charging by friction: When two different materials are rubbed together, electrons transfer from one to the other. The material that gains electrons becomes negatively charged; the one that loses electrons becomes positively charged.

Charging by contact: A charged object touches a neutral object, transferring charge directly. Both objects end up with the same sign of charge.

Charging by induction: A charged object is brought near (but does not touch) a neutral conductor. The conductor is earthed while the charged object is nearby, then the earth connection is removed. The conductor is left with a charge opposite to that of the inducing object.

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In induction, the total charge of the isolated conductor remains zero during the process. The earth connection allows charge to flow, so when the earth is disconnected, the conductor retains a net charge.

Coulomb's Law

The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them:

$F = \frac{kQ_1 Q_2}{r^2}$

Where:

• $F$ = electrostatic force (N)
• $Q_1$, $Q_2$ = magnitudes of the charges (C)
• $r$ = distance between the charges (m)
• $k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 \mathrm{ N m}^2\mathrm{/C}^2$ (Coulomb's constant)
• $\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{ F/m}$ (permittivity of free space)

In vector form (for the force on $Q_2$ due to $Q_1$):

$\vec{F}_{12} = \frac{kQ_1 Q_2}{r^2} \hat{r}_{12}$

Where $\hat{r}_{12}$ is the unit vector pointing from $Q_1$ to $Q_2$. A negative value of $F$ indicates attraction; a positive value indicates repulsion.

Electric Field

An electric field is a region of space where a charged particle experiences an electrostatic force. The electric field strength is defined as the force per unit positive charge:

$E = \frac{F}{q}$

Where $E$ is measured in newtons per coulomb (N/C) or volts per metre (V/m).

Electric field of a point charge:

$E = \frac{kQ}{r^2} = \frac{Q}{4\pi\varepsilon_0 r^2}$

The field points away from a positive charge and towards a negative charge.

Electric Field Lines

Charges and Fields

Place positive and negative charges to visualise the resulting electric field lines and equipotential lines.

Electric field lines are a visual representation of the electric field:

• Field lines start on positive charges and end on negative charges
• The density of field lines indicates the strength of the field
• Field lines never cross
• Field lines are perpendicular to the surface of a conductor at equilibrium

Field line patterns:

Configuration	Field Line Pattern
Single positive charge	Radial lines pointing outward
Single negative charge	Radial lines pointing inward
Two equal positive charges	Lines repel, zero field point between them
Parallel plates	Uniform field between plates (fringing at edges)

Uniform Electric Field Between Parallel Plates

When a potential difference $V$ is applied across two parallel plates separated by distance $d$, a uniform electric field is produced between the plates:

$E = \frac{V}{d}$

This is only valid for the region between the plates (away from the edges).

Work Done in an Electric Field

When a charge $q$ moves through a potential difference $V$, the work done is:

$W = qV = qEd$

Where $E$ is the uniform field strength and $d$ is the distance moved in the direction of the field.

If the charge moves against the field, work is done on the charge (it gains electric potential energy). If it moves with the field, work is done by the charge (it loses electric potential energy).

Relationship Between E, V, and d

For a uniform field:

$E = \frac{V}{d} \quad \Longrightarrow \quad V = Ed$

The electric field points from high potential to low potential. A positive charge accelerates in the direction of the field (from high to low potential), while a negative charge accelerates opposite to the field.

Worked Example 1

Two point charges $Q_1 = 3 \times 10^{-6} \mathrm{ C}$ and $Q_2 = -5 \times 10^{-6} \mathrm{ C}$ are separated by $0.2 \mathrm{ m}$. Find the electrostatic force between them.

Solution

$F = \frac{kQ_1 Q_2}{r^2} = \frac{(8.99 \times 10^9)(3 \times 10^{-6})(5 \times 10^{-6})}{(0.2)^2}$

$F = \frac{8.99 \times 10^9 \times 15 \times 10^{-12}}{0.04} = \frac{0.1349}{0.04} = 3.37 \mathrm{ N}$

The negative sign of $Q_2$ means the force is attractive.

Worked Example 1b

A point charge of $+5 \times 10^{-8} \mathrm{ C}$ is placed in a uniform electric field of $2000 \mathrm{ V/m}$. Find the force on the charge and the work done to move it $0.3 \mathrm{ m}$ against the field.

Solution

$F = qE = (5 \times 10^{-8})(2000) = 1.0 \times 10^{-4} \mathrm{ N}$

$W = qEd = (5 \times 10^{-8})(2000)(0.3) = 3.0 \times 10^{-5} \mathrm{ J}$

The force acts in the direction of the field (for a positive charge). Moving against the field requires work to be done on the charge.

Worked Example 2

A uniform electric field of strength $5000 \mathrm{ V/m}$ exists between two parallel plates separated by $4 \mathrm{ cm}$. Find the potential difference and the force on an electron placed between the plates.

Solution

$V = Ed = 5000 \times 0.04 = 200 \mathrm{ V}$

$F = qE = (1.6 \times 10^{-19})(5000) = 8.0 \times 10^{-16} \mathrm{ N}$

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Current Electricity

Electric Current

Electric current is the rate of flow of electric charge:

$I = \frac{dQ}{dt}$

Where $I$ is measured in amperes (A). One ampere is one coulomb per second.

Conventional current flows from positive to negative (the direction positive charges would move). Electron flow is in the opposite direction, from negative to positive.

For a steady current: $I = \frac{Q}{t}$

The charge carried by $n$ electrons: $Q = ne$

Ammeter and Voltmeter

Instrument	Connection	Ideal Resistance	Purpose
Ammeter	In series	Zero	Measures current
Voltmeter	In parallel	Infinite	Measures potential difference

A real ammeter has very low resistance so it does not significantly affect the current in the circuit. A real voltmeter has very high resistance so it draws negligible current from the circuit.

Potential Difference

The potential difference (voltage) between two points is the energy transferred per unit charge as charge moves between those points:

$V = \frac{W}{Q} = \frac{E}{Q}$

Where $V$ is measured in volts (V). One volt is one joule per coulomb.

Resistance and Ohm's Law

Resistance is a measure of opposition to current flow:

$R = \frac{V}{I}$

Where $R$ is measured in ohms ($\Omega$).

Ohm's Law states that for a metallic conductor at constant temperature, the current through it is directly proportional to the potential difference across it:

$V = IR$

A component that obeys Ohm's law is called ohmic. Its I-V graph is a straight line through the origin.

Resistivity

The resistance of a uniform conductor depends on its dimensions and the material:

$R = \frac{\rho L}{A}$

Where:

• $R$ = resistance ($\Omega$)
• $\rho$ = resistivity ($\Omega \mathrm{ m}$)
• $L$ = length of conductor (m)
• $A$ = cross-sectional area (m$^2$)

Common resistivities at $20^\circ\mathrm{C}$:

Material	Resistivity ($\Omega \mathrm{ m}$)
Silver	$1.59 \times 10^{-8}$
Copper	$1.68 \times 10^{-8}$
Gold	$2.44 \times 10^{-8}$
Aluminium	$2.65 \times 10^{-8}$
Iron	$9.71 \times 10^{-8}$
Nichrome	$1.10 \times 10^{-6}$
Glass	$\sim 10^{10} - 10^{14}$
Rubber	$\sim 10^{13} - 10^{16}$

Factors Affecting Resistance

• Length: Resistance is directly proportional to length ($R \propto L$)
• Cross-sectional area: Resistance is inversely proportional to area ($R \propto 1/A$)
• Temperature: For metals, resistance increases with temperature. For semiconductors, resistance decreases with temperature
• Material: Determined by the resistivity $\rho$

Thermistor

A thermistor is a temperature-dependent resistor.

NTC (Negative Temperature Coefficient) thermistor: Resistance decreases as temperature increases. The I-V characteristic is non-linear and shows that at higher currents (which heat the thermistor), the resistance drops.

PTC (Positive Temperature Coefficient) thermistor: Resistance increases as temperature increases.

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NTC thermistors are commonly used in temperature sensing circuits. As the temperature rises, the resistance falls, causing more current to flow, which can be measured or used to trigger a circuit.

Light-Dependent Resistor (LDR)

An LDR is a component whose resistance decreases with increasing light intensity.

• In darkness: resistance is very high ($\sim \mathrm{M}\Omega$)
• In bright light: resistance is low ($\sim \mathrm{k}\Omega$)

LDRs are used in light-sensitive switches, automatic street lights, and camera exposure meters.

