Year 12 PhysicsModule 6⏱ ~25 min5 MC · 3 Short AnswerLesson 21 of 21
Module Consolidation
You'll be able to connect IQ1–IQ4 into a coherent picture of electromagnetism.
Consolidate the complete Module 6 story: fields exert forces, changing flux induces emf, machines transfer energy, and power networks manage voltage, current and loss. The emphasis is choosing models with the right signs, angles, units and conditions.
Big picture: Picture an illustrative 1930s scenario in the style of the Sydney Harbour Bridge's original arc lighting: AC generated at a city power station, stepped up for transmission, and stepped down by a bank of local transformers for the lamps. (The station, voltage and transformer count used here are illustrative and are not drawn from an archival source.) Can you trace the complete M6 physics chain, from Faraday's law at the generator right through to the current in those arc lamps, in a single connected explanation?
0/5TASKS
Learn0 of 6 steps complete
1
Diagnose recall across all four inquiry questions and set the shared notation conventions.
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Without looking at your notes, name one formula from each of the four Module 6 inquiry questions.
Write them here, then check the formula grid below to see how you went.
Warm-up, which of the following is not a formula you need for Module 6?
Learning Intentions
goals
Know, All Key Formulas
Recall all twelve Module 6 formulas and the conditions each applies
Identify which inquiry question each formula belongs to
Understand, Common Pitfalls
Explain the five most common Module 6 examination errors
Distinguish between flux, rate of change of flux, and induced EMF
Can Do, Synthesise
Connect IQ1–IQ4 into a coherent picture of electromagnetism
Self-identify weak areas and target revision accordingly
Key Terms, Module 6 Vocabulary
vocab
Magnetic flux ($\Phi$)The product of magnetic field strength and the area it passes through perpendicularly: $\Phi = BA\cos\theta$. Measured in webers (Wb).
Faraday's LawThe induced EMF in a coil equals the negative rate of change of flux linkage: $\varepsilon = -N\Delta\Phi/\Delta t$.
Lenz's LawThe direction of induced current opposes the change in flux that produced it. Ensures energy conservation.
Back EMFThe EMF induced in a motor's rotating coil that opposes the applied voltage. Increases with motor speed.
Torque ($\tau$)The rotational force on a current-carrying coil in a magnetic field. Given by $\tau = nBAI\cos\theta$.
Slip (induction motor)The difference between the stator's rotating field speed and the rotor's actual speed. Necessary for rotor current induction.
Cross-lesson links: L01–L20 covered the full electromagnetism strand. L21 consolidates all topics, electric fields, magnetic forces, motors, generators, transformers, and applications, into an integrated review. Every HSC Extended Response on electromagnetism draws simultaneously from multiple lessons; L21 is your synthesis practice.
Core Content
2
Map each formula to its symbols, reference direction, units and applicability conditions.
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Key Formulas, All Four Inquiry Questions
+5 XP
The twelve equations you must know for the HSC examination
Follow a single unit of electrical energy from its birth at a generator turbine: the spinning coil's changing flux induces a voltage (Faraday's law, L13); that voltage is stepped up by a transformer (L15); it travels hundreds of kilometres at 330 kV with small I²R losses (L20); a substation steps it down to 11 kV, then to 230 V; it enters your home and drives a motor (L09–L10) or heats a coil (eddy currents, L19). Module 6 contains more formulas per topic than any other Year 12 module, but they all describe different stages of that one journey. The table below groups them by inquiry question so the physics story holds them together.
IQ1, Charged Particles in Fields
Coulomb's Law
$F = k\dfrac{q_1 q_2}{r^2}$ ($k = 8.99 \times 10^9$ N m² C⁻²)
$P_{\text{loss}} = I^2R$ (use $I = P/V$ to find current first)
IQ1 magnitudes: $F=|q|E$, $F_B=|q|v_\perp B$, $r=mv_\perp/(|q|B)$; use the charge sign only for direction. IQ2: $F=BIl$, $\tau=nBAI\cos\theta$ when $\theta$ is the coil-plane angle. IQ3: $\Phi=BA\cos\theta$ for the area-normal angle, $\varepsilon=-\Delta(N\Phi)/\Delta t$, $\varepsilon_0=NBA\omega$ for the ideal generator. IQ4: $V_p/V_s=N_p/N_s$, $I=(V-\varepsilon_\text{back})/R$, $P_\text{loss}=I^2R$.
