IB Physics flashcards · SL and HL · first assessment 2025
D.4 Induction flashcards: IB Physics HL
Revision flashcards for D.4 Induction, written for the IB Diploma Physics course first assessed in 2025. This is a Higher Level topic, so every card is HL only. Use study mode to test yourself one card at a time, or read the full list below with every answer.
Magnetic flux and flux linkage, Faraday's and Lenz's laws, motional emf and the a.c. generator.
- 70 cards
- Definition: 16
- Equation: 13
- Concept/explain: 13
- HL-only cards marked
All 70 D.4 Induction cards
- DefinitionHL onlyData booklet: Yes
Define magnetic flux.
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The magnetic flux Φ through a surface is the product of the magnetic flux density B and the area A of the surface perpendicular to the field, Φ = BA cosθ, where θ is the angle between B and the normal to the area. Exam tip: the mark is lost if you write 'the angle between B and the plane of the coil' — IB defines θ from the NORMAL, so the flux is a maximum (θ = 0) when B is perpendicular to the plane of the coil and zero when B lies in the plane. Scalar quantity; SI unit weber (Wb) = T m² = V s.
- DefinitionHL only
Define the weber.
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One weber is the magnetic flux through a surface of area 1 m² placed perpendicular to a uniform magnetic field of flux density 1 T. Equivalently, 1 Wb = 1 T m² = 1 V s, so a flux linkage changing at 1 Wb s⁻¹ induces an emf of 1 V. Exam tip: students who quote 'Wb = T m²' but cannot state the perpendicular condition lose the definition mark; always include 'normal to (perpendicular to) the field'. Magnetic flux is a scalar, so flux from opposite-sense field regions is subtracted, not added as vectors.
- DefinitionHL only
Define magnetic flux density and state its unit.
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Magnetic flux density B is the magnetic flux per unit area passing perpendicular through a surface, B = Φ/A; operationally it is defined from F = BIL sinθ as the force per unit current per unit length on a conductor placed perpendicular to the field. Exam tip: do not call B the 'magnetic field strength' in a definition question and do not confuse it with flux Φ — B is per unit area. Vector quantity; SI unit tesla (T) = Wb m⁻² = N A⁻¹ m⁻¹ = kg s⁻² A⁻¹. Typical values: Earth ≈ 5 × 10⁻⁵ T, laboratory magnet ≈ 0.1 T.
- DefinitionHL only
Define magnetic flux linkage.
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Flux linkage is the product of the magnetic flux through one turn of a coil and the number of turns, flux linkage = NΦ = NBA cosθ. It is the total flux 'linked' by the whole circuit. Exam tip: IB mark schemes accept the unit weber (Wb) or weber-turns (Wb turns); the commonest error is to forget N when a question gives a coil rather than a single loop, which scales every subsequent emf answer wrongly. Scalar. Faraday's law is properly stated in terms of the rate of change of flux LINKAGE, not of flux alone.
- DefinitionHL onlyData booklet: Yes
State Faraday's law of electromagnetic induction.
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The magnitude of the emf induced in a circuit is directly proportional to (equal to) the rate of change of magnetic flux linkage through the circuit: ε = −N ΔΦ/Δt. Exam tip: 'rate of change of flux linkage' is the marking point — 'proportional to the flux' or 'proportional to the change in flux' scores zero because it omits the rate. A steady flux, however large, induces no emf. The minus sign is Lenz's law and is normally omitted when only a magnitude is required. Unit of ε: volt (V) = Wb s⁻¹; scalar.
- DefinitionHL only
State Lenz's law.
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The direction of an induced emf (and hence induced current) is such that the effects it produces oppose the change in magnetic flux that is causing it. Exam tip: the two marking points are 'opposes' and 'the CHANGE in flux' — writing 'opposes the magnetic field' or 'opposes the motion of the flux' is not accepted. Lenz's law is a consequence of conservation of energy: if the induced current aided the change, the system would accelerate itself and generate energy from nothing. It is represented by the negative sign in ε = −N ΔΦ/Δt.
- DefinitionHL only
Define electromagnetic induction and induced emf.
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Electromagnetic induction is the production of an emf across a conductor whenever the magnetic flux linkage through the circuit changes — either because the field changes, the area changes, or the orientation changes. The induced emf is the electrical energy transferred per unit charge driven round the circuit by this process. Exam tip: an emf exists even in an open circuit where no current flows; students who say 'a current is induced' lose the mark when the circuit is broken. Unit V = J C⁻¹; scalar. Nature of Science link: Faraday's 1831 experiments unified electricity and magnetism and underpin all modern generation.
- DefinitionHL onlyData booklet: Yes
Define motional emf and explain its microscopic origin.