I-V Characteristics

Ohmic conductor (e.g., a fixed resistor at constant temperature):

• I-V graph is a straight line through the origin
• Resistance is constant at all voltages

Filament lamp:

• I-V graph is a curve that flattens at higher currents
• Resistance increases with temperature (as current increases, the filament heats up)
• The graph is steeper at low voltages and shallower at high voltages

Semiconductor diode:

• Allows current to flow easily in one direction (forward bias) but blocks current in the opposite direction (reverse bias)
• In forward bias: current increases rapidly above a threshold voltage ($\sim 0.7 \mathrm{ V}$ for silicon)
• In reverse bias: almost no current flows until breakdown voltage is reached

Worked Example 3

A copper wire has a length of $10 \mathrm{ m}$, a diameter of $0.5 \mathrm{ mm}$, and a resistivity of $1.68 \times 10^{-8} \Omega \mathrm{ m}$. Find its resistance.

Solution

$A = \pi r^2 = \pi \left(\frac{0.5 \times 10^{-3}}{2}\right)^2 = \pi (2.5 \times 10^{-4})^2 = \pi \times 6.25 \times 10^{-8} = 1.963 \times 10^{-7} \mathrm{ m}^2$

$R = \frac{\rho L}{A} = \frac{(1.68 \times 10^{-8})(10)}{1.963 \times 10^{-7}} = \frac{1.68 \times 10^{-7}}{1.963 \times 10^{-7}} = 0.856 \Omega$

Worked Example 3b

A nichrome wire of length $2.0 \mathrm{ m}$ has resistance $10 \Omega$. If the wire is stretched to $3.0 \mathrm{ m}$ (keeping the volume constant), what is its new resistance?

Solution

Since volume is constant: $A_1 L_1 = A_2 L_2$

$\frac{A_2}{A_1} = \frac{L_1}{L_2} = \frac{2.0}{3.0} = \frac{2}{3}$

$R_2 = \frac{\rho L_2}{A_2} = \frac{\rho L_2}{A_1 \times (L_1/L_2)} = R_1 \times \left(\frac{L_2}{L_1}\right)^2 = 10 \times \left(\frac{3.0}{2.0}\right)^2 = 10 \times 2.25 = 22.5 \Omega$

If you get this wrong, revise: Resistivity formula and the relationship between resistance, length, and cross-sectional area.

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Series and Parallel Circuits

Series Circuits

In a series circuit, components are connected end to end along a single path.

• Current: The same current flows through all components: $I = I_1 = I_2 = I_3 = \cdots$
• Voltage: The total voltage equals the sum of the individual voltages: $V = V_1 + V_2 + V_3 + \cdots$
• Resistance: The total resistance equals the sum of the individual resistances: $R_{\mathrm{total}} = R_1 + R_2 + R_3 + \cdots$

Parallel Circuits

In a parallel circuit, components are connected across the same two points, providing multiple paths for current.

• Voltage: The same voltage is across all branches: $V = V_1 = V_2 = V_3 = \cdots$
• Current: The total current equals the sum of the branch currents: $I = I_1 + I_2 + I_3 + \cdots$
• Resistance: The reciprocal of the total resistance equals the sum of the reciprocals of individual resistances:

$\frac{1}{R_{\mathrm{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$

For two resistors in parallel:

$R_{\mathrm{total}} = \frac{R_1 R_2}{R_1 + R_2}$

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In a parallel circuit, the total resistance is always less than the smallest individual resistance. Adding more parallel branches always decreases the total resistance.

Combined Series and Parallel Circuits

For circuits with both series and parallel combinations, simplify step by step:

1. Identify parallel groups and find their equivalent resistance
2. Add series resistances
3. Repeat until the circuit is reduced to a single equivalent resistance

Potential Divider

A potential divider (voltage divider) consists of two or more resistors in series across a supply voltage. It produces a fraction of the input voltage.

For two resistors $R_1$ and $R_2$ in series across input voltage $V_{\mathrm{in}}$:

$V_{\mathrm{out}} = V_{\mathrm{in}} \times \frac{R_2}{R_1 + R_2}$

The output voltage is taken across $R_2$ (the lower resistor in the divider).

Loaded potential divider: When a load resistance $R_L$ is connected across $R_2$, the effective resistance of the lower branch changes. The loaded output voltage is:

$V_{\mathrm{out}} = V_{\mathrm{in}} \times \frac{R_2 \parallel R_L}{R_1 + (R_2 \parallel R_L)}$

Where $R_2 \parallel R_L = \frac{R_2 R_L}{R_2 + R_L}$.

The load reduces the output voltage (the divider is "loaded down"). For the output voltage to remain stable, $R_L$ should be much larger than $R_2$.

Potentiometer

A potentiometer is a variable resistor with three terminals that acts as an adjustable potential divider. A sliding contact moves along a resistive track, allowing continuous adjustment of the output voltage from $0$ to $V_{\mathrm{in}}$.

Internal Resistance of Real Sources

A real power source (battery, cell) has an internal resistance $r$. The EMF ($\varepsilon$) of the source is the total energy supplied per unit charge. The terminal PD ($V$) is the voltage measured across the terminals when current flows.

$V = \varepsilon - Ir$

Where:

• $V$ = terminal potential difference (V)
• $\varepsilon$ = electromotive force (V)
• $I$ = current (A)
• $r$ = internal resistance ($\Omega$)
• $Ir$ = "lost volts" (voltage lost across the internal resistance)

The maximum terminal PD occurs when no current flows (open circuit): $V = \varepsilon$.

The terminal PD decreases as the current increases.

Measuring EMF and Internal Resistance

Method 1: Open-circuit and closed-circuit measurements

1. Measure the terminal PD with no load (open circuit): $V_{\mathrm{open}} = \varepsilon$
2. Measure the terminal PD with a known load resistance $R$: $V = \varepsilon - Ir$
3. Since $I = V/R$, solve for $r$: $r = \frac{\varepsilon - V}{V} \times R = R\left(\frac{\varepsilon}{V} - 1\right)$

Method 2: Graphical method

Measure the terminal PD for several different load resistances and plot $V$ against $I$:

$V = \varepsilon - Ir$

This is a straight line with:

• $y$-intercept = $\varepsilon$
• Gradient = $-r$
• $x$-intercept = $\varepsilon / r$ (short-circuit current)

Worked Example 4

A battery has an EMF of $12 \mathrm{ V}$ and internal resistance of $0.5 \Omega$. It is connected to an external resistance of $5.5 \Omega$. Find the current, terminal PD, and power dissipated in the external resistance.

Solution

$I = \frac{\varepsilon}{R + r} = \frac{12}{5.5 + 0.5} = \frac{12}{6} = 2.0 \mathrm{ A}$

$V = \varepsilon - Ir = 12 - 2.0 \times 0.5 = 12 - 1.0 = 11.0 \mathrm{ V}$

$P = I^2 R = (2.0)^2 \times 5.5 = 4 \times 5.5 = 22.0 \mathrm{ W}$

Worked Example 5

A potential divider consists of $R_1 = 4 \mathrm{ k}\Omega$ and $R_2 = 6 \mathrm{ k}\Omega$ connected across a $12 \mathrm{ V}$ supply. Find the output voltage across $R_2$. A load resistance of $10 \mathrm{ k}\Omega$ is then connected across $R_2$. Find the new output voltage.

Solution

Unloaded:

$V_{\mathrm{out}} = 12 \times \frac{6}{4 + 6} = 12 \times \frac{6}{10} = 7.2 \mathrm{ V}$

Loaded:

$R_2 \parallel R_L = \frac{6 \times 10}{6 + 10} = \frac{60}{16} = 3.75 \mathrm{ k}\Omega$

$V_{\mathrm{out}} = 12 \times \frac{3.75}{4 + 3.75} = 12 \times \frac{3.75}{7.75} = 12 \times 0.4839 = 5.81 \mathrm{ V}$

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Electrical Energy and Power

Electrical Power

Power is the rate of energy transfer:

$P = IV$

Using Ohm's law ($V = IR$), this can be rewritten as:

$P = I^2 R = \frac{V^2}{R}$

Where $P$ is measured in watts (W).

Formula	Use When
$P = IV$	Current and voltage are known
$P = I^2 R$	Current and resistance are known
$P = V^2/R$	Voltage and resistance are known

Electrical Energy

Energy transferred is power multiplied by time:

$E = Pt = VIt = I^2 Rt = \frac{V^2}{R} t$

Where $E$ is measured in joules (J).

The Kilowatt-Hour

The kilowatt-hour (kWh) is a practical unit of electrical energy used for billing:

$1 \mathrm{ kWh} = 1000 \mathrm{ W} \times 3600 \mathrm{ s} = 3.6 \times 10^6 \mathrm{ J} = 3.6 \mathrm{ MJ}$

Electricity cost calculation:

$\mathrm{Cost} = \mathrm{Energy (kWh)} \times \mathrm{Rate (\$/kWh)}$

Efficiency

Efficiency is the ratio of useful energy (or power) output to total energy (or power) input:

$\mathrm{Efficiency} = \frac{\mathrm{Useful output}}{\mathrm{Total input}} \times 100\%$

For electrical devices:

$\mathrm{Efficiency} = \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \times 100\% = \frac{V_{\mathrm{out}} I_{\mathrm{out}}}{V_{\mathrm{in}} I_{\mathrm{in}}} \times 100\%$

Worked Example 6

A $2 \mathrm{ kW}$ electric heater is used for $3$ hours per day. If the electricity rate is USD 0.90 per kWh, find the daily cost and the total energy consumed in one month (30 days).

