Electromagnetic Induction (HL) — IB Physics HL HL Study Guide

For: IB Physics HL candidates sitting IB Physics HL.

Covers: Magnetic flux and flux linkage, Faraday's and Lenz's laws, AC generators, transformers, capacitor charging/discharging in RC circuits, and RMS values for AC circuit calculations.

You should already know: IGCSE Physics, basic algebra and calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Electromagnetic Induction (HL)?

Electromagnetic induction is the physical phenomenon where a changing magnetic field induces an electromotive force (emf, or voltage) in a nearby conductor, with induced current flowing if the conductor forms a complete circuit. It is the foundational principle behind all grid-scale electrical power generation, consumer voltage regulation via transformers, and many common electronic circuits including timing systems and power supplies. This topic makes up 8-12% of your total IB Physics HL exam marks, and is tested in both Paper 2 (structured response) and Paper 3 (practical data analysis).

2. Magnetic flux and flux linkage

Magnetic flux ($\Phi$) measures the total magnetic field passing through a defined area, with units of Webers (Wb, where $1\text{Wb}=1{\text{Tcdotpm}}^2$). It is calculated using: $$\Phi =BA\cos \theta$$ where $B$ = magnetic flux density (T), $A$ = cross-sectional area of the surface (${\text{m}}^2$), and $\theta$ = the angle between the magnetic field vector and the normal line (perpendicular) to the surface.

For a coil of wire with $N$ identical turns, flux linkage is the total magnetic flux cutting all turns of the coil, calculated as $N\Phi$, with units of Wb-turns (often simplified to Wb for exam purposes).

Worked Example

A 75-turn rectangular coil of area $0.03{\text{m}}^2$ is placed in a uniform $0.5\text{T}$ magnetic field, with the normal to the coil at $25^{\circ }$ to the field direction. Calculate the total flux linkage.

1. Calculate single-turn flux: $\Phi =0.5\times 0.03\times \cos (25^{\circ })=0.0136\text{Wb}$
2. Multiply by number of turns: $N\Phi =75\times 0.0136=1.02\text{Wb}$, rounded to $1.0\text{Wb}$ (2 significant figures, per IB convention)

Exam tip: Examiners frequently give you the angle between the magnetic field and the plane of the coil instead of the normal. If this is the case, subtract the given angle from $90^{\circ }$ before substituting into the flux formula, or use $\Phi =BA\sin (\text{angle to plane})$ to avoid errors.

3. Faraday's and Lenz's laws

Faraday’s Law of Induction defines the magnitude of the induced emf, while Lenz’s Law defines its direction:

1. Faraday’s Law: The magnitude of the induced emf is equal to the negative rate of change of flux linkage. For average emf over a time interval $\Delta t$: $$\epsilon =-N\frac{\Delta \Phi }{\Delta t}$$ For instantaneous emf using calculus (required for HL): $$\epsilon =-N\frac{d\Phi }{dt}$$
2. Lenz’s Law: The direction of the induced emf and induced current is such that it creates a magnetic field that opposes the change in magnetic flux that caused it. This is the source of the negative sign in Faraday’s Law, and is a consequence of conservation of energy (if the induced current aided the original flux change, you would generate infinite free energy).

A common application of Lenz’s Law is eddy currents: induced currents in solid metal objects that oppose motion relative to a magnetic field, used in magnetic braking systems for trains and rollercoasters.

Worked Example

The coil from the previous example is rotated $65^{\circ }$ in $0.2\text{s}$ so that its normal is now perpendicular to the magnetic field. Calculate the average induced emf.

1. Initial flux linkage = $1.02\text{Wb}$, final flux linkage = $0$ (since $\cos (90^{\circ })=0$)
2. $\Delta (N\Phi )=0-1.02=-1.02\text{Wb}$
3. Substitute into Faraday’s Law: $\epsilon =-\left(\frac{-1.02}{0.2}\right)=5.1\text{V}$
4. Direction: The induced current creates a magnetic field in the same direction as the original external field, to oppose the decrease in flux.

Exam tip: If asked for the direction of induced current, always explicitly state Lenz’s Law in your answer to earn full marks; do not just write the final direction.

4. AC generators and transformers

AC Generators

AC generators convert mechanical kinetic energy to electrical energy by rotating a coil in a uniform magnetic field, creating a sinusoidally changing flux linkage. The instantaneous induced emf is given by: $$\epsilon ={\epsilon }_0\sin (\omega t)$$ where ${\epsilon }_0=NBA\omega$ = peak emf, $\omega =2\pi f$ = angular velocity of rotation, and $f$ = frequency of rotation in Hz. Slip rings and brushes maintain electrical contact with the rotating coil to output alternating current to the external circuit.

Transformers

Transformers step up or step down AC voltage using mutual induction between two coiled wires wrapped around a shared soft iron core. For an ideal transformer (100% efficiency, no flux leakage, zero coil resistance): $$\frac{V_p}{V_s}=\frac{N_p}{N_s},\frac{I_p}{I_s}=\frac{N_s}{N_p},V_p{I}_p=V_s{I}_s$$ where subscripts $p$ = primary coil (input) and $s$ = secondary coil (output).