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Motional emf is the emf induced across a conductor of length L moving with velocity v perpendicular to a magnetic field B, ε = BvL. Microscopically, each free electron in the rod experiences a magnetic force F = qvB along the rod; charge accumulates at the ends until the electric force qE balances it, giving E = vB and ε = EL = BvL. Exam tip: only the component of v perpendicular to B and to L contributes — a rod moving parallel to B gives zero emf. It is fully equivalent to Faraday's law because the circuit area swept per second is Lv. Unit V; scalar.
- DefinitionHL only
Define eddy currents.
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Eddy currents are circulating (looping) induced currents set up within the body of a bulk conductor when the magnetic flux through it changes. By Lenz's law they oppose the change, producing a retarding force, and they dissipate energy as heat because of the resistance of the metal. Exam tip: say 'currents induced in the BULK/body of the conductor', not merely 'currents in a wire'. Useful applications: induction hobs, electromagnetic (eddy-current) braking on trains, metal detectors. Unwanted in transformer cores, where a laminated core of thin insulated sheets increases the resistance of the eddy-current paths and reduces the loss.
- DefinitionHL onlyData booklet: Yes
Define the root-mean-square value of an alternating current.
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The rms current is the value of a steady direct current that would dissipate thermal energy in a given resistor at the same average rate as the alternating current does. For a sinusoidal current, I_rms = I₀/√2 ≈ 0.707 I₀. Exam tip: the mark is for the comparison with a DIRECT current dissipating the same MEAN power in the same resistance — answers such as 'the average current' score zero (the mean of a sine wave over a cycle is zero). Unit ampere (A); scalar. Quoted mains and meter readings are always rms values unless stated otherwise.
- DefinitionHL onlyData booklet: Yes
Define the root-mean-square value of an alternating potential difference, and the peak value.
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The rms potential difference is the steady dc voltage that would deliver the same average power to a given resistor as the alternating supply, V_rms = V₀/√2. The peak (maximum) value V₀ is the amplitude of the sinusoid, i.e. the largest instantaneous value reached each cycle. Exam tip: for 230 V mains, V₀ = 230 × √2 ≈ 325 V — insulation and component ratings must withstand the PEAK, a frequently tested point. Peak-to-peak = 2V₀, often what an oscilloscope trace shows; halve it before applying V_rms = V₀/√2. Unit volt (V); scalar.
- DefinitionHL onlyData booklet: No – derive
Define an alternating current generator (ac generator) and describe how it produces an emf.
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An ac generator is a device that converts mechanical (kinetic) energy into electrical energy by rotating a coil of N turns and area A at constant angular velocity ω in a uniform magnetic field B. The flux linkage NBA cosωt varies sinusoidally, so by Faraday's law a sinusoidal emf ε = NBAω sinωt is induced; slip rings and brushes take the alternating output to the external circuit. Exam tip: state slip rings (ac) — split-ring commutator would give dc. The emf is zero when the coil plane is perpendicular to B (flux maximum) and maximum when the coil plane contains B (flux zero).
- DefinitionHL onlyData booklet: Yes
Define an ideal transformer and state its assumptions.
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An ideal transformer is one in which no energy is dissipated, so the output power equals the input power: ε_p I_p = ε_s I_s, giving ε_p/ε_s = N_p/N_s = I_s/I_p. The assumptions are: zero resistance in the primary and secondary windings, no flux leakage (all flux links both coils), no eddy-current heating and no hysteresis loss in the core. Exam tip: the efficiency mark is for stating power in = power out; do not claim an ideal transformer 'increases power' — stepping voltage UP steps current DOWN in the same ratio. Transformers work only with alternating current, since a steady flux induces no emf.
- DefinitionHL onlyData booklet: Yes
Define the turns ratio of a transformer and distinguish step-up from step-down.
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The turns ratio is N_s/N_p, the number of secondary turns divided by the number of primary turns; for an ideal transformer ε_s/ε_p = N_s/N_p. A step-up transformer has N_s > N_p and raises the voltage while reducing the current; a step-down transformer has N_s < N_p and lowers the voltage while raising the current. Exam tip: the classic lost mark is inverting the ratio — always write the equation as a proportion and check that the larger coil corresponds to the larger voltage. Dimensionless. Grid link: step-up before transmission, step-down at substations and at the consumer.
- DefinitionHL only
Define hysteresis loss in a transformer core.
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Hysteresis loss is the energy dissipated as heat each cycle when the ferromagnetic core is repeatedly magnetised and demagnetised by the alternating primary current, because the magnetic domains do not return along the same path — energy proportional to the area of the B–H hysteresis loop is lost per cycle. Exam tip: it is reduced by using a soft magnetic material such as soft iron or silicon steel with a NARROW hysteresis loop; students frequently confuse it with eddy-current loss (reduced by lamination) — the mark scheme expects the correct remedy paired with the correct loss. Unit of the energy loss: joule per cycle (J), or watt (W) as a power loss.