Solution

Daily energy: $E = Pt = 2 \times 3 = 6 \mathrm{ kWh}$

Daily cost: \mathrm{Cost} = 6 \times 0.90 = \5.40$

Monthly energy: $6 \times 30 = 180 \mathrm{ kWh}$

Monthly cost: 180 \times 0.90 = \162$

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Magnetic Fields

Magnetic Field Around a Current-Carrying Wire

A long straight wire carrying current $I$ produces a concentric circular magnetic field. The magnitude of the magnetic flux density at distance $r$ from the wire is:

$B = \frac{\mu_0 I}{2\pi r}$

Where:

• $B$ = magnetic flux density (T, tesla)
• $\mu_0 = 4\pi \times 10^{-7} \mathrm{ T m/A}$ (permeability of free space)
• $I$ = current (A)
• $r$ = perpendicular distance from the wire (m)

The direction of the field is given by the right-hand grip rule: if the thumb points in the direction of conventional current, the fingers curl in the direction of the magnetic field.

Magnetic Field Inside a Solenoid

A solenoid is a long coil of wire. When current flows, it produces a nearly uniform magnetic field inside and a weak field outside. The magnetic flux density inside an ideal solenoid is:

$B = \mu_0 n I$

Where $n$ is the number of turns per unit length ($n = N/L$, where $N$ is total turns and $L$ is the length of the solenoid).

The field inside the solenoid is uniform and parallel to the axis. Outside, the field is weak and diverges at the ends.

Force on a Current-Carrying Wire in a Magnetic Field

A wire of length $L$ carrying current $I$ in a magnetic field of flux density $B$ experiences a force:

$F = BIL \sin\theta$

Where $\theta$ is the angle between the direction of the current and the direction of the magnetic field.

• When $\theta = 90^\circ$ (wire perpendicular to field): $F = BIL$ (maximum force)
• When $\theta = 0^\circ$ (wire parallel to field): $F = 0$ (no force)

The direction of the force is given by Fleming's Left-Hand Rule:

• First finger: direction of the magnetic field ($B$)
• Second finger: direction of conventional current ($I$)
• Thumb: direction of the force ($F$)

Force Between Parallel Current-Carrying Wires

Two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ exert forces on each other. The force per unit length is:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

• If the currents flow in the same direction, the wires attract each other
• If the currents flow in opposite directions, the wires repel each other

Worked Example 7

A wire carrying a current of $5 \mathrm{ A}$ is placed in a magnetic field of flux density $0.3 \mathrm{ T}$. The wire is $0.4 \mathrm{ m}$ long and makes an angle of $30^\circ$ with the field. Find the force on the wire.

Solution

$F = BIL \sin\theta = 0.3 \times 5 \times 0.4 \times \sin 30^\circ$

$F = 0.6 \times 0.5 = 0.30 \mathrm{ N}$

Worked Example 8

Two long parallel wires separated by $0.1 \mathrm{ m}$ carry currents of $10 \mathrm{ A}$ and $15 \mathrm{ A}$ in the same direction. Find the force per unit length between them.

Solution

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(10)(15)}{2\pi \times 0.1}$

$\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 150}{0.2\pi} = \frac{6 \times 10^{-5}}{0.2} = 3.0 \times 10^{-4} \mathrm{ N/m}$

Since the currents are in the same direction, the force is attractive.

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Electromagnetic Induction

Magnetic Flux

Magnetic flux through a surface is defined as:

$\Phi = BA \cos\theta$

Where:

• $\Phi$ = magnetic flux (Wb, weber)
• $B$ = magnetic flux density (T)
• $A$ = area of the surface (m$^2$)
• $\theta$ = angle between the magnetic field and the normal to the surface

When $\theta = 0^\circ$ (field perpendicular to surface): $\Phi = BA$ (maximum flux)

When $\theta = 90^\circ$ (field parallel to surface): $\Phi = 0$ (no flux)

Faraday's Law of Electromagnetic Induction

The magnitude of the induced EMF is equal to the rate of change of magnetic flux linkage:

$\varepsilon = -\frac{d\Phi}{dt}$

For a coil with $N$ turns:

$\varepsilon = -N\frac{d\Phi}{dt}$

The negative sign indicates the direction of the induced EMF (see Lenz's Law).

An EMF can be induced by:

1. Changing the magnetic field strength $B$
2. Changing the area $A$ of the coil (e.g., moving a wire into or out of a field)
3. Changing the angle $\theta$ between the field and the coil (e.g., rotating a coil)
4. Moving a magnet towards or away from a coil

Lenz's Law

The direction of the induced EMF is such that it opposes the change producing it.

This is a consequence of conservation of energy. The induced current creates a magnetic field that opposes the change in flux that produced it.

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To determine the direction of the induced current using Lenz's Law:

1. Identify whether the flux through the coil is increasing or decreasing
2. If flux is increasing, the induced current creates a field to oppose the increase (opposing the external field)
3. If flux is decreasing, the induced current creates a field to oppose the decrease (reinforcing the external field)

AC Generator

An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A coil is rotated at constant angular velocity in a uniform magnetic field.

The induced EMF is:

$\varepsilon = \varepsilon_0 \sin(\omega t) = NAB\omega \sin(\omega t)$

Where:

• $\varepsilon_0 = NAB\omega$ is the peak EMF
• $\omega = 2\pi f$ is the angular frequency
• $N$ = number of turns
• $A$ = area of the coil
• $B$ = magnetic flux density
• $f$ = frequency of rotation

The output is sinusoidal, alternating between $+\varepsilon_0$ and $-\varepsilon_0$.

Transformers

A transformer changes the voltage of an AC supply. It consists of a primary coil and a secondary coil wound around a soft iron core.

Transformer equation:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where:

• $V_p$, $V_s$ = primary and secondary voltages
• $N_p$, $N_s$ = number of turns on primary and secondary coils

If $N_s \gt N_p$: step-up transformer ($V_s \gt V_p$)

If $N_s \lt N_p$: step-down transformer ($V_s \lt V_p$)

Conservation of energy (ideal transformer):

$V_p I_p = V_s I_s$

$\frac{I_s}{I_p} = \frac{N_p}{N_s}$

A step-up transformer increases voltage but decreases current, and vice versa.

Efficiency of transformers:

$\mathrm{Efficiency} = \frac{V_s I_s}{V_p I_p} \times 100\%$

Real transformers have losses due to:

• Copper losses (resistive heating in the windings)
• Eddy current losses (induced currents in the iron core)
• Hysteresis losses (repeated magnetisation and demagnetisation of the core)
• Flux leakage (not all magnetic flux links both coils)

Transformers only work with AC. A changing current in the primary is needed to produce a

changing magnetic flux, which induces an EMF in the secondary. DC produces a steady field and no induced EMF.

Worked Example 9

A transformer has 200 turns on the primary coil and 1000 turns on the secondary coil. The primary voltage is $220 \mathrm{ V}$ and the primary current is $5 \mathrm{ A}$. Find the secondary voltage, secondary current, and power.

Solution

$V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{1000}{200} = 220 \times 5 = 1100 \mathrm{ V}$

$I_s = I_p \times \frac{N_p}{N_s} = 5 \times \frac{200}{1000} = 5 \times 0.2 = 1.0 \mathrm{ A}$

$P = V_p I_p = 220 \times 5 = 1100 \mathrm{ W}$

(Checking: $V_s I_s = 1100 \times 1.0 = 1100 \mathrm{ W}$, which confirms conservation of energy.)

Worked Example 9b

A rectangular coil of 100 turns, each of area $0.02 \mathrm{ m}^2$, is rotated at $50 \mathrm{ rev/s}$ in a uniform magnetic field of flux density $0.5 \mathrm{ T}$. Find the peak EMF.

Solution

$\omega = 2\pi f = 2\pi \times 50 = 100\pi \mathrm{ rad/s}$

$\varepsilon_0 = NAB\omega = 100 \times 0.02 \times 0.5 \times 100\pi = 100\pi = 314 \mathrm{ V}$

If you get this wrong, revise: Faraday's law and the AC generator EMF formula.