Note: Transformers only operate with alternating current (AC), as direct current (DC) produces a constant magnetic flux in the primary coil, so no emf is induced in the secondary coil. This is a frequently tested trick question in multiple-choice papers.

Worked Example

An ideal step-down transformer converts $230\text{V}$ RMS mains voltage to $18\text{V}$ RMS for a portable drill that draws $2.5\text{A}$ RMS current. If the primary coil has 3800 turns, calculate (a) the number of secondary turns, (b) the primary RMS current.

1. (a) $N_s=N_p\times \frac{V_s}{V_p}=3800\times \frac{18}{230}=297$, rounded to 300 turns (2 sig figs)
2. (b) $I_p=I_s\times \frac{V_s}{V_p}=2.5\times \frac{18}{230}=0.20\text{A}$ RMS

5. Capacitors — charging/discharging through resistors

A resistor-capacitor (RC) circuit is a series circuit containing a resistor of resistance $R$ and capacitor of capacitance $C$. The time constant $\tau =RC$ (units: seconds) is the time taken for the capacitor charge/voltage to fall to $1/e\approx 37\%$ of its initial value during discharging, or rise to $1-1/e\approx 63\%$ of its maximum value during charging.

Charging formulas (capacitor connected to DC voltage source $V_0$):

$$Q(t)=Q_0\left(1-e^{-t/\tau }\right),V_c(t)=V_0\left(1-e^{-t/\tau }\right),I(t)=I_0{e}^{-t/\tau }$$ where $Q_0=C{V}_0$ = maximum charge, $I_0=V_0/R$ = initial charging current.

Discharging formulas (fully charged capacitor disconnected from source, connected to resistor):

$$Q(t)=Q_0{e}^{-t/\tau },V_c(t)=V_0{e}^{-t/\tau },I(t)=-I_0{e}^{-t/\tau }$$ The negative sign indicates current flows in the opposite direction to charging current.

Worked Example

A $470\mu \text{F}$ capacitor is charged through a $1.2\text{kΩ}$ resistor from a $9\text{V}$ battery. Calculate (a) the time constant, (b) the voltage across the capacitor after 2 seconds.

1. (a) $\tau =RC=1200\times 470\times 10^{-6}=0.564\text{s}$, rounded to $0.56\text{s}$
2. (b) $V_c=9\left(1-e^{-2/0.564}\right)=9\left(1-e^{-3.55}\right)=9\times 0.971=8.7\text{V}$

6. RMS values

The root mean square (RMS) value of an alternating current or voltage is the equivalent constant DC value that dissipates the same average power in a pure resistive load. Since the average value of a sinusoidal AC signal over a full cycle is zero, RMS values are used for all practical AC power calculations.

Key formulas: $$V_{\text{rms}}=\frac{V_0}{\sqrt{2}},I_{\text{rms}}=\frac{I_0}{\sqrt{2}}$$ $$P_{\text{avg}}=V_{\text{rms}}{I}_{\text{rms}}=I_{\text{rms}}^{2}R=\frac{V_{\text{rms}}^2}{R}$$ where $V_0$ and $I_0$ are peak voltage and peak current respectively.

Worked Example

A household oven is rated for $2.2\text{kW}$ average power when connected to a $220\text{V}$ RMS mains supply. Calculate (a) the RMS current drawn by the oven, (b) the peak voltage of the supply.

1. (a) $I_{\text{rms}}=\frac{P_{\text{avg}}}{V_{\text{rms}}}=\frac{2200}{220}=10\text{A}$ RMS
2. (b) $V_0=V_{\text{rms}}\times \sqrt{2}=220\times 1.414=311\text{V}$, rounded to $310\text{V}$ (2 sig figs)

Exam tip: Never use peak voltage or current values to calculate average power, unless explicitly asked for peak power. All household appliance power ratings are given for RMS input voltage.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Using the wrong angle in magnetic flux calculations Wrong move: Using the angle between the magnetic field and the plane of the coil instead of the normal. Why it happens: Misreading the question or misremembering the formula. Correct move: If given the angle to the plane, subtract it from $90^{\circ }$ before substituting into $\Phi =BA\cos \theta$, or use $\Phi =BA\sin (\text{angle to plane})$ explicitly.
• Pitfall 2: Ignoring Lenz's Law when asked for induced current direction Wrong move: Only calculating the magnitude of induced emf and guessing the direction. Why it happens: Rushing through calculations to save time. Correct move: Always state "By Lenz's Law, the induced current opposes the change in flux" before explaining the direction to earn full marks.
• Pitfall 3: Applying the ideal transformer current ratio to non-ideal transformers Wrong move: Using $\frac{I_p}{I_s}=\frac{N_s}{N_p}$ when the question states the transformer has efficiency < 100%. Why it happens: Memorizing formulas without context. Correct move: For non-ideal transformers, use $P_{\text{out}}=\eta \times P_{\text{in}}$ (where $\eta$ = efficiency) to calculate currents, only the voltage ratio $\frac{V_p}{V_s}\approx \frac{N_p}{N_s}$ remains approximately valid.
• Pitfall 4: Using linear decay instead of exponential decay for RC circuits Wrong move: Calculating capacitor voltage as $V_c=V_0(1-t/\tau )$ instead of the exponential formula. Why it happens: Confusing RC decay with linear motion or radioactive half-life (which is exponential but often approximated for 1-2 half-lives). Correct move: Always use the exponential formulas, and remember that after $3\tau$ the capacitor is ~95% charged/discharged, and after $5\tau$ it is effectively fully charged/discharged.
• Pitfall 5: Using peak values for AC power calculations Wrong move: Substituting $V_0$ instead of $V_{\text{rms}}$ into power formulas. Why it happens: Forgetting the definition of RMS values. Correct move: If you are given a peak value, convert it to RMS first before calculating average power, unless explicitly asked for peak power.