- DefinitionHL only
Define flux leakage and state one further real-transformer loss.
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Flux leakage is the fraction of the magnetic flux produced by the primary coil that does not pass through (link) the secondary coil, so the secondary emf is smaller than the ideal-transformer prediction; it is reduced by winding both coils on the same limb of a continuous, closed, high-permeability core. A further loss is resistive (Joule/copper) heating I²R in the windings, reduced by using thick low-resistance copper wire. Exam tip: when asked to 'outline why a transformer is not 100% efficient', give named losses AND their remedies — flux leakage, eddy currents (laminated core), hysteresis (soft iron), resistive heating (thick copper).
- EquationHL onlyData booklet: Yes
State the equation for magnetic flux and define every symbol.
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Φ = BA cosθ. Φ = magnetic flux (Wb = T m² = V s); B = magnetic flux density (T); A = area of the surface/loop (m²); θ = angle between B and the NORMAL to the area (degrees or rad). Valid for a uniform field over a plane area. Data booklet: yes. Common misuse: taking θ as the angle between B and the plane of the coil, which swaps sin for cos and gives zero flux when it should be a maximum. Sanity check: B = 0.20 T, A = 5.0 × 10⁻³ m², θ = 60° → Φ = 0.20 × 5.0 × 10⁻³ × 0.50 = 5.0 × 10⁻⁴ Wb.
- EquationHL onlyData booklet: No – memorise
State the expression for magnetic flux linkage and its rearrangements.
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Flux linkage = NΦ = NBA cosθ. N = number of turns (dimensionless); Φ = flux through one turn (Wb); B (T); A (m²); θ measured from the normal. Unit: Wb (or Wb turns). Data booklet: the definition of Φ is given, the factor N must be remembered. Rearranged: B = NΦ/(NA cosθ) = Φ/(A cosθ); for a rotating coil NΦ = NBA cosωt. Common misuse: omitting N, which under-predicts the induced emf by a factor of N. Sanity check: 250 turns, B = 0.15 T, A = 4.0 × 10⁻³ m², θ = 0 → NΦ = 250 × 0.15 × 4.0 × 10⁻³ = 0.15 Wb.
- EquationHL onlyData booklet: Yes
State Faraday's law as an equation and define every symbol.
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ε = −N ΔΦ/Δt. ε = induced emf (V); N = number of turns; ΔΦ = change in magnetic flux through one turn (Wb); Δt = time over which the change occurs (s); the minus sign is Lenz's law (direction opposes the change). Valid for any cause of flux change; ΔΦ/Δt is the gradient of a flux–time graph. Data booklet: yes. Common misuse: using Φ instead of ΔΦ, or forgetting N. Sanity check: a 400-turn coil in which the flux falls from 8.0 × 10⁻⁴ Wb to zero in 0.020 s gives ε = 400 × 8.0 × 10⁻⁴/0.020 = 16 V.
- EquationHL onlyData booklet: Yes
State the equation for the emf induced in a straight conductor moving through a magnetic field.
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ε = BvL, and ε = BvLN for a coil of N turns. ε = induced emf (V); B = magnetic flux density (T); v = speed of the conductor (m s⁻¹); L = length of conductor in the field (m); N = number of turns. Valid only when B, v and L are mutually perpendicular; otherwise use the perpendicular components (ε = BvL sinθ for v at angle θ to B). Data booklet: yes. Common misuse: using the full length of the rod rather than the length inside the field region. Sanity check: B = 0.050 T, v = 12 m s⁻¹, L = 0.80 m → ε = 0.48 V.
- EquationHL onlyData booklet: No – derive
Give the expression for the emf of a coil rotating in a uniform magnetic field.
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ε = NBAω sinωt (taking ε = 0 at t = 0, i.e. coil plane perpendicular to B). ε = instantaneous emf (V); N = turns; B = flux density (T); A = coil area (m²); ω = angular frequency (rad s⁻¹), ω = 2πf; t = time (s). Derived from NΦ = NBA cosωt and ε = −N dΦ/dt. Data booklet: not printed — derive it from Faraday's law. Common misuse: using f instead of ω, which under-predicts ε by 2π. Sanity check: N = 100, B = 0.20 T, A = 0.010 m², f = 50 Hz → ω = 314 rad s⁻¹, ε₀ = 100 × 0.20 × 0.010 × 314 = 63 V.
- EquationHL onlyData booklet: No – derive
State the expression for the peak emf of an ac generator and its rearrangements.