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AC Circuits

AC Voltage and Current

In an AC circuit, the voltage and current vary sinusoidally with time:

$V = V_0 \sin(\omega t) = V_0 \sin(2\pi ft)$

$I = I_0 \sin(\omega t) = I_0 \sin(2\pi ft)$

Where:

• $V_0$, $I_0$ = peak (maximum) values
• $f$ = frequency (Hz)
• $\omega = 2\pi f$ = angular frequency (rad/s)
• $T = 1/f$ = period (s)

Root Mean Square (RMS) Values

The RMS value of an AC quantity is the equivalent DC value that would produce the same heating effect in a resistor.

For a sinusoidal waveform:

$V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0$

$I_{\mathrm{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$

Power in an AC circuit:

$P = V_{\mathrm{rms}} I_{\mathrm{rms}} = \frac{V_0 I_0}{2}$

Mains Electricity

In Hong Kong (DSE context), mains electricity is supplied as:

• Voltage: $220 \mathrm{ V}$ (RMS)
• Frequency: $50 \mathrm{ Hz}$

The peak voltage is:

$V_0 = 220\sqrt{2} = 311 \mathrm{ V}$

The period is:

$T = \frac{1}{f} = \frac{1}{50} = 0.02 \mathrm{ s} = 20 \mathrm{ ms}$

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Summary Table

Topic	Key Formula	Key Concept
Coulomb's Law	$F = kQ_1 Q_2/r^2$	Inverse square law for electric forces
Electric field	$E = F/q = kQ/r^2$	Force per unit charge
Uniform field	$E = V/d$	Parallel plates
Ohm's Law	$V = IR$	Linear I-V for ohmic conductors
Resistivity	$R = \rho L/A$	Material and geometry dependence
Series resistance	$R = R_1 + R_2 + \cdots$	Same current
Parallel resistance	$1/R = 1/R_1 + 1/R_2 + \cdots$	Same voltage
Potential divider	$V_{\mathrm{out}} = V_{\mathrm{in}} R_2/(R_1 + R_2)$	Voltage fraction
Internal resistance	$V = \varepsilon - Ir$	Lost volts
Power	$P = IV = I^2 R = V^2/R$	Rate of energy transfer
Magnetic field (wire)	$B = \mu_0 I/(2\pi r)$	Right-hand grip rule
Magnetic field (solenoid)	$B = \mu_0 n I$	Uniform field inside
Force on wire	$F = BIL \sin\theta$	Fleming's Left-Hand Rule
Faraday's Law	$\varepsilon = -N d\Phi/dt$	Changing flux induces EMF
Transformer	$V_s/V_p = N_s/N_p$	AC only, energy conservation
AC RMS	$V_{\mathrm{rms}} = V_0/\sqrt{2}$	Equivalent DC heating effect

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Exam Tips

• Always distinguish between EMF (total energy per unit charge) and terminal PD (energy per unit charge delivered to the external circuit).
• In circuit problems, start by finding the total resistance, then the total current, then work out individual voltages and currents.
• When using Coulomb's Law, use the magnitudes of the charges and determine the direction of the force separately.
• For electromagnetic induction problems, clearly identify what is changing (flux, area, angle) and apply Faraday's Law accordingly.
• Remember that transformers only work with AC; always check whether the source is AC or DC.
• In potential divider problems with a load, first find the parallel combination before applying the divider formula.
• When drawing magnetic field lines, remember: lines go from N to S outside a magnet, and from S to N inside.
• For Lenz's Law, always state that the induced current opposes the change in flux, not the flux itself.

Exam-Style Practice Questions

Question 1: Two small charged spheres, each carrying a charge of $+4 \times 10^{-8} \mathrm{ C}$, are separated by $0.05 \mathrm{ m}$ in a vacuum. Calculate the electrostatic force between them.

Solution

$F = \frac{kQ_1 Q_2}{r^2} = \frac{(8.99 \times 10^9)(4 \times 10^{-8})^2}{(0.05)^2}$

$F = \frac{8.99 \times 10^9 \times 16 \times 10^{-16}}{0.0025} = \frac{1.4384 \times 10^{-5}}{0.0025} = 5.75 \times 10^{-3} \mathrm{ N}$

The force is repulsive since both charges are positive.

Question 2: A $2000 \mathrm{ W}$ heater, a $100 \mathrm{ W}$ lamp, and a $500 \mathrm{ W}$ fan are connected in parallel across a $220 \mathrm{ V}$ mains supply. Find the total current drawn and the total cost of running them for 5 hours at USD 0.90 per kWh.

Solution

Total power: $P = 2000 + 100 + 500 = 2600 \mathrm{ W} = 2.6 \mathrm{ kW}$

Total current: $I = \frac{P}{V} = \frac{2600}{220} = 11.82 \mathrm{ A}$

Energy consumed: $E = Pt = 2.6 \times 5 = 13 \mathrm{ kWh}$

Total cost: \mathrm{Cost} = 13 \times 0.90 = \11.70$

Question 3: In a potential divider circuit, $R_1 = 10 \mathrm{ k}\Omega$ and $R_2 = 20 \mathrm{ k}\Omega$ are connected in series across a $9 \mathrm{ V}$ battery. A voltmeter of resistance $50 \mathrm{ k}\Omega$ is connected across $R_2$. Find the reading on the voltmeter.

Solution

The voltmeter acts as a load across $R_2$:

$R_2 \parallel R_V = \frac{20 \times 50}{20 + 50} = \frac{1000}{70} = 14.29 \mathrm{ k}\Omega$

$V_{\mathrm{out}} = 9 \times \frac{14.29}{10 + 14.29} = 9 \times \frac{14.29}{24.29} = 9 \times 0.5884 = 5.30 \mathrm{ V}$

Question 4: A cell of EMF $6 \mathrm{ V}$ and internal resistance $0.8 \Omega$ is connected to an external circuit consisting of a $4.2 \Omega$ resistor in series with a parallel combination of two $6 \Omega$ resistors. Find the current from the cell and the terminal PD.

Solution

Equivalent resistance of the parallel combination:

$R_p = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3 \Omega$

Total external resistance: $R = 4.2 + 3 = 7.2 \Omega$

Total circuit resistance: $R_{\mathrm{total}} = 7.2 + 0.8 = 8.0 \Omega$

Current: $I = \frac{\varepsilon}{R_{\mathrm{total}}} = \frac{6}{8.0} = 0.75 \mathrm{ A}$

Terminal PD: $V = \varepsilon - Ir = 6 - 0.75 \times 0.8 = 6 - 0.6 = 5.4 \mathrm{ V}$

Question 5: A rectangular coil of 100 turns, each of area $0.02 \mathrm{ m}^2$, is rotated at $50 \mathrm{ rev/s}$ in a uniform magnetic field of flux density $0.5 \mathrm{ T}$. The axis of rotation is perpendicular to the field. Find the peak EMF and the EMF when the plane of the coil makes an angle of $60^\circ$ with the field.

Solution

Angular frequency: $\omega = 2\pi f = 2\pi \times 50 = 100\pi \mathrm{ rad/s}$

Peak EMF: $\varepsilon_0 = NAB\omega = 100 \times 0.02 \times 0.5 \times 100\pi = 100\pi = 314 \mathrm{ V}$

When the plane of the coil makes $60^\circ$ with the field, the angle between the normal to the coil and the field is $30^\circ$. The EMF is:

$\varepsilon = \varepsilon_0 \sin 30^\circ = 314 \times 0.5 = 157 \mathrm{ V}$

Question 6: A step-down transformer with 2000 primary turns and 100 secondary turns is used to reduce the mains voltage of $220 \mathrm{ V}$ RMS to supply a device. If the device draws a current of $10 \mathrm{ A}$ and the transformer is $90\%$ efficient, find the secondary voltage, the primary current, and the power loss.

Solution

Secondary voltage: $V_s = 220 \times \frac{100}{2000} = 220 \times 0.05 = 11 \mathrm{ V}$

Output power: $P_{\mathrm{out}} = V_s I_s = 11 \times 10 = 110 \mathrm{ W}$

Input power: $P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{0.90} = \frac{110}{0.90} = 122.2 \mathrm{ W}$

Primary current: $I_p = \frac{P_{\mathrm{in}}}{V_p} = \frac{122.2}{220} = 0.556 \mathrm{ A}$

Power loss: $P_{\mathrm{loss}} = P_{\mathrm{in}} - P_{\mathrm{out}} = 122.2 - 110 = 12.2 \mathrm{ W}$

Question 7: An electron is accelerated from rest through a potential difference of $5000 \mathrm{ V}$ between two parallel plates separated by $2 \mathrm{ cm}$. Find the electric field strength, the force on the electron, and the velocity of the electron as it reaches the positive plate.