8. Practice Questions (IB Physics HL Style)

Question 1

A square coil of 120 turns, side length 6 cm, is placed in a uniform $0.7\text{T}$ magnetic field with its plane parallel to the field. The coil is rotated at a constant frequency of 60 Hz around an axis perpendicular to the magnetic field. (a) Calculate the peak induced emf in the coil. [3 marks] (b) State the position of the coil when the induced emf is zero. [1 mark] (c) Calculate the RMS value of the induced emf. [1 mark]

Worked Solution

(a) Step 1: Calculate coil area $A=(0.06{)}^2=0.0036{\text{m}}^2$. Step 2: Angular velocity $\omega =2\pi f=2\pi \times 60=377\text{rad/s}$. Step 3: Peak emf ${\epsilon }_0=NBA\omega =120\times 0.7\times 0.0036\times 377=114\text{V}$, rounded to $110\text{V}$ (2 sig figs). (b) Emf is zero when the rate of change of flux is zero, i.e., when the normal to the coil is parallel to the magnetic field (plane of the coil is perpendicular to the field). (c) $V_{\text{rms}}=\frac{114}{\sqrt{2}}=81\text{V}$ (2 sig figs).

Question 2

A non-ideal transformer has an efficiency of 92%, 3000 primary turns, and 150 secondary turns. The primary is connected to a $240\text{V}$ RMS supply, and the secondary is connected to a $8\Omega$ resistive load. (a) Calculate the secondary RMS voltage, assuming negligible flux leakage. [1 mark] (b) Calculate the average power delivered to the load. [2 marks] (c) Calculate the primary RMS current. [2 marks]

Worked Solution

(a) $V_s=V_p\times \frac{N_s}{N_p}=240\times \frac{150}{3000}=12\text{V}$ RMS. (b) Power to load $P_{\text{out}}=\frac{V_s^2}{R}=\frac{12^2}{8}=18\text{W}$. (c) Input power $P_{\text{in}}=\frac{P_{\text{out}}}{\eta }=\frac{18}{0.92}=19.57\text{W}$. Primary current $I_p=\frac{P_{\text{in}}}{V_p}=\frac{19.57}{240}=0.082\text{A}$ RMS (2 sig figs).

Question 3

A $330\mu \text{F}$ capacitor is fully charged to $12\text{V}$, then disconnected from the power supply and connected across a $2.2\text{kΩ}$ resistor. (a) Calculate the time constant of the circuit. [1 mark] (b) Calculate the current flowing in the circuit 1 second after the circuit is connected. [2 marks] (c) Calculate the time taken for the capacitor charge to drop to 10% of its initial value. [2 marks]

Worked Solution

(a) $\tau =RC=2200\times 330\times 10^{-6}=0.726\text{s}$, rounded to $0.73\text{s}$. (b) Initial discharge current $I_0=\frac{V_0}{R}=\frac{12}{2200}=0.00545\text{A}$. $I(t)=I_0{e}^{-t/\tau }=0.00545\times e^{-1/0.726}=0.00545\times 0.252=0.0014\text{A}$, or $1.4\text{mA}$ (2 sig figs). (c) $\frac{Q}{Q_0}=0.1=e^{-t/\tau }⟹\ln (0.1)=-t/\tau ⟹t=\tau \times \ln (10)=0.726\times 2.303=1.7\text{s}$ (2 sig figs).

9. Quick Reference Cheatsheet

10. What's Next

Electromagnetic induction is a foundational topic that connects directly to two later HL syllabus areas: first, Topic 11: Electromagnetic Waves, where oscillating electric and magnetic fields induce each other to propagate through space without a medium; and second, Topic 12: Quantum and Nuclear Physics, where induced emf is used to accelerate charged particles in particle accelerators and measure photoelectric current. Mastery of this topic is also required for Paper 3 practical exams, as 30-40% of HL practical data analysis questions test RC circuit or AC generator data interpretation.

If you struggle with any of the concepts, calculation techniques, or exam tricks covered in this guide, you can ask Ollie for personalized explanations, additional practice questions, or step-by-step walkthroughs of official past exam problems at any time. Head to the homepage] to access your personalized study dashboard, where you can take targeted quizzes on electromagnetic induction to test your mastery and identify knowledge gaps before your exam.