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ε₀ = NBAω = 2πfNBA. ε₀ = peak (maximum) emf (V); N = turns; B = flux density (T); A = area of the coil (m²); ω = angular frequency (rad s⁻¹); f = rotation frequency (Hz). Valid for a coil rotating at constant ω in a uniform field. Data booklet: not printed — derive from ε = NBAω sinωt. Rearranged: B = ε₀/(NAω), ω = ε₀/(NBA). Common misuse: quoting the rms output when the question asks for the peak, or vice versa — ε_rms = ε₀/√2. Sanity check: doubling f doubles ε₀ AND doubles the frequency of the output waveform.
- EquationHL onlyData booklet: Yes
State the relationship between rms and peak current for a sinusoidal alternating current.
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I_rms = I₀/√2 ≈ 0.707 I₀, equivalently I₀ = √2 I_rms. I_rms = root-mean-square current (A); I₀ = peak (maximum) current (A). Valid ONLY for a sinusoidal waveform — a square or triangular wave has a different factor. Data booklet: yes. Common misuse: applying the √2 factor twice, or treating I_rms as a time-average of I (the mean of a sine over one cycle is zero; it is the mean of I² that matters, ⟨I²⟩ = I₀²/2). Sanity check: an ammeter reading 3.0 A on ac corresponds to a peak current of 3.0 × 1.41 = 4.2 A.
- EquationHL onlyData booklet: Yes
State the relationship between rms and peak potential difference, with a worked check.
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V_rms = V₀/√2, so V₀ = √2 V_rms. V_rms = root-mean-square pd (V); V₀ = peak pd (V). Sinusoidal supplies only. Data booklet: yes. Peak-to-peak V_pp = 2V₀, so from an oscilloscope V_rms = V_pp/(2√2). Common misuse: reading V_pp from a CRO trace and putting it straight into V_rms = V₀/√2, which doubles the answer. Sanity check: mains quoted as 230 V rms has V₀ = 230√2 = 325 V and V_pp = 650 V — the value the insulation must withstand.
- EquationHL onlyData booklet: Yes
State the resistance relationships for a resistor in an ac circuit.
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R = V₀/I₀ = V_rms/I_rms. R = resistance (Ω); V₀, I₀ = peak pd (V) and peak current (A); V_rms, I_rms = rms values. Valid for a purely resistive (ohmic) load, where current and voltage are in phase; the √2 factors cancel, so peak and rms values give the same R. Data booklet: yes. Common misuse: mixing a peak voltage with an rms current in the same division, giving an answer wrong by √2. Sanity check: V₀ = 12 V, I₀ = 0.50 A → R = 24 Ω; equivalently V_rms = 8.5 V and I_rms = 0.35 A → R = 24 Ω.
- EquationHL onlyData booklet: Yes
State the equations for maximum and average power dissipated in a resistor by a sinusoidal ac supply.
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P_max = I₀V₀ and P̄ = ½I₀V₀. P_max = peak instantaneous power (W); P̄ = mean power over a whole number of cycles (W); I₀ = peak current (A); V₀ = peak pd (V). Equivalent forms: P̄ = I_rms V_rms = I_rms²R = V_rms²/R = ½I₀²R = P_max/2. Resistive loads only. Data booklet: yes. Common misuse: using peak values in P = I²R and forgetting the factor ½, doubling the mean power. Sanity check: V₀ = 325 V across R = 100 Ω gives I₀ = 3.25 A, P_max = 1.06 kW, P̄ = 528 W = (230)²/100.
- EquationHL onlyData booklet: Yes
State the ideal transformer equation and define every symbol.
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ε_p/ε_s = N_p/N_s = I_s/I_p. ε_p, ε_s = primary and secondary emf (V); N_p, N_s = number of primary and secondary turns (dimensionless); I_p, I_s = primary and secondary currents (A). Valid for an ideal (100% efficient) transformer operating on ac; the current ratio is INVERTED relative to the voltage ratio because ε_p I_p = ε_s I_s. Data booklet: yes. Common misuse: writing I_p/I_s = N_p/N_s. Sanity check: 230 V, N_p = 1000, N_s = 50 → ε_s = 11.5 V; if the secondary draws 2.0 A then I_p = 2.0 × 50/1000 = 0.10 A, and 230 × 0.10 = 11.5 × 2.0 = 23 W each side.
- EquationHL onlyData booklet: Yes
State the equation for power loss in transmission cables and explain the use of high voltage.
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P_loss = I²R, where P_loss = power dissipated as heat in the lines (W); I = rms current in the cable (A); R = total resistance of the transmission line (Ω). For a fixed transmitted power P = IV, raising V by a factor n reduces I by n and P_loss by n². Data booklet: yes (P = I²R). Common misuse: substituting the supply voltage into P = V²/R — the V in that form is the pd ACROSS THE CABLE, not the transmission voltage. Sanity check: 10 MW at 20 kV gives I = 500 A and P_loss = 500² × 5.0 = 1.3 MW in a 5.0 Ω line; at 400 kV, I = 25 A and P_loss = 3.1 kW.