Pause, copy the formula groups (organised by IQ) into your notes before moving on.
An electron moves at $3.0 \times 10^6$ m/s perpendicular to $B = 0.10$ T. What is the magnetic force on the electron?
3
Connect fields, forces, induction and electromagnetic technologies across IQ1–IQ4.
2
Inquiry Question Summaries
+5 XP
The conceptual story behind each of the four inquiry questions
We just saw the formula set organised by inquiry question. That raises a question: what is the connecting conceptual story behind all four IQs? This card answers it → IQ1 and IQ2 are about forces on charges/conductors; IQ3 and IQ4 are about changing flux inducing emf and its applications.
IQ1, Charged Particles in Electric and Magnetic Fields
Electric fields exert forces on charges: $F = qE$. Positive charges move with the field; negative charges move against it.
Magnetic-force magnitude is $F_B=|q|vB$ when $v\perp B$; the signed cross product $q(\vec v\times\vec B)$ sets direction.
With a uniform $E$ field and constant transverse acceleration, charged-particle motion is parabolic; other entry geometry must be resolved into components.
For perpendicular entry into a uniform $B$ field, the circular radius is $r=mv/(|q|B)$.
Crossed $E$ and $B$ fields can be used as a velocity selector: $v = E/B$.
IQ2, The Motor Effect
A current-carrying conductor in a magnetic field experiences force: $F = BIl$.
Two parallel current-carrying conductors attract (same direction) or repel (opposite direction).
A coil in a magnetic field experiences torque: $\tau = nBAI\cos\theta$.
A radial magnetic field and split-ring commutator enable smooth continuous rotation in a DC motor.
Back EMF opposes the applied voltage and increases with motor speed.
IQ3, Electromagnetic Induction
Magnetic flux: $\Phi = BA\cos\theta$. Changing flux induces EMF.
Lenz's Law: induced current opposes the change that produced it. This ensures conservation of energy.
AC generators produce alternating EMF by rotating a coil in a magnetic field.
Transformers change AC voltage using electromagnetic induction between coupled coils.
IQ4, Applications of Electromagnetism
DC motors: split-ring commutator, radial field, back EMF, starting resistors.
AC induction motors: rotating stator field induces rotor currents, no brushes, slip is necessary.
Mass analysis: a crossed-field selector is optional; detector radius measures $m/|q|$ only with known speed preparation and charge state (enrichment context).
Eddy currents: useful for magnetic braking and induction cooking; unwanted in transformer cores (reduced by lamination).
Power transmission: high voltage reduces current and $I^2R$ losses. Transformers enable voltage conversion.
IQ1: $E$ and $B$ forces on charges → circular path, velocity selector. IQ2: motor effect, torque, back emf. IQ3: Faraday, Lenz, AC generator, transformer. IQ4: induction motor (slip), eddy currents, lamination, power transmission. IQ1+IQ2 → force/motion; IQ3+IQ4 → induction/energy transfer.
Pause, write one keyword per IQ (four keywords total) as a quick-recall cue into your book before moving on.
Lenz's Law ensures that electromagnetic induction is consistent with conservation of energy.
A large constant magnetic flux through a coil produces a large induced EMF.
When a motor stalls, the back EMF equals the applied voltage.
4
Repair common sign, angle, flux-linkage, power and reference-frame errors.
3
Common Examination Pitfalls
+5 XP
Five errors that cost marks in HSC Physics Module 6 responses
We just saw the conceptual story linking all four IQs. That raises a question: where do students actually lose marks? This card answers it → five recurrent errors in marking guidelines: wrong trig for torque, forgetting back emf at stall, misusing transmission voltage, confusing flux with rate of flux change, and Lenz's Law direction errors.
Each of these pitfalls appears regularly in HSC marking guidelines, recognising them is the first step to avoiding them.
Pitfall 1, Confusing sin and cos for torque
Torque on a coil: $\tau = nBAI\cos\theta$ where $\theta$ is the angle between the plane of the coil and $B$. Many students use sin incorrectly or confuse this with the angle to the normal. Maximum torque occurs when the coil is parallel to $B$ (i.e. $\theta = 0°$, $\cos\theta = 1$).