Solution

Electric field: $E = \frac{V}{d} = \frac{5000}{0.02} = 2.5 \times 10^5 \mathrm{ V/m}$

Force on electron: $F = eE = 1.6 \times 10^{-19} \times 2.5 \times 10^5 = 4.0 \times 10^{-14} \mathrm{ N}$

Using energy conservation: $eV = \frac{1}{2}m_e v^2$

$v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 5000}{9.11 \times 10^{-31}}}$

$v = \sqrt{\frac{1.6 \times 10^{-15}}{9.11 \times 10^{-31}}} = \sqrt{1.756 \times 10^{15}} = 4.19 \times 10^7 \mathrm{ m/s}$

Question 8: A straight wire of length $0.5 \mathrm{ m}$ carries a current of $8 \mathrm{ A}$ and lies perpendicular to a uniform magnetic field of flux density $0.4 \mathrm{ T}$. Calculate the force on the wire. If the wire is rotated so that it makes an angle of $45^\circ$ with the field, what is the new force?

Solution

Force when perpendicular ($\theta = 90^\circ$):

$F = BIL \sin 90^\circ = 0.4 \times 8 \times 0.5 \times 1 = 1.6 \mathrm{ N}$

Force when at $45^\circ$:

$F = BIL \sin 45^\circ = 0.4 \times 8 \times 0.5 \times 0.7071 = 1.131 \mathrm{ N}$

Question 9: A cell is connected to a variable resistor $R$. A voltmeter across the cell reads $1.5 \mathrm{ V}$ when $R = 8 \Omega$ and $1.2 \mathrm{ V}$ when $R = 4 \Omega$. Find the EMF and internal resistance of the cell.

Solution

When $R = 8 \Omega$: $I_1 = \frac{V_1}{R_1} = \frac{1.5}{8} = 0.1875 \mathrm{ A}$

$\varepsilon = V_1 + I_1 r = 1.5 + 0.1875r \quad \cdots (1)$

When $R = 4 \Omega$: $I_2 = \frac{V_2}{R_2} = \frac{1.2}{4} = 0.3 \mathrm{ A}$

$\varepsilon = V_2 + I_2 r = 1.2 + 0.3r \quad \cdots (2)$

Equating (1) and (2):

$1.5 + 0.1875r = 1.2 + 0.3r$

$0.3 = 0.1125r$

$r = \frac{0.3}{0.1125} = 2.67 \Omega$

$\varepsilon = 1.2 + 0.3 \times 2.67 = 1.2 + 0.8 = 2.0 \mathrm{ V}$

Question 10: A solenoid of length $0.3 \mathrm{ m}$ has 600 turns and carries a current of $4 \mathrm{ A}$. A straight wire running parallel to the axis of the solenoid at a distance of $0.02 \mathrm{ m}$ from it carries a current of $5 \mathrm{ A}$ in the same direction. Find the magnetic flux density inside the solenoid and the force per unit length on the wire.

Solution

Magnetic flux density inside the solenoid:

$n = \frac{N}{L} = \frac{600}{0.3} = 2000 \mathrm{ turns/m}$

$B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(4) = 4\pi \times 10^{-7} \times 8000 = 0.01005 \mathrm{ T}$

The wire runs parallel to the axis of the solenoid, so the current in the wire is parallel to the magnetic field inside the solenoid. Since the angle between the current and the field is $0^\circ$:

$F = BIL \sin 0^\circ = 0$

The force per unit length on the wire is zero.

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Problem Set

Problem 1: Electric Field of a Point Charge

A point charge of $+6 \times 10^{-9} \mathrm{ C}$ creates an electric field. Find the field strength at a distance of $0.05 \mathrm{ m}$ from the charge and the force on a $-2 \times 10^{-9} \mathrm{ C}$ charge placed at that point.

Solution

$E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(6 \times 10^{-9})}{(0.05)^2} = \frac{53.94}{0.0025} = 21576 \mathrm{ V/m}$

$F = qE = (2 \times 10^{-9})(21576) = 4.32 \times 10^{-5} \mathrm{ N}$

The force is attractive (towards the positive charge).

If you get this wrong, revise: Coulomb's law, electric field of a point charge, and the relationship $E = F/q$.

Problem 2: Series-Parallel Circuit

A $12 \mathrm{ V}$ battery (internal resistance $0.5 \Omega$) is connected to a $3 \Omega$ resistor in series with a parallel combination of $6 \Omega$ and $12 \Omega$. Find the total current and the power dissipated in the $6 \Omega$ resistor.

Solution

$R_p = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \Omega$

$R_{\mathrm{total}} = 0.5 + 3 + 4 = 7.5 \Omega$

$I = \frac{12}{7.5} = 1.6 \mathrm{ A}$

Voltage across the parallel combination:

$V_p = 12 - 1.6(0.5 + 3) = 12 - 5.6 = 6.4 \mathrm{ V}$

Current through $6 \Omega$: $I_6 = \frac{6.4}{6} = 1.067 \mathrm{ A}$

$P_6 = I_6^2 R = (1.067)^2 \times 6 = 1.139 \times 6 = 6.83 \mathrm{ W}$

If you get this wrong, revise: Series and parallel resistance combinations and internal resistance.

Problem 3: Current from Charge Flow

A wire carries a current of $3 \mathrm{ A}$. How many electrons pass through a cross-section of the wire in $10$ seconds?

Solution

$Q = It = 3 \times 10 = 30 \mathrm{ C}$

$n = \frac{Q}{e} = \frac{30}{1.6 \times 10^{-19}} = 1.875 \times 10^{20} \mathrm{ electrons}$

If you get this wrong, revise: The relationship between current, charge, and the elementary charge $Q = ne = It$.

Problem 4: Resistivity — Finding Length

A tungsten wire has a resistance of $5 \Omega$ and a diameter of $0.2 \mathrm{ mm}$. The resistivity of tungsten is $5.6 \times 10^{-8} \Omega \mathrm{ m}$. Find the length of the wire.

Solution

$A = \pi r^2 = \pi(1.0 \times 10^{-4})^2 = \pi \times 10^{-8} = 3.14 \times 10^{-8} \mathrm{ m}^2$

$R = \frac{\rho L}{A} \implies L = \frac{RA}{\rho} = \frac{5 \times 3.14 \times 10^{-8}}{5.6 \times 10^{-8}} = \frac{15.7}{5.6} = 2.80 \mathrm{ m}$

If you get this wrong, revise: Resistivity formula $R = \rho L / A$ and rearranging to find length.

Problem 5: Potential Divider with Thermistor

A potential divider consists of a $10 \mathrm{ k}\Omega$ fixed resistor and an NTC thermistor in series across a $9 \mathrm{ V}$ supply. At $20^\circ\mathrm{C}$, the thermistor resistance is $10 \mathrm{ k}\Omega$. At $80^\circ\mathrm{C}$, the thermistor resistance drops to $2 \mathrm{ k}\Omega$. Find the output voltage across the thermistor at each temperature.

Solution

At $20^\circ\mathrm{C}$:

$V_{\mathrm{out}} = 9 \times \frac{10}{10 + 10} = 9 \times 0.5 = 4.5 \mathrm{ V}$

At $80^\circ\mathrm{C}$:

$V_{\mathrm{out}} = 9 \times \frac{2}{10 + 2} = 9 \times \frac{2}{12} = 1.5 \mathrm{ V}$

The output voltage decreases as temperature increases, which is the principle behind temperature sensing circuits.

If you get this wrong, revise: Potential divider formula and how NTC thermistor resistance varies with temperature.

Problem 6: Kirchhoff's Laws — Two-Loop Circuit

A $12 \mathrm{ V}$ battery is connected to a $4 \Omega$ and $6 \Omega$ resistor in parallel. Find the current through each resistor and the total current from the battery.

Solution

Both resistors have $12 \mathrm{ V}$ across them (parallel):

$I_4 = \frac{12}{4} = 3.0 \mathrm{ A}$

$I_6 = \frac{12}{6} = 2.0 \mathrm{ A}$

$I_{\mathrm{total}} = 3.0 + 2.0 = 5.0 \mathrm{ A}$

If you get this wrong, revise: Kirchhoff's current law and parallel circuit analysis.

Problem 7: Force on a Moving Charge

An electron ($m_e = 9.11 \times 10^{-31} \mathrm{ kg}$) moves at $2.0 \times 10^6 \mathrm{ m/s}$ perpendicular to a magnetic field of flux density $0.5 \mathrm{ T}$. Find the force on the electron and the radius of its circular path.

Solution

$F = qvB = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.5) = 1.6 \times 10^{-13} \mathrm{ N}$

For circular motion: $F = \frac{mv^2}{r}$

$r = \frac{mv}{qB} = \frac{(9.11 \times 10^{-31})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.5)} = \frac{1.822 \times 10^{-24}}{8.0 \times 10^{-20}} = 2.28 \times 10^{-5} \mathrm{ m}$

If you get this wrong, revise: Force on a moving charge in a magnetic field ($F = qvB$) and circular motion of charged particles.