- EquationHL onlyData booklet: Yes
State the equation for transformer efficiency and typical real values.
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efficiency = P_out/P_in = ε_s I_s/(ε_p I_p), expressed as a fraction or ×100%. P_out = useful power delivered to the secondary circuit (W); P_in = power supplied to the primary (W). Valid for any real transformer; an ideal transformer has efficiency 1 (100%). Data booklet: efficiency = useful output/total input is printed; the transformer form must be constructed. Common misuse: assuming turns ratio equals power ratio. Real large grid transformers exceed 98%. Sanity check: 240 V, 5.0 A in and 12 V, 96 A out gives 1152/1200 = 0.96, so 48 W is lost as heat in the core and windings.
- Graph/diagramHL only
Describe how to obtain the induced emf–time graph from a magnetic flux linkage–time graph.
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Axes: flux linkage NΦ / Wb (vertical) against time t / s (horizontal), and beneath it ε / V against t / s. The induced emf at any instant is minus the GRADIENT of the flux-linkage graph, ε = −N ΔΦ/Δt. A horizontal section (constant flux) gives ε = 0; a straight sloping section gives a constant emf equal to minus the slope; a steeper slope gives a larger emf; a maximum or minimum of NΦ gives ε = 0. A sinusoidal NΦ = NBA cosωt yields a sinusoidal ε = NBAω sinωt, a quarter-cycle (90°) out of phase. Extract ε from a tangent's rise/run in Wb s⁻¹ = V.
- Graph/diagramHL only
Describe the emf–time graph produced when a bar magnet falls straight through a flat coil.
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Axes: ε / mV (vertical) against t / s (horizontal). Two pulses of OPPOSITE sign separated by a brief near-zero region as the mid-point of the magnet passes the coil. The second pulse is larger in magnitude (the magnet has accelerated, so dΦ/dt is greater) and narrower in time; the two pulses have EQUAL areas because area = ∫ε dt = ΔΦ linkage, and the flux rises and falls by the same amount. Frequently tested point: the sign reversal is Lenz's law — the coil repels the approaching pole and attracts the receding one, so the magnet's downward acceleration is less than g. Dropping from a greater height makes both pulses taller and narrower with unchanged area.
- Graph/diagramHL only
Describe the flux linkage and emf graphs for a coil rotating at constant angular velocity in a uniform field.
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Axes: NΦ / Wb and ε / V (vertical) against t / s (horizontal). NΦ = NBA cosωt is a cosine of amplitude NBA; ε = NBAω sinωt is a sine of amplitude NBAω, lagging the flux graph by a quarter period (90°, π/2 rad). ε = 0 when the coil plane is perpendicular to B (flux a maximum, gradient zero); ε is a maximum when the coil plane is parallel to B (flux zero, gradient steepest). Extract ω from the period, T = 2π/ω, and hence B from ε₀ = NBAω. Halving ω halves both the emf amplitude and the frequency of the output.
- Graph/diagramHL only
Describe the linearised graph used to determine B from the peak emf of a rotating search coil.
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Axes: peak emf ε₀ / V (vertical) against angular frequency ω / rad s⁻¹ (horizontal), with N and A fixed. Since ε₀ = NBAω, the graph is a straight line through the ORIGIN of gradient NBA, so B = gradient/(NA). Obtain ε₀ from an oscilloscope trace (half the peak-to-peak deflection) and ω = 2π/T from the same trace or from a stroboscope. Draw maximum and minimum gradient lines through the error bars to find the uncertainty in B. A non-zero intercept signals a systematic error, e.g. a CRO y-shift or a residual background field; plotting against f instead of ω changes the gradient to 2πNBA.
- Graph/diagramHL only
Describe the graph of magnetic flux against cosθ for a coil in a uniform field.
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Axes: flux Φ / Wb (vertical) against cosθ (horizontal, dimensionless, running from −1 to +1), where θ is the angle between B and the coil normal. Because Φ = BA cosθ, the graph is a straight line through the origin of gradient BA; hence B = gradient/A. This is the standard linearisation of the non-linear Φ–θ curve. Extract B by measuring the coil area with calipers and reading the gradient in Wb. Increasing B or A steepens the line; the line passes through Φ = 0 at cosθ = 0 (θ = 90°, coil plane containing B). Plotting Φ against θ instead gives a cosine curve from which no gradient can be read directly.
- Graph/diagramHL only
Describe the graph of secondary voltage against number of secondary turns for a transformer.
