AQA A-Level Physics: Magnetic Fields and Electromagnetic Induction Complete Guide
AQA A-Level Physics: Magnetic Fields and Electromagnetic Induction Complete Guide
Magnetic Fields and Induction is Section 3.7.5 of the AQA A-Level Physics specification — the third member of the "fields trilogy" that began with gravity and electricity. It is also where the analogy that worked so cleanly between gravitational and electric fields starts to break down. Magnetic forces do not act in the direction of the field; they act perpendicular to both the field and the velocity of the charge. Magnetic fields cannot be derived from a scalar potential. And the most powerful idea in the section — electromagnetic induction — has no gravitational or electrostatic analogue. Once you accept that magnetism is genuinely different, the topic becomes one of the most rewarding on the course because the physics is directly visible in transformers, motors, generators, and the National Grid.
This guide covers the whole of Section 3.7.5 in the order a Paper 2 question is likely to test it: the force on a current-carrying wire, the force on a moving charge, charged-particle motion in magnetic fields, electromagnetic induction, Faraday's and Lenz's laws, alternating currents, transformers and power transmission. It ends with the field-comparisons synthesis that examiners reward heavily and a walkthrough of Required Practical 11.
Magnetic Flux Density and the Force on a Current-Carrying Wire
The strength of a magnetic field is measured by the magnetic flux density B, in tesla (T). One tesla is the field that produces a force of one newton on a one-metre length of wire carrying one amp at right angles to the field. The defining equation is:
F = BIL sin θ
where I is the current in the wire, L is the length of wire inside the field, and θ is the angle between the wire and the field. When the wire is perpendicular to the field (θ = 90°), F = BIL. When it is parallel, F = 0.
The direction of the force is given by Fleming's left-hand rule: the thumb, first finger and second finger represent the thrust (force), field, and current respectively, all mutually perpendicular. The rule applies to conventional current — positive charge flow — so when dealing with electrons you must reverse the direction.
A neat consequence is that two parallel wires carrying currents in the same direction attract each other, while two wires carrying opposite currents repel. This is the basis of the now-retired SI definition of the ampere.
Force on a Moving Charge
A current is just charge in motion, so an isolated moving charge in a magnetic field also experiences a force. The equation is:
F = BQv sin θ
where Q is the charge and v is its speed. For θ = 90° this reduces to F = BQv. The force is always perpendicular to the velocity — so the magnetic force does no work on the charge (since work = F·d cosα and α = 90°). A magnetic force can change the direction of a charged particle but not its kinetic energy.
This perpendicularity has a striking consequence: a charged particle entering a uniform magnetic field at right angles to B moves in a circle. The magnetic force provides the centripetal force:
BQv = mv² / r
so:
r = mv / (BQ)
The radius increases with momentum and decreases with the strength of the field or the charge of the particle. This single equation underlies the design of mass spectrometers, cyclotrons, the cathode-ray tubes of older oscilloscopes, the bending magnets in particle accelerators, and modern medical imaging devices.
The Cyclotron and Mass Spectrometer
In a cyclotron, charged particles spiral outward through a region of uniform magnetic field. The radius increases each time the particles are accelerated across a gap, but — because v and r both increase — the period T = 2πm / (BQ) is independent of speed. This means the particles can be repeatedly accelerated by an alternating voltage of fixed frequency until they reach the edge of the apparatus.
In a mass spectrometer, ions of known charge and speed are deflected through circular arcs whose radii depend only on their masses: r = mv / (BQ). By measuring r you measure m/Q. Mass spectrometry is one of the central tools of modern chemistry and is also widely tested in synoptic exam questions linking physics to chemistry.
Electromagnetic Induction: Flux and Flux Linkage
Magnetic flux Φ through a flat surface of area A at angle θ to a uniform field B is:
Φ = BA cos θ
Units are webers (Wb). One weber is one tesla square metre. The flux is the count of field lines passing through the surface, weighted by their direction relative to the surface normal.
For a coil of N turns enclosing flux Φ, the flux linkage is NΦ. It is the flux linkage rather than the flux itself that determines the size of the induced EMF.
Faraday's and Lenz's Laws
Faraday's law states that the induced EMF in a circuit is equal to the rate of change of flux linkage:
ε = −dΦ_linked / dt = −d(NΦ) / dt
The minus sign expresses Lenz's law: the induced EMF (and any induced current) acts in such a direction as to oppose the change in flux that produced it. The negative sign is not optional — it is a direct statement of energy conservation. If the induced current did not oppose the change, you could extract energy from the magnetic field indefinitely without doing any work.
A good way to apply Lenz's law in practice: identify the direction of the magnetic flux change (is the flux through the coil increasing or decreasing?), then work out the direction of the induced current that opposes that change, using the right-hand grip rule.
Worked Example: A Wire Moving in a Field
A straight conducting rod of length L moves at speed v through a uniform magnetic field B, with the velocity perpendicular to both the rod and the field. The rod sweeps out an area Lv per second, so the rate of change of flux is BLv. The induced EMF is therefore:
ε = BLv
If the rod forms part of a complete circuit of resistance R, the induced current is I = BLv / R. The force on this current-carrying rod is F = BIL = B²L²v / R, opposing its motion — exactly as Lenz's law requires.