Pitfall 2, Forgetting back EMF at stall
When a motor stalls, back EMF = 0 and current surges to $V/R$. This is the maximum current, not the running current. Students often quote running current when asked about stall conditions.
Pitfall 3, Using $P = V^2/R$ for transmission loss
Power loss in transmission lines is $P_{\text{loss}} = I^2R$, not $V^2/R$. The $V$ in $P = V^2/R$ is the voltage drop across the line, not the transmission voltage. Always find $I = P/V$ first, then calculate $I^2R$.
Pitfall 4, Confusing flux and rate of change of flux
Large flux does not mean large induced EMF. A coil can have large constant flux and zero induced EMF. Faraday's Law says EMF depends on how quickly flux changes ($\Delta\Phi/\Delta t$), not on the flux value itself.
Pitfall 5, Lenz's Law direction errors
Always identify the external field direction first, then determine if flux is increasing or decreasing. The induced field opposes the change, not the field itself. A common error: "flux is into the page, so induced field is out of the page", only correct if flux is increasing.
5 exam traps: (1) $\tau = nBAI\cos\theta$, use cos, $\theta$ = coil-to-B (max at $\theta=0°$). (2) Stall: back emf = 0, $I = V/R$ (max). (3) Transmission: $P_\text{loss}=I^2R$, not $V^2/R$, find $I = P/V$ first. (4) EMF ∝ $\Delta\Phi/\Delta t$, not $\Phi$. (5) Lenz's Law opposes the change in flux, not flux itself.
Pause, list all five pitfalls (with brief fix for each) into your exam notes before moving on.
600 MW of power is transmitted at 300 kV through lines with total resistance 3.0 Ω. The power loss in the lines is:
5
Build the energy-and-fields concept map linking the whole module.
4
Synthesis, The Big Picture
+5 XP
How the four inquiry questions connect into one story
We just saw the five common pitfalls. That raises a question: how do all four IQs fit into a single coherent argument for an extended response? This card answers it → two master ideas unify everything: fields exert forces on charges (IQ1+IQ2) and changing flux induces emf (IQ3+IQ4).
Module 6 is built on two fundamental ideas that every exam response should reflect.
Fundamental idea 1, Charges in fields
Electric and magnetic fields exert forces on charges (IQ1). Understanding these forces explains everything from cathode rays to mass spectrometers. This is the foundation of IQ2 (motor effect): forces on current-carrying conductors are the same physics applied to bulk charge flow.
Fundamental idea 2, Changing flux induces EMF
Faraday's Law (IQ3) connects magnetism to electricity. A changing magnetic flux induces an EMF, this drives generators, transformers, induction motors, and wireless technology. IQ4 is the engineering application of IQ3: how we harness induction to generate, transmit, and use electrical power efficiently.
IQ2 and IQ4 are the bridges showing how forces on currents produce motion (motor), and how induction produces the electricity that powers civilisation (generator → transmission → transformer → load).
Two master ideas: (1) Fields exert forces on charges → IQ1 (particle paths) + IQ2 (motor, torque). (2) Changing flux induces emf → IQ3 (Faraday, Lenz, generator) + IQ4 (transformer, transmission, induction motor, eddy currents). All extended responses on M6 should reference which master idea applies.
Pause, write the two master ideas and which IQs they cover into your book. This is your extended-response frame.
Three of these statements about Module 6 are correct. Pick the odd one out.
Conditions, standards and data interpretation
Before using a formula: declare the reference direction, field uniformity, relevant angle, whether $q$ is signed or a magnitude, and whether the model is non-relativistic/ideal. Keep $\Phi$ (Wb) separate from $N\Phi$ (Wb-turn), and use nominal 230/400 V for current Australian low-voltage distribution. The current SI ampere fixes the elementary charge; the parallel-wire definition is historical.
For graphs: state what a gradient or area means with units: gradient of $V$–$x$ is $-E_x$; gradient of $v$–$t$ is acceleration and its area is displacement; gradient of $N\Phi$–$t$ is $-\varepsilon$; gradient of $F/l$ against $I_1I_2/d$ tests $\mu_0/(2\pi)$. Quote measurement resolution, repeated-reading spread, uncertainty and model limitations when evaluating experimental evidence.