Problem 8: Faraday's Law — Moving Wire

A straight wire of length $0.3 \mathrm{ m}$ moves perpendicular to a uniform magnetic field of $0.8 \mathrm{ T}$ at a speed of $5 \mathrm{ m/s}$. Find the induced EMF.

Solution

$\varepsilon = BLv = (0.8)(0.3)(5) = 1.2 \mathrm{ V}$

If you get this wrong, revise: Faraday's law for a straight conductor moving through a magnetic field ($\varepsilon = BLv$).

Problem 9: AC Power Calculation

A $220 \mathrm{ V}$ RMS mains supply is connected to a $100 \Omega$ resistor. Find the peak current, the peak power, and the average power dissipated.

Solution

$I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{R} = \frac{220}{100} = 2.2 \mathrm{ A}$

$I_0 = I_{\mathrm{rms}}\sqrt{2} = 2.2 \times 1.414 = 3.11 \mathrm{ A}$

$V_0 = V_{\mathrm{rms}}\sqrt{2} = 220 \times 1.414 = 311 \mathrm{ V}$

$P_{\mathrm{peak}} = V_0 I_0 = 311 \times 3.11 = 967 \mathrm{ W}$

$P_{\mathrm{average}} = V_{\mathrm{rms}} I_{\mathrm{rms}} = 220 \times 2.2 = 484 \mathrm{ W}$

Note: $P_{\mathrm{average}} = \frac{P_{\mathrm{peak}}}{2}$ as expected.

If you get this wrong, revise: RMS values, peak values, and power in AC circuits.

Problem 10: Transformer Efficiency

A step-up transformer has 500 turns on the primary and 2500 turns on the secondary. The primary voltage is $110 \mathrm{ V}$ and the primary current is $10 \mathrm{ A}$. If the transformer is $95\%$ efficient, find the secondary voltage, the output power, and the secondary current.

Solution

$V_s = V_p \times \frac{N_s}{N_p} = 110 \times \frac{2500}{500} = 110 \times 5 = 550 \mathrm{ V}$

$P_{\mathrm{in}} = V_p I_p = 110 \times 10 = 1100 \mathrm{ W}$

$P_{\mathrm{out}} = 0.95 \times 1100 = 1045 \mathrm{ W}$

$I_s = \frac{P_{\mathrm{out}}}{V_s} = \frac{1045}{550} = 1.90 \mathrm{ A}$

If you get this wrong, revise: Transformer equation and efficiency calculations.

Problem 11: Internal Resistance Graph

A battery with unknown EMF and internal resistance is connected to a variable resistor. The following measurements are obtained: at $I = 0.5 \mathrm{ A}$, $V = 11.5 \mathrm{ V}$; at $I = 1.5 \mathrm{ A}$, $V = 10.5 \mathrm{ V}$. Find $\varepsilon$ and $r$.

Solution

From $V = \varepsilon - Ir$:

At $I = 0.5$: $11.5 = \varepsilon - 0.5r \quad \cdots (1)$

At $I = 1.5$: $10.5 = \varepsilon - 1.5r \quad \cdots (2)$

Subtracting (2) from (1):

$1.0 = 1.0r$

$r = 1.0 \Omega$

Substituting into (1):

$\varepsilon = 11.5 + 0.5(1.0) = 12.0 \mathrm{ V}$

If you get this wrong, revise: Internal resistance and the equation $V = \varepsilon - Ir$.

Problem 12: Magnetic Field at Centre of Circular Loop

A circular loop of wire of radius $0.1 \mathrm{ m}$ carries a current of $5 \mathrm{ A}$. Find the magnetic flux density at the centre of the loop.

Solution

$B = \frac{\mu_0 I}{2r} = \frac{(4\pi \times 10^{-7})(5)}{2 \times 0.1} = \frac{6.283 \times 10^{-6}}{0.2} = 3.14 \times 10^{-5} \mathrm{ T}$

If you get this wrong, revise: Magnetic field at the centre of a circular current loop ($B = \mu_0 I / 2r$).

Problem 13: Energy to Heat Water

An electric kettle rated at $1500 \mathrm{ W}$, $220 \mathrm{ V}$ is used to boil $1.2 \mathrm{ kg}$ of water from $20^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$. Specific heat capacity of water is $4200 \mathrm{ J/(kg }^\circ\mathrm{C)}$. If the kettle is $85\%$ efficient, find the time taken.

Solution

$Q = mc\Delta T = 1.2 \times 4200 \times 80 = 403200 \mathrm{ J}$

Energy that must be supplied: $E = \frac{403200}{0.85} = 474353 \mathrm{ J}$

$t = \frac{E}{P} = \frac{474353}{1500} = 316.2 \mathrm{ s} = 5.27 \mathrm{ minutes}$

If you get this wrong, revise: Electrical energy, specific heat capacity, and efficiency.

Problem 14: Superposition of Electric Fields

Two point charges, $Q_1 = +4 \times 10^{-9} \mathrm{ C}$ at the origin and $Q_2 = +9 \times 10^{-9} \mathrm{ C}$ at $x = 0.1 \mathrm{ m}$, lie along the x-axis. Find the point on the x-axis between the charges where the electric field is zero.

Solution

At a point $x$ metres from $Q_1$ (between the charges, so $0 \lt x \lt 0.1$):

$E_1 = E_2$

$\frac{kQ_1}{x^2} = \frac{kQ_2}{(0.1 - x)^2}$

$\frac{4}{x^2} = \frac{9}{(0.1 - x)^2}$

$\frac{2}{x} = \frac{3}{0.1 - x}$

$2(0.1 - x) = 3x$

$0.2 - 2x = 3x$

$5x = 0.2$

$x = 0.04 \mathrm{ m}$

The electric field is zero at $x = 0.04 \mathrm{ m}$ from $Q_1$ (or $0.06 \mathrm{ m}$ from $Q_2$).

If you get this wrong, revise: Superposition of electric fields and solving for the null point.

Problem 15: Eddy Currents and Lenz's Law

A bar magnet is dropped through a copper tube. Explain why it falls more slowly than through a plastic tube and describe the direction of the induced currents.

Solution

As the magnet falls through the copper tube, the changing magnetic flux through the tube induces eddy currents in the copper walls (by Faraday's law).

By Lenz's law, these eddy currents create magnetic fields that oppose the change in flux:

• Above the magnet (where flux is increasing): the induced field opposes the magnet's field, producing an upward force
• Below the magnet (where flux is decreasing): the induced field reinforces the magnet's field, also producing an upward force

Both effects create a retarding force on the magnet, slowing its fall. A plastic tube is an insulator, so no eddy currents are induced, and the magnet falls freely under gravity.

If you get this wrong, revise: Lenz's law, eddy currents, and electromagnetic braking.

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tip

Diagnostic Test Ready to test your understanding of Electricity and Magnetism? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Electricity and Magnetism with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

danger

Common Pitfalls

• Confusing series and parallel circuit rules: In series, current is the SAME through all components but voltage is shared. In parallel, voltage is the SAME across all branches but current is shared. Students frequently apply the wrong rule -- remember: series = shared voltage, parallel = shared current.

• Using the wrong formula for transformer efficiency: The ideal transformer equation (Vp/Vs = Np/Ns) assumes 100% efficiency. Real transformers have energy losses, so the power output is less than the power input. If efficiency is given, use: efficiency = (Vs * Is) / (Vp * Ip), and account for the losses.

• Forgetting that magnetic field lines form closed loops: Electric field lines start on positive charges and end on negative charges, but magnetic field lines always form CLOSED LOOPS with no beginning or end. There are no magnetic monopoles. This is why cutting a bar magnet in half produces two smaller magnets, not isolated north and south poles.

• Misidentifying the direction of induced current: When applying Lenz's law, first determine whether the flux through the coil is increasing or decreasing. If increasing, the induced current creates a magnetic field that OPPOSES the increase. If decreasing, the induced current creates a field that TRIES TO MAINTAIN the flux. Then use the right-hand grip rule to find the current direction.

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Derivations

Derivation: Electric Field Strength from Coulomb's Law

A point charge $Q$ produces an electric field. A test charge $q$ placed at distance $r$ from $Q$ experiences a force given by Coulomb's law:

$F = \frac{kQq}{r^2} = \frac{Qq}{4\pi\varepsilon_0 r^2}$

The electric field strength is defined as the force per unit positive charge:

$E = \frac{F}{q} = \frac{kQ}{r^2} = \frac{Q}{4\pi\varepsilon_0 r^2}$

For a uniform field between parallel plates (potential difference $V$, separation $d$):

$E = \frac{V}{d}$

Derivation: Capacitor Energy

A capacitor of capacitance $C$ charged to potential difference $V$ stores charge $Q = CV$.