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Axes: secondary rms voltage V_s / V (vertical) against number of secondary turns N_s (horizontal, dimensionless), for a fixed primary supply V_p and fixed N_p. For an ideal transformer V_s = (V_p/N_p)N_s, a straight line through the origin of gradient V_p/N_p (volts per turn), so N_p can be found from N_p = V_p/gradient. In practice the measured points fall slightly BELOW the ideal line, and increasingly so at large N_s, because of flux leakage and winding resistance; the graph may also curve if the core approaches magnetic saturation. Raising V_p steepens the line proportionally.
- Graph/diagramHL only
Describe the power–time graph for a resistor connected to a sinusoidal ac supply.
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Axes: instantaneous power P / W (vertical) against time t / s (horizontal). P = I₀V₀ sin²ωt, so the curve is always POSITIVE (power is dissipated on both halves of the cycle), touching zero twice per cycle and peaking at P_max = I₀V₀; its frequency is DOUBLE that of the current or voltage waveform. The mean value is the horizontal line at P̄ = ½I₀V₀ = I_rms V_rms, which is exactly halfway between 0 and P_max because sin²ωt averages to ½. The area under the curve over an interval gives the energy transferred in joules. Doubling V₀ quadruples both P_max and P̄.
- Concept/explainHL onlyData booklet: Yes
Explain what magnetic flux through a coil means and why the angle term appears in Φ = BA cosθ.
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- Magnetic flux is the product of the magnetic flux density and the area normal to (perpendicular to) the field
- Φ = BA cosθ, where θ is the angle between B and the NORMAL to the plane of the loop
- only the component B cosθ along the normal threads the area
- Φ is maximum (BA) when the plane of the coil is perpendicular to B, and zero when the plane contains B
- flux is a scalar; unit weber, 1 Wb = 1 T m² = 1 V s
- flux can be visualised as the number of field lines passing through the loop. Exam tip: the commonest error is measuring θ from the plane of the coil, giving sinθ instead of cosθ.
- Concept/explainHL onlyData booklet: Yes
Distinguish between magnetic flux and magnetic flux linkage, and explain why Faraday's law uses flux linkage.
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- Flux Φ = BA cosθ applies to a single loop, unit Wb
- flux linkage = NΦ = NBA cosθ for a coil of N turns, unit Wb turns (weber-turns)
- each turn is a separate conducting loop in series, so each turn has the same emf induced in it and the emfs add
- hence total emf ε = −N ΔΦ/Δt, i.e. the rate of change of flux LINKAGE
- doubling N doubles the emf for the same rate of change of flux
- a graph of NΦ against t has gradient equal to −ε. Exam tip: students routinely omit N, or quote the unit of flux linkage as Wb; both cost marks in "determine" questions.
- Concept/explainHL onlyData booklet: Yes
State and explain Faraday's law of electromagnetic induction.
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- The magnitude of the induced emf is proportional to (equal to) the rate of change of magnetic flux linkage: ε = −N ΔΦ/Δt
- flux can change by changing B, changing the area A, or changing the orientation θ of the coil
- a relative motion or a changing current in a nearby circuit is therefore needed — a steady flux gives zero emf
- the emf exists whether or not the circuit is complete; current flows only if there is a complete circuit
- on a flux-linkage–time graph the induced emf is the gradient
- the negative sign is Lenz's law. Exam tip: incomplete answers say "emf is proportional to the flux" instead of to its RATE OF CHANGE.
- Concept/explainHL only
Explain Lenz's law and show that it is a statement of conservation of energy.
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- The induced emf/current acts in a direction such that it opposes the CHANGE in flux that produced it
- hence the induced current creates a magnetic field opposing the increase or decrease of flux
- e.g. a north pole approaching a coil induces a current making the near face a north pole, repelling the magnet
- work must therefore be done against this opposing force to keep the magnet moving
- that mechanical work is converted into electrical energy and then to internal energy in the coil's resistance
- if the induced effect helped the change, the system would accelerate and generate energy from nothing, violating conservation of energy
- this is the origin of the minus sign in Faraday's law. Exam tip: "opposes the flux" is not accepted — it opposes the CHANGE in flux.
- Concept/explainHL onlyData booklet: Yes
Explain the origin of the motional emf ε = BvL for a rod moving perpendicular to a magnetic field.
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- Free electrons in the rod move with velocity v, so each experiences a magnetic force F = qvB along the rod
- charge accumulates at the ends, making one end negative and the other positive
- this separation sets up an electric field E along the rod, opposing further separation
- equilibrium when qE = qvB, so E = vB and the pd across length L is ε = EL = BvL
- for a coil of N turns ε = BvLN
- equivalently, the rod sweeps area LvΔt per time Δt so ΔΦ/Δt = BLv, which is Faraday's law
- valid only when v, B and L are mutually perpendicular. Exam tip: state the force on the CHARGES, not just "flux is cut".
- Concept/explainHL only
Outline what eddy currents are, one useful application and one way of reducing them.