6
Self-assess the module and teach one formula with its conditions before mixed practice.
Activity 1, Self-Assessment Checklist
EvaluateBand 4
Rate each skill honestly, then revisit weak areas before your exam
Calculate electric force and field strength for point charges using Coulomb's Law.
Determine the direction of magnetic force on a moving charge or current (right-hand rule).
Calculate the radius of a charged particle's circular path in a uniform $B$-field.
Calculate torque on a current-carrying coil at any angle using $\tau = nBAI\cos\theta$.
Explain how a DC motor produces continuous rotation (commutator, radial field).
Calculate back EMF and motor current at different operating speeds.
Calculate magnetic flux and induced EMF using Faraday's Law.
Apply Lenz's Law to determine the direction of induced current.
Calculate transformer voltages and currents using the turns ratio.
Calculate power loss in transmission lines and explain why high voltage is used.
Explain the operation of an AC induction motor and why slip is necessary.
Explain how laminating a transformer core reduces eddy current losses.
For any items you rated 2/5 or below, note the lesson number below and revisit that lesson.
Activity 2, Formula Deep-Dive
AnalyseBand 5
Choose one formula and explain it completely
Select one formula from the formula grid in Card 1 that you find most challenging. In your response, explain: (a) what each variable means and its units, (b) the physical situation the formula applies to, (c) a worked calculation example with realistic numbers, and (d) one common mistake students make with it.
A transformer with $N_p = 100$ turns and $N_s = 400$ turns is connected to $V_p = 50$ V AC. The secondary voltage $V_s$ (in V) is _____.
Analyse7 marks
A conducting rod slides at constant speed along two horizontal rails connected by a resistor, in a vertical magnetic field. Analyse the emf, the current and the forces involved, and account for the energy transfers. (7 marks)
1
Unpack the verb
Analyse is this dot point's verb and it asks for energy transfers explicitly. This is a consolidation card: it uses flux, Faraday, Lenz and the motor effect in one connected argument.
Plan: find the flux change from the geometry, get the emf, get the current, get the force on the rod from the motor effect, then balance the energy.
Each stage supplies the input to the next, so the order is forced. Getting the order right is most of the difficulty in this classic question.
2
Find the flux change from the geometry
The rod, the rails and the resistor enclose a circuit whose area increases as the rod slides. With the field perpendicular to that area, the flux is $\Phi = BA$, so a rod of length $l$ moving at speed $v$ sweeps out area at a rate $lv$ and the flux increases at a rate $Blv$.
This is the changing-area route to induction — the field and the orientation are fixed and only the area varies.
3
Get the emf and the current
By Faraday's Law the induced emf is $\varepsilon = \Delta\Phi/\Delta t = Blv$ for a single-turn circuit. If the circuit has total resistance $R$, a current $I = Blv/R$ flows around it.
The emf is constant here because the speed is constant, which is what makes the rest of the analysis a steady state rather than a changing one.
4
Get the force on the rod
The rod now carries current $I$ in the magnetic field, so by the motor effect it experiences a force $F = BIl = B^2l^2v/R$. By Lenz's Law this force opposes the motion producing it, so it acts backwards along the rails, retarding the rod.
The rod is both the source of the emf and a current-carrying conductor in the field. Seeing it play both roles is the key to the whole problem.
5
Impose the constant-speed condition
Because the rod moves at constant speed its acceleration is zero, so the applied force must exactly balance the magnetic retarding force: $F_{\text{applied}} = B^2l^2v/R$.
The question says constant speed, and that is a physical constraint with a consequence, not just scene-setting.
6
Balance the energy
The applied force does work at a rate $P = F_{\text{applied}}v = B^2l^2v^2/R$. The resistor dissipates power at $I^2R = (Blv/R)^2R = B^2l^2v^2/R$ — the same quantity. So all the mechanical work done against the magnetic force is converted into electrical energy and dissipated as heat, with none stored, because the kinetic energy is constant.
The two expressions being identical is the evidence, not the assertion. Note the kinetic energy is unchanged, so every joule of work has to go to the resistor.