Alternating Current: rms vs Peak
An alternating current (AC) varies sinusoidally with time:
I = I₀ sin(2πft)
with peak current I₀ and frequency f. In the UK domestic supply, f = 50 Hz and the peak voltage of the 230 V mains supply is V₀ = √2 × 230 ≈ 325 V.
The reason the "230 V" number is √2 times smaller than the peak is that 230 V is the root-mean-square voltage. For a sinusoidal waveform:
- I_rms = I₀ / √2
- V_rms = V₀ / √2
The rms values are the DC equivalents — an AC supply of V_rms across a resistor dissipates the same average power as a DC supply of V_rms across the same resistor. The average power dissipated in a resistor by an AC supply is P_av = V_rms × I_rms = (1/2) V₀ I₀.
Examiners often ask candidates to read peak values from an oscilloscope trace. The peak-to-peak amplitude is 2V₀, and the period is read from the time-base. Be ready to convert period to frequency (f = 1/T) and to convert peak to rms by dividing by √2.
Transformers and the Turns Ratio
A transformer changes the voltage of an alternating supply. It consists of a primary coil of N_p turns and a secondary coil of N_s turns wound on a common laminated soft-iron core. The varying current in the primary produces a varying flux in the core, which induces an EMF in the secondary. For an ideal transformer:
V_s / V_p = N_s / N_p
If N_s > N_p, the transformer is step-up and the secondary voltage is higher than the primary. If N_s < N_p, it is step-down and the secondary voltage is lower.
Conservation of energy gives the current ratio for an ideal transformer:
I_s / I_p = N_p / N_s
A step-up transformer that raises the voltage also reduces the current in the same ratio, and vice versa. Real transformers have efficiencies of 95-99% — losses come from resistive heating in the windings (I²R loss), eddy currents in the core (reduced by lamination), hysteresis in the iron (reduced by soft magnetic material), and incomplete flux linkage between the coils (reduced by wrapping both coils on the same core).
Power Transmission at High Voltage
Electricity is transmitted across the National Grid at very high voltage — 275 kV or 400 kV on the main transmission lines — for a single overwhelming reason: resistive losses in the cables scale as I², so reducing the current reduces the losses dramatically.
For a given power P = VI, halving I means doubling V; the resistive power loss in the cables, P_loss = I²R, then quarters. Stepping up to ten times the original voltage cuts the resistive loss to 1% of its original value.
At each end of the grid the voltage is transformed: stepped up at the generating station, stepped down at substations near homes. Local distribution is at 11 kV, with final step-down to 230 V at street-level transformers. A common Paper 2 calculation asks you to compare power losses at 230 V transmission with those at 400 kV transmission for the same delivered power and cable resistance; the ratio is (400 000 / 230)² ≈ 3 × 10⁶, an enormous factor.
AC is used in transmission because transformers only work with alternating current. DC transmission is becoming more common for very long-distance lines (where its lower reactive losses outweigh the conversion cost), but for the national grid the historical case for AC is overwhelming.
Field Comparisons: Gravitational, Electric and Magnetic
A standard Paper 2 question asks candidates to "compare gravitational, electric and magnetic fields." A complete answer mentions:
| Property | Gravitational | Electric | Magnetic |
|---|---|---|---|
| Source | Mass | Charge | Moving charge / current |
| Field direction | Towards source mass | From + towards − (or radially out from +) | Tangent to field lines (N → S outside magnet) |
| Force on test object | F = mg | F = QE | F = BQv (only on moving charge) |
| Force direction | Along field | Along field | Perpendicular to field AND velocity |
| Field-strength inverse-square law | Yes (point mass) | Yes (point charge) | No (more complex geometry) |
| Sign | Always attractive | Attractive or repulsive | Determined by Fleming's left-hand rule |
| Does work on charge | Yes | Yes | No (force always perpendicular to v) |
| Potential | Scalar, V = −GM/r | Scalar, V = Q/(4πε₀r) | No simple scalar potential |
| Induced by changing other field | No | Yes (changing B field) | Yes (changing E field, or moving charge) |
This table is worth memorising in one sitting. Examiners reward candidates who can pick exactly the right comparison points without grasping at vague similarities.
Required Practical 11: Force on a Current-Carrying Wire
Required Practical 11 asks you to investigate the force F = BIL between a current-carrying conductor and a permanent magnet. The standard setup uses a top-pan balance, a horseshoe magnet placed on the balance, and a length of stiff copper wire passing horizontally between the poles of the magnet at right angles to the field.
The procedure is:
- Zero the balance with the wire in place but no current flowing.
- Pass a known current I through the wire and record the apparent change in mass Δm on the balance.
- The downward force on the magnet is F = Δmg (by Newton's third law, equal and opposite to the upward force on the wire).
- Repeat for a range of currents and plot F against I. The graph is a straight line of gradient BL, allowing B to be calculated if L is known.
Key sources of uncertainty:
- The wire must be exactly perpendicular to the field for F = BIL to apply.