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- Eddy currents are circulating induced currents in the body of a conductor that experiences a changing magnetic flux
- by Lenz's law they oppose the change, producing a retarding force on relative motion
- they dissipate energy as internal energy through the conductor's resistance (I²R)
- applications: electromagnetic (eddy-current) braking on trains and roller-coasters, induction hobs, induction furnaces, metal detectors
- disadvantage: energy loss and heating in transformer and motor cores
- reduced by laminating the core — thin sheets separated by insulating varnish break the current paths and raise resistance — or by using ferrite/high-resistivity cores. Exam tip: "eddy currents heat the core" alone is not enough; link the loss explicitly to Lenz's law and to I²R dissipation.
- Concept/explainHL onlyData booklet: No – derive
Explain how a simple AC generator produces a sinusoidal emf.
Show answer
- A coil of N turns and area A rotates with angular speed ω in a uniform field B
- flux linkage NΦ = NBA cos(ωt), so by Faraday's law ε = NBAω sin(ωt)
- the emf is therefore sinusoidal with peak value ε₀ = NBAω
- emf is ZERO when the plane of the coil is perpendicular to B (flux maximum, rate of change zero) and MAXIMUM when the plane is parallel to B (flux zero, rate of change greatest)
- slip rings and brushes take the alternating output to the external circuit (a split-ring commutator would rectify it)
- the emf and flux-linkage graphs are 90° out of phase. Exam tip: students confuse maximum flux with maximum emf — the emf is the GRADIENT of the flux-linkage graph.
- Concept/explainHL onlyData booklet: No – derive
Explain how the emf–time graph of an AC generator changes when the rotation speed is doubled.
Show answer
- Peak emf ε₀ = NBAω, so doubling ω doubles the peak emf
- the period T = 2π/ω halves and the frequency doubles, so twice as many cycles appear in the same time
- the graph therefore has twice the amplitude and half the period — both change together, which is the key marking point
- the shape stays sinusoidal and the emf is still zero at the same coil orientations
- rms emf also doubles since ε_rms = ε₀/√2
- mean power in a fixed resistor ∝ ε₀², so it becomes four times as large
- the area under each half-cycle, equal to the change in flux linkage 2NBA, is unchanged
- by contrast doubling B or A doubles the amplitude only and leaves the period the same. Exam tip: a very common sketch error is to double the amplitude but keep the period unchanged, or vice versa; both features must be shown.
- Concept/explainHL onlyData booklet: Yes
Explain what is meant by the rms value of an alternating current and why I_rms = I₀/√2.
Show answer
- The rms current is the value of the steady direct current that would dissipate the same average power in the same resistor
- instantaneous power P = I²R varies as sin²(ωt); the mean of sin² over a cycle is ½
- so P̄ = ½I₀²R = (I₀/√2)²R, giving I_rms = I₀/√2 and similarly V_rms = V₀/√2
- hence P̄ = ½I₀V₀ = I_rms V_rms, while P_max = I₀V₀ = 2P̄
- resistance is unchanged: R = V₀/I₀ = V_rms/I_rms
- mains quoted as 230 V means 230 V rms, peak ≈ 325 V. Exam tip: the mean CURRENT over a cycle is zero — that is why rms, not mean, is used; power is never negative because I² is positive.
- Concept/explainHL onlyData booklet: Yes
Explain how an ideal transformer works and why it cannot operate on a steady DC supply.
Show answer
- Alternating current in the primary produces an alternating magnetic flux in the soft-iron core
- the laminated core links (channels) essentially all this flux through the secondary coil
- the changing flux linkage induces an alternating emf in the secondary by Faraday's law
- since each turn links the same flux, ε_p/ε_s = N_p/N_s
- for an ideal (100% efficient) transformer input power = output power, so ε_pI_p = ε_sI_s and ε_p/ε_s = I_s/I_p
- N_s > N_p is a step-up transformer (voltage up, current down)
- with steady DC the flux is constant, dΦ/dt = 0 and no emf is induced — only at switch-on/off. Exam tip: say "changing flux linkage", not just "the current passes to the secondary".
- Concept/explainHL only
Outline the causes of energy loss in a real transformer and how each is reduced.
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- Eddy currents induced in the core dissipate energy — reduced by laminating the core with insulated thin sheets, or using a high-resistivity ferrite core
- hysteresis: energy is used repeatedly re-magnetising the core each cycle — reduced by using a soft magnetic material (soft iron) with a narrow hysteresis loop
- resistive (Joule/copper) heating I²R in the windings — reduced by using thick, low-resistivity copper wire, or cooling oil
- flux leakage: not all primary flux links the secondary — reduced by a continuous closed core and by winding the coils on the same limb, one over the other
- real efficiencies are typically 95–99%. Exam tip: "heat is lost" earns nothing; name the mechanism AND the specific remedy.