7
The full-mark answer
The rod, rails and resistor enclose a circuit whose area grows as the rod slides. With the field perpendicular to that area the flux is $\Phi = BA$, and a rod of length $l$ moving at speed $v$ sweeps area at a rate $lv$, so the flux increases at a rate $Blv$. By Faraday's Law the induced emf is therefore $\varepsilon = Blv$, and with total circuit resistance $R$ a current $I = Blv/R$ flows. The rod now carries that current in the magnetic field, so by the motor effect it experiences a force $F = BIl = B^2l^2v/R$, and by Lenz's Law that force opposes the motion producing it and retards the rod. Since the rod moves at constant speed its acceleration is zero, so the applied force must balance this exactly at $B^2l^2v/R$. The applied force therefore does work at a rate $B^2l^2v^2/R$, while the resistor dissipates power at $I^2R = B^2l^2v^2/R$ — the same rate. The energy account is closed: because the kinetic energy is constant, every joule of mechanical work done against the magnetic force is converted to electrical energy and then to heat in the resistor.
Seven marks: the changing area, the flux rate, the emf, the current, the retarding force with Lenz's Law, the equilibrium condition, and an energy balance proved by two expressions agreeing rather than asserted.
Compare6 marks
The motor effect and electromagnetic induction are sometimes described as converse processes. Compare them, using a DC motor and a DC generator as your examples. (6 marks)
1
Unpack the verb
Compare needs both processes in every point. This is the consolidation move for the module: the same components, run in opposite directions.
Plan: state the input and output of each, then the governing law of each, then what they share structurally, then the sense in which they are converse.
The word "converse" is a claim to be tested, not a label to repeat. The answer should end by saying exactly in what sense it is true.
2
Compare inputs and outputs
In the motor effect, electrical energy is supplied and mechanical energy is produced: a current in a field experiences a force. In electromagnetic induction, mechanical energy is supplied and electrical energy is produced: motion in a field produces an emf.
Setting the two energy flows side by side in one sentence each is what makes the converse relationship visible immediately.
3
Compare the governing relationships
The motor effect is described by $F = BIl\sin\theta$, relating force to current. Induction is described by $\varepsilon = -N\,\Delta\Phi/\Delta t$, relating emf to the rate of change of flux. The first has no time derivative in it; the second is entirely about a rate.
The structural difference between the two laws is worth noting: a steady current gives a steady force, but a steady flux gives no emf at all.
4
Give the shared apparatus
The two processes use the same components. A simple DC motor and a simple DC generator are built identically — a coil, a magnetic field, a split-ring commutator and brushes — and the same machine will act as either, depending on whether it is supplied with current or turned by hand.
This is the strongest evidence for the converse claim, because it is a physical fact about one device rather than an analogy between two.
5
Say in what sense they are converse, and where the description breaks
They are converse in the direction of the energy conversion, and they are linked by Lenz's Law, which makes each one produce the other in opposition: a motor generates a back emf that opposes its supply, and a generator experiences a retarding force that opposes its rotation. They are not inverses in a strict mathematical sense, because the force law involves current and the induction law involves a rate of change of flux, and both processes occur together in any real machine rather than one at a time.
Naming the limits of the word "converse" is the top-band move. Both effects run simultaneously in every real motor and generator, which a strict converse would not allow.
6
The full-mark answer
In the motor effect electrical energy is supplied and mechanical energy produced: a current-carrying conductor in a magnetic field experiences a force $F = BIl\sin\theta$. In electromagnetic induction mechanical energy is supplied and electrical energy produced: relative motion changes the flux through a circuit and induces an emf $\varepsilon = -N\,\Delta\Phi/\Delta t$. The two laws differ structurally — the force law contains no rate of change, so a steady current gives a steady force, while the induction law is entirely a rate, so a steady flux gives no emf at all. The apparatus, however, is shared: a simple DC motor and a simple DC generator are built from the same coil, magnet, commutator and brushes, and one machine will act as either depending on whether current is supplied to it or it is turned by hand. They are converse in the direction of energy conversion, and Lenz's Law links them so that each produces the other in opposition: a motor develops a back emf opposing its supply, and a generator experiences a retarding force opposing its rotation. They are not strict inverses, though, because the two relationships involve different quantities and because both effects are present simultaneously in every real machine.