- The wire must lie wholly inside the uniform-field region between the poles; otherwise L is hard to define.
- The wire heats up at higher currents, so readings should be taken quickly.
Examiners reward candidates who explicitly state that the force on the magnet is equal and opposite to the force on the wire (Newton's third law) and who explain why the gradient gives B and not BIL.
How to Study This Topic
Magnetic fields is one of the more conceptually rich topics on the A-Level course, and rote memorisation alone will not get you to the top grades. The most useful single tactic is to build the three-column field-comparisons table from memory and use it as the spine of your revision.
- Memorise the four central equations: F = BIL, F = BQv, r = mv/(BQ), ε = −d(NΦ)/dt. The transformer turns-ratio and rms relations can be derived from energy conservation.
- Drill Fleming's left-hand rule and the right-hand grip rule until they are automatic. AQA mark schemes routinely credit "direction" marks separately, and getting the rule wrong loses both the direction mark and any subsequent magnitude marks for opposing currents.
- Build the three-column comparison table weekly.
- Practise Required Practical 11 with a fake dataset, focusing on linearising F = BIL into a plot of F against I.
- Drill grid-transmission problems: identify the power loss scaling as I², explain why step-up at the station and step-down at the user, calculate efficiency.
Related LearningBro Courses
LearningBro's AQA A-Level Physics: Magnetic Fields and Induction course covers all of the above in eight lessons:
- Magnetic Fields and Forces — flux density, Fleming's left-hand rule, F = BIL.
- Charged Particles in Magnetic Fields — F = BQv, r = mv/(BQ), cyclotrons and mass spectrometers.
- Electromagnetic Induction — flux, flux linkage, induced EMF from a moving rod.
- Faraday's and Lenz's Laws — ε = −d(NΦ)/dt and the sign convention.
- Alternating Currents — peak and rms values; oscilloscope reading.
- Transformers and Power Transmission — turns ratio, efficiency, why grid voltage is high.
- Field Comparisons Synthesis — gravitational, electric and magnetic side by side.
- Required Practical 11: Force on a Current-Carrying Wire — method, F-against-I graph, error analysis.
Common Exam Pitfalls
This unit has a small number of high-frequency mistakes that account for most of the lost marks.
- Hand rules used in the wrong direction. Fleming's left-hand rule is for the force on a current-carrying conductor in a magnetic field (motor effect). The right-hand rule is for the induced current (generator effect). Mixing them up gives a sign error that propagates through every subsequent step. Mark the rule on your equation in the exam.
- Flux versus flux linkage. Φ = B A is the flux through a single turn; N Φ is the flux linkage through an N-turn coil. Faraday's law in AQA's preferred form is ε = -d(N Φ)/dt, with the N inside the derivative. Forgetting the factor of N is a one-mark drop on every induction question.
- rms versus peak. For a sinusoidal alternating supply, V_rms = V_peak / sqrt 2 and I_rms = I_peak / sqrt 2. The mains supply quoted as "230 V" is the rms value; the peak voltage is about 325 V. Power dissipation must be computed with rms values, not peak.
- Transformer efficiency confused with turns ratio. The ideal turns-ratio relation V_s / V_p = N_s / N_p holds for an ideal transformer; the current ratio is the inverse. Real transformers are 95-99% efficient and the power balance V_p I_p eta = V_s I_s should be used when efficiency is given.
- The minus sign on Lenz's law. The negative sign in ε = -d(N Φ)/dt is a directional statement: the induced current opposes the change in flux. A question asking for the direction of the induced current cannot be answered from Faraday's law alone; you must invoke Lenz's law to fix the sign.
- Charged-particle radius formula. r = m v / (B q) for a particle moving perpendicular to a magnetic field. The cyclotron frequency f = B q / (2 pi m) is independent of v — which is why a cyclotron can accelerate particles to ever-higher speeds at the same drive frequency, provided relativistic effects are negligible.
- AC mains frequency assumptions. UK mains is 50 Hz, not 60 Hz. The period is 20 ms; the peak-to-peak interval on an oscilloscope screen is 20 ms divided by the number of cycles displayed.
Recommended Three-Week Revision Schedule
Magnetic fields and induction is a small unit by mark allocation but pays back well to focused revision because the calculations are short and repeatable.
- Week 1 (concepts and equations). Build a one-side-of-A4 summary sheet covering F = B I L sin theta, F = B q v, r = m v / (B q), Φ = B A, ε = -d(N Φ)/dt, the rms relations, and the transformer equations. Work two or three short questions on each.
- Week 2 (synoptic problems). Combine magnetic fields with circular motion (cyclotron problems), with electric fields (velocity selectors and mass spectrometers), and with AC (transformer efficiency questions). Two past-paper questions per topic.
- Week 3 (Paper 2 exam practice). A full Paper 2 question on induction every other day, marked against the official mark scheme. Time yourself: AQA expect about 1.1 minutes per mark, so a 6-mark induction question should be completed in under seven minutes.
Keep the required practical (RP11, force on a current-carrying wire) at the front of mind throughout — it sits behind any practical-skills 6-mark question on magnetic forces.