The work done to add a small charge $dq$ when the capacitor is at potential $v$ is:

$dW = v\, dq = \frac{q}{C}\, dq$

The total work to charge from $0$ to $Q$:

$W = \int_0^Q \frac{q}{C}\, dq = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$

This energy is stored in the electric field between the plates.

Derivation: Time Constant of an RC Circuit

In an RC circuit, when a capacitor $C$ is charged through a resistor $R$ from a supply voltage $V_0$, the charge on the capacitor at time $t$ is:

$Q = Q_0(1 - e^{-t/RC})$

where $Q_0 = CV_0$ is the maximum charge.

The time constant $\tau = RC$ is the time for the charge to reach $(1 - 1/e) \approx 63.2\%$ of its maximum value. At $t = \tau$: $Q = Q_0(1 - e^{-1}) = 0.632Q_0$.

At $t = 5\tau$: $Q = Q_0(1 - e^{-5}) = 0.993Q_0$ (effectively fully charged).

During discharge: $Q = Q_0 e^{-t/RC}$, so at $t = \tau$: $Q = 0.368Q_0$.

Derivation: Force on a Charge in Combined Electric and Magnetic Fields

A charge $q$ moving with velocity $v$ in a region with both electric field $E$ and magnetic field $B$ experiences:

$\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}$

For the velocity selector (used in mass spectrometers), $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the velocity. The forces balance when:

$qE = qvB \implies v = \frac{E}{B}$

Only charges with this specific velocity pass through undeflected.

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Experimental Methods

Investigating Ohm's Law

Apparatus: A variable resistor (rheostat), an ammeter, a voltmeter, a test resistor, a battery, and connecting wires.

Procedure:

1. Connect the test resistor in series with the ammeter and rheostat. Connect the voltmeter in parallel across the test resistor.
2. Adjust the rheostat to obtain different values of current.
3. Record pairs of voltage $V$ and current $I$.
4. Plot $I$ (y-axis) versus $V$ (x-axis).
5. A straight line through the origin confirms Ohm's law. The gradient equals $1/R$.

Precautions:

• Use a low voltage to avoid heating the resistor.
• Wait for readings to stabilise before recording.
• Ensure good electrical connections to minimise contact resistance.

Measuring Capacitance Using a Reed Switch

Apparatus: A capacitor, a reed switch driven by an AC signal of known frequency $f$, a resistor, a DC power supply, and a voltmeter.

Procedure:

1. The reed switch alternately charges the capacitor (from the supply voltage $V$) and discharges it (through a microammeter) at frequency $f$.
2. The charge per cycle: $Q = CV$.
3. The average current through the microammeter: $I = Qf = CVf$.
4. Calculate: $C = I/(Vf)$.

Determining the Charge-to-Mass Ratio of an Electron

Apparatus: An electron deflection tube, a power supply for the electron gun, Helmholtz coils to produce a uniform magnetic field, and measuring equipment.

Procedure:

1. Accelerate electrons through a known potential difference $V_a$.
2. The electrons enter a region of uniform magnetic field $B$ and travel in a circular arc.
3. Measure the radius $r$ of the circular path.
4. The kinetic energy: $\frac{1}{2}mv^2 = eV_a$.
5. The magnetic force provides the centripetal force: $evB = \frac{mv^2}{r}$.
6. Combining: $\frac{e}{m} = \frac{2V_a}{B^2 r^2}$.

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Data Analysis and Uncertainty

Uncertainty in RC Circuit Measurements

When measuring the time constant $\tau = RC$:

$\frac{\Delta\tau}{\tau} = \sqrt{\left(\frac{\Delta R}{R}\right)^2 + \left(\frac{\Delta C}{C}\right)^2}$

Example: $R = (10.0 \pm 0.5) \mathrm{ k}\Omega$, $C = (100 \pm 5) \mathrm{ \mu F}$:

$\tau = RC = 10.0 \times 10^3 \times 100 \times 10^{-6} = 1.00 \mathrm{ s}$

$\frac{\Delta\tau}{\tau} = \sqrt{(0.05)^2 + (0.05)^2} = \sqrt{0.005} = 0.0707 = 7.1\%$

$\Delta\tau = 0.071 \times 1.00 = 0.07 \mathrm{ s}$

$\tau = (1.00 \pm 0.07) \mathrm{ s}$

Analysing Exponential Decay Data

When verifying the discharge equation $Q = Q_0 e^{-t/RC}$:

• Plot $\ln Q$ (y-axis) versus $t$ (x-axis).
• A straight line with negative gradient confirms exponential decay.
• The gradient equals $-1/(RC)$.
• The y-intercept equals $\ln Q_0$.

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Additional Worked Examples

Worked Example 11

A $200 \mathrm{ \mu F}$ capacitor is charged to $50 \mathrm{ V}$ and then disconnected from the supply. It is then connected across an uncharged $300 \mathrm{ \mu F}$ capacitor.

(a) Calculate the final common voltage across both capacitors.

(b) Calculate the energy stored before and after connection, and explain the energy loss.

Solution

(a) Charge on the charged capacitor: $Q = C_1 V_1 = 200 \times 10^{-6} \times 50 = 0.010 \mathrm{ C}$

When connected in parallel, charge is conserved: $Q = (C_1 + C_2)V_f$

$V_f = \frac{Q}{C_1 + C_2} = \frac{0.010}{200 \times 10^{-6} + 300 \times 10^{-6}} = \frac{0.010}{500 \times 10^{-6}} = 20 \mathrm{ V}$

(b) Energy before: $E_i = \frac{1}{2}C_1 V_1^2 = \frac{1}{2}(200 \times 10^{-6})(2500) = 0.250 \mathrm{ J}$

Energy after: $E_f = \frac{1}{2}(C_1 + C_2)V_f^2 = \frac{1}{2}(500 \times 10^{-6})(400) = 0.100 \mathrm{ J}$

Energy lost: $\Delta E = 0.250 - 0.100 = 0.150 \mathrm{ J}$

This energy is dissipated as heat in the connecting wires and as electromagnetic radiation during the transient current flow when the capacitors are connected.

Worked Example 12

A $470 \mathrm{ \mu F}$ capacitor is charged through a $100 \mathrm{ k}\Omega$ resistor from a $12 \mathrm{ V}$ supply.

(a) Calculate the time constant.

(b) Calculate the charge on the capacitor after one time constant.

(c) Calculate the time for the capacitor to reach $95\%$ of its maximum charge.

(d) Calculate the current in the circuit at $t = 0$ and at $t = \tau$.

Solution

(a) $\tau = RC = 100 \times 10^3 \times 470 \times 10^{-6} = 47.0 \mathrm{ s}$

(b) After one time constant: $Q = Q_0(1 - e^{-1}) = 0.632 Q_0$

$Q_0 = CV_0 = 470 \times 10^{-6} \times 12 = 5.64 \times 10^{-3} \mathrm{ C}$

$Q = 0.632 \times 5.64 \times 10^{-3} = 3.56 \times 10^{-3} \mathrm{ C}$

(c) $Q/Q_0 = 0.95 = 1 - e^{-t/\tau}$

$e^{-t/\tau} = 0.05 \implies -t/\tau = \ln(0.05) = -3.00$

$t = 3.00\tau = 3.00 \times 47.0 = 141 \mathrm{ s}$

(d) At $t = 0$: $I_0 = V_0/R = 12/(100 \times 10^3) = 1.20 \times 10^{-4} \mathrm{ A} = 120 \mathrm{ \mu A}$

At $t = \tau$: $I = I_0 e^{-1} = 120 \times 0.368 = 44.1 \mathrm{ \mu A}$

Worked Example 13

An electron is accelerated from rest through a potential difference of $2000 \mathrm{ V}$ and then enters a uniform magnetic field of $5.0 \times 10^{-3} \mathrm{ T}$ perpendicular to its velocity. Calculate the radius of the circular path.

(Electron charge $= 1.6 \times 10^{-19} \mathrm{ C}$, electron mass $= 9.11 \times 10^{-31} \mathrm{ kg}$)

Solution

Kinetic energy gained: $\frac{1}{2}mv^2 = eV$

$v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.11 \times 10^{-31}}} = \sqrt{\frac{6.4 \times 10^{-16}}{9.11 \times 10^{-31}}} = \sqrt{7.03 \times 10^{14}} = 2.65 \times 10^7 \mathrm{ m/s}$

$r = \frac{mv}{eB} = \frac{9.11 \times 10^{-31} \times 2.65 \times 10^7}{1.6 \times 10^{-19} \times 5.0 \times 10^{-3}} = \frac{2.41 \times 10^{-23}}{8.0 \times 10^{-22}} = 0.0301 \mathrm{ m} = 3.01 \mathrm{ cm}$

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Exam-Style Questions

Question 1 (DSE Structured)

(a) Define capacitance.