Six marks: the energy directions, the two laws, the structural difference between them, the shared apparatus as evidence, the Lenz link, and the limits of calling them converse.
Practicemixed bank + 3 SAQs
7
Apply the formula-condition map to mixed Module 6 retrieval and short answers.
Module 6 consolidation, mixed question set
+5 XP
Five questions drawn from the lesson bank spanning all four inquiry questions. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
Short Answer, 9 marks
+5 XP
ApplyBand 4(3 marks) 1. A coil with 30 turns, area $4.0 \times 10^{-3}$ m², carries 2.0 A in a magnetic field of 0.25 T. Calculate (a) the maximum torque on the coil, and (b) the torque when the coil plane makes a 30° angle with the field.
1 mark: correct application of $\tau = nBAI\cos\theta$ for max torque ($\theta = 0°$) · 1 mark: correct numerical answer for (a) · 1 mark: correct answer for (b) with $\cos 30°$
UnderstandBand 4(3 marks) 2. Explain why an AC induction motor cannot reach synchronous speed. In your answer, refer to the role of electromagnetic induction and Lenz's Law.
1 mark: identifies that at synchronous speed there would be no relative motion between rotor and rotating field · 1 mark: no relative motion means no changing flux through rotor conductors (Faraday) · 1 mark: Lenz's Law, induced rotor currents are needed to produce the torque that drives the motor; without flux change, no currents, no torque
EvaluateBand 6(3 marks) 3. A student claims: "Using a higher transmission voltage reduces power loss because $P = V^2/R$ shows that higher voltage means more power dissipated." Evaluate this claim and correct any errors in the student's reasoning.
1 mark: identifies the error, $P = V^2/R$ applies to the voltage drop across the line, not the transmission voltage · 1 mark: correct reasoning, higher transmission voltage reduces current ($I = P_{\text{power}}/V$) for the same power delivered · 1 mark: correct formula for line losses $P_{\text{loss}} = I^2R$, and explains that lower current dramatically reduces losses
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Q2 (3 marks): At synchronous speed, the rotor would rotate at exactly the same rate as the stator's magnetic field, so there would be no relative motion between them (1 mark). No relative motion means the magnetic flux through the rotor conductors is constant, by Faraday's Law, no changing flux means no induced EMF and no induced currents in the rotor (1 mark). By Lenz's Law, it is these induced rotor currents that interact with the stator field to produce the torque driving the motor. Without the currents, there is no torque, and the motor cannot sustain synchronous speed. Therefore slip (rotor lagging behind the stator field) is a necessary condition for operation (1 mark).
Q3 (3 marks): The student has made an error: $P = V^2/R$ applies to the voltage drop across the transmission line itself, not to the grid transmission voltage (1 mark). The correct analysis uses $P_{\text{loss}} = I^2R$, where $I = P_{\text{delivered}}/V_{\text{transmission}}$. Increasing the transmission voltage reduces the current for the same delivered power (1 mark). Since $P_{\text{loss}} = I^2R \propto 1/V^2$, doubling the transmission voltage reduces line losses by a factor of 4, the opposite of what the student claimed (1 mark).
Reviewcumulative retrieval + reflection
8
Complete cumulative retrieval and identify the next targeted remediation step.
Check what actually stuck
Take the full module quiz
quiz
You have completed all 21 lessons of Module 6, Electromagnetism. Well done!
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
At the start you were asked to trace the complete M6 physics chain from Faraday's law at the generator to the current in the arc lamps of that illustrative 1930s bridge-lighting scenario.
Here is a model answer: The generator's spinning coil experienced continuously changing magnetic flux → Faraday's Law induced AC at a generation voltage of order 10 kV → step-up transformers raised voltage for low-loss transmission → step-down transformers at the bridge reduced voltage to lamp operating level → current through each arc lamp's electrodes produced light, with local transformers setting the voltage each lamp needed.
Notice how IQ3 (induction), IQ4 (transformers, transmission), and even IQ2 (the generator itself uses the motor-effect principle in reverse) all appear in that single sentence. This synthesis is what the HSC Extended Response on electromagnetism demands.