(b) A $100 \mathrm{ \mu F}$ capacitor is connected in series with a $1.0 \mathrm{ M}\Omega$ resistor and a $6.0 \mathrm{ V}$ battery.

(i) Calculate the time constant.

(ii) Calculate the maximum energy stored in the capacitor.

(iii) Calculate the voltage across the capacitor after $30 \mathrm{ s}$.

(iv) Sketch a graph showing how the current in the circuit varies with time.

Solution

(a) Capacitance is the ratio of the charge stored on a capacitor to the potential difference across it: $C = Q/V$. The SI unit is the farad (F), where $1 \mathrm{ F} = 1 \mathrm{ C/V}$.

(b) (i) $\tau = RC = 1.0 \times 10^6 \times 100 \times 10^{-6} = 100 \mathrm{ s}$

(ii) $E_{\max} = \frac{1}{2}CV^2 = \frac{1}{2}(100 \times 10^{-6})(36) = 1.8 \times 10^{-3} \mathrm{ J} = 1.8 \mathrm{ mJ}$

(iii) $V_C = V_0(1 - e^{-t/\tau}) = 6.0(1 - e^{-30/100}) = 6.0(1 - e^{-0.30}) = 6.0(1 - 0.741) = 6.0 \times 0.259 = 1.55 \mathrm{ V}$

(iv) The current starts at $I_0 = V_0/R = 6.0/10^6 = 6.0 \mathrm{ \mu A}$ and decays exponentially: $I = I_0 e^{-t/\tau}$. The graph is an exponential decay curve starting at $6.0 \mathrm{ \mu A}$ and approaching zero. At $t = \tau = 100 \mathrm{ s}$, the current has fallen to $6.0 \times 0.368 = 2.21 \mathrm{ \mu A}$.

Question 2 (DSE Structured)

(a) State Coulomb's law.

(b) Two point charges, $q_1 = +3.0 \mathrm{ \mu C}$ and $q_2 = -5.0 \mathrm{ \mu C}$, are separated by $0.20 \mathrm{ m}$ in a vacuum.

(i) Calculate the force between them. Is it attractive or repulsive?

(ii) Calculate the electric field strength at the midpoint between the two charges.

(iii) At what point on the line joining the charges is the electric field strength zero?

Solution

(a) Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: $F = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}$. The force is attractive for opposite charges and repulsive for like charges.

(b) (i) $F = \frac{|q_1 q_2|}{4\pi\varepsilon_0 r^2} = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-6} \times 5.0 \times 10^{-6}}{(0.20)^2} = \frac{8.99 \times 10^9 \times 1.5 \times 10^{-11}}{0.04} = \frac{0.1349}{0.04} = 3.37 \mathrm{ N}$

The force is attractive (opposite charges).

(ii) At the midpoint ($r = 0.10 \mathrm{ m}$ from each charge):

$E_1$ (from $+3.0 \mathrm{ \mu C}$, pointing away): $E_1 = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-6}}{(0.10)^2} = \frac{2.697 \times 10^4}{0.01} = 2.70 \times 10^6 \mathrm{ V/m}$

$E_2$ (from $-5.0 \mathrm{ \mu C}$, pointing towards it): $E_2 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{(0.10)^2} = \frac{4.495 \times 10^4}{0.01} = 4.50 \times 10^6 \mathrm{ V/m}$

Since both fields point in the same direction (from $+q_1$ towards $-q_2$):

$E_{\mathrm{net}} = E_1 + E_2 = 2.70 \times 10^6 + 4.50 \times 10^6 = 7.20 \times 10^6 \mathrm{ V/m}$

(iii) The zero-field point must be outside the charges, on the side of the smaller charge ($q_1 = +3.0 \mathrm{ \mu C}$), since the larger charge dominates near the midpoint.

Let the point be distance $x$ from $q_1$ (beyond $q_1$). Distance from $q_2 = x + 0.20$.

$\frac{kq_1}{x^2} = \frac{kq_2}{(x + 0.20)^2}$

$\frac{3.0}{x^2} = \frac{5.0}{(x + 0.20)^2}$

$3.0(x + 0.20)^2 = 5.0x^2$

$3.0(x^2 + 0.40x + 0.04) = 5.0x^2$

$3x^2 + 1.2x + 0.12 = 5x^2$

$2x^2 - 1.2x - 0.12 = 0$

$x = \frac{1.2 \pm \sqrt{1.44 + 0.96}}{4} = \frac{1.2 \pm \sqrt{2.40}}{4} = \frac{1.2 \pm 1.549}{4}$

Taking the positive root: $x = \frac{1.2 + 1.549}{4} = 0.687 \mathrm{ m}$ from $q_1$.

Question 3 (DSE Structured)

(a) Explain the difference between electric potential and electric field strength.

(b) Two parallel metal plates are separated by $5.0 \mathrm{ cm}$ with a potential difference of $500 \mathrm{ V}$ across them.

(i) Calculate the electric field strength between the plates.

(ii) An electron enters the field midway between the plates, moving perpendicular to the field with speed $3.0 \times 10^7 \mathrm{ m/s}$. Calculate the vertical deflection of the electron as it exits the plates (plate length $= 10 \mathrm{ cm}$).

(iii) Calculate the angle at which the electron leaves the field.

Solution

(a) Electric potential ($V$) is the electric potential energy per unit charge at a point: $V = E_p/q$. It is a scalar quantity with unit volt (V).

Electric field strength ($E$) is the force per unit positive charge at a point: $E = F/q$. It is a vector quantity with unit $\mathrm{ V/m}$ or $\mathrm{ N/C}$. The field points from high potential to low potential.

(b) (i) $E = V/d = 500/0.05 = 10000 \mathrm{ V/m} = 10.0 \mathrm{ kV/m}$

(ii) The electron enters horizontally. Time to cross the plates:

$t = \frac{L}{v_x} = \frac{0.10}{3.0 \times 10^7} = 3.33 \times 10^{-9} \mathrm{ s}$

Vertical acceleration: $a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 10000}{9.11 \times 10^{-31}} = \frac{1.6 \times 10^{-15}}{9.11 \times 10^{-31}} = 1.76 \times 10^{15} \mathrm{ m/s}^2$

Vertical deflection: $y = \frac{1}{2}at^2 = \frac{1}{2}(1.76 \times 10^{15})(3.33 \times 10^{-9})^2 = 0.5 \times 1.76 \times 10^{15} \times 1.11 \times 10^{-17} = 9.76 \times 10^{-3} \mathrm{ m} = 9.76 \mathrm{ mm}$

(iii) Vertical velocity at exit: $v_y = at = 1.76 \times 10^{15} \times 3.33 \times 10^{-9} = 5.86 \times 10^6 \mathrm{ m/s}$

$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{5.86 \times 10^6}{3.0 \times 10^7}\right) = \tan^{-1}(0.195) = 11.0^\circ$

Question 4 (DSE Structured)

(a) Describe the structure and operation of a simple AC generator.

(b) A rectangular coil of 100 turns, dimensions $0.05 \mathrm{ m} \times 0.08 \mathrm{ m}$, rotates at $50 \mathrm{ rev/s}$ in a magnetic field of $0.3 \mathrm{ T}$.

(i) Calculate the peak EMF.

(ii) Write the expression for the EMF as a function of time.

(iii) Calculate the RMS EMF.

(iv) The coil is connected to a $50 \Omega$ resistor. Calculate the average power dissipated.

Solution

(a) An AC generator consists of a rectangular coil rotating in a uniform magnetic field. As the coil rotates, the magnetic flux through the coil changes sinusoidally, inducing an EMF by Faraday's law. Slip rings and brushes provide continuous electrical contact with the external circuit. The EMF varies sinusoidally, producing alternating current.

(b) (i) $\omega = 2\pi \times 50 = 314.2 \mathrm{ rad/s}$

$A = 0.05 \times 0.08 = 0.004 \mathrm{ m}^2$

$\varepsilon_0 = NBA\omega = 100 \times 0.3 \times 0.004 \times 314.2 = 37.7 \mathrm{ V}$

(ii) $\varepsilon = 37.7\sin(314.2t) \mathrm{ V}$

(assuming the EMF is zero at $t = 0$)

(iii) $\varepsilon_{\mathrm{RMS}} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{37.7}{1.414} = 26.7 \mathrm{ V}$

(iv) Average power: $P = \frac{\varepsilon_{\mathrm{RMS}}^2}{R} = \frac{(26.7)^2}{50} = \frac{712.9}{50} = 14.3 \mathrm{ W}$