- Concept/explainHL only
Explain why electrical energy is transmitted over long distances at high voltage.
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- For a given transmitted power P = IV, raising V lowers the current I in the cables
- power dissipated in the transmission line is P_loss = I²R, so it falls with the SQUARE of the current
- e.g. raising the voltage by ×10 cuts the current by 10 and the loss by 100
- step-up transformers at the power station and step-down transformers near consumers make this possible, which requires AC
- thinner, cheaper, lighter cables can then be used for the same loss
- drawbacks: expensive insulation, tall pylons, safety and corona discharge, land use and public concern. Exam tip: never use P = V²/R for the cable loss — V there would be the pd ACROSS the cable, not the transmission voltage; use I²R.
- Concept/explainHL onlyData booklet: No – derive
A conducting rod slides at constant velocity along frictionless rails in a magnetic field, completing a circuit. Explain why a force must be applied and where the energy goes.
Show answer
- The rod cuts flux, inducing emf ε = BvL and a current I = BvL/R in the circuit
- by Lenz's law the induced current in the rod experiences a magnetic force F = BIL opposing the motion (backwards)
- so an external applied force of equal magnitude is needed for constant velocity (zero net force, Newton's first law)
- the mechanical power supplied is P = Fv = B²L²v²/R
- this equals the electrical power generated εI = I²R, dissipated as internal energy in the resistance
- so mechanical work → electrical energy → internal energy, with no energy created: Lenz's law is conservation of energy
- if the applied force is removed the rod decelerates exponentially. Exam tip: state that the OPPOSING force follows from Lenz's law.
- Worked problemHL onlyData booklet: Yes
A flat coil of 200 turns and area 4.0 × 10⁻³ m² is placed in a uniform field of flux density 0.15 T, with the normal to the coil at 60° to the field. Calculate the flux and the flux linkage.
Show answer
Use Φ = BA cosθ with θ measured from the NORMAL. Φ = 0.15 T × 4.0 × 10⁻³ m² × cos60° = 0.15 × 4.0 × 10⁻³ × 0.500 = 3.0 × 10⁻⁴ Wb. Flux linkage NΦ = 200 × 3.0 × 10⁻⁴ = 6.0 × 10⁻² Wb turns. Answers to 2 s.f. as the data are given to 2 s.f. Check/Trap: if the question had said the PLANE of the coil makes 60° with the field, the normal would be at 30° and you would use cos30° = 0.866, giving 5.2 × 10⁻⁴ Wb — always identify the angle relative to the normal before substituting.
- Worked problemHL onlyData booklet: Yes
The 200-turn coil of area 4.0 × 10⁻³ m² lies with its plane perpendicular to a 0.15 T field. It is rotated through 90° in 0.20 s. Determine the average induced emf.
Show answer
Initially the normal is parallel to B: Φ₁ = BA = 0.15 × 4.0 × 10⁻³ = 6.0 × 10⁻⁴ Wb. Finally the plane contains B, so Φ₂ = 0. ΔΦ = 6.0 × 10⁻⁴ Wb. Change in flux linkage NΔΦ = 200 × 6.0 × 10⁻⁴ = 0.12 Wb turns. Faraday: |ε| = NΔΦ/Δt = 0.12 / 0.20 = 0.60 V (2 s.f.). Check/Trap: this is the AVERAGE emf over the quarter turn — the instantaneous emf varies sinusoidally and its peak here would be larger. Do not divide by 90 or convert to radians: only the initial and final flux and the time interval matter for an average value.
- Worked problemHL onlyData booklet: Yes
A 150-turn coil of area 2.0 × 10⁻² m² lies perpendicular to a field that increases uniformly from 0.20 T to 0.50 T in 1.5 s. The coil has total resistance 12 Ω. Determine the induced emf and current.
Show answer
The area and orientation are constant, so ΔΦ = ΔB × A = (0.50 − 0.20) × 2.0 × 10⁻² = 0.30 × 2.0 × 10⁻² = 6.0 × 10⁻³ Wb. Flux linkage change = 150 × 6.0 × 10⁻³ = 0.90 Wb turns. |ε| = NΔΦ/Δt = 0.90 / 1.5 = 0.60 V. Current I = ε/R = 0.60 / 12 = 0.050 A = 50 mA (2 s.f.). Check/Trap: use ΔB, not the final value of B — using 0.50 T gives 1.0 V, a classic error. The emf is constant here because B changes at a constant rate; if B–t were curved the emf would vary as the gradient.
- Worked problemHL onlyData booklet: Yes
An aircraft of wingspan 60 m flies horizontally at 250 m s⁻¹ where the vertical component of Earth's magnetic field is 4.0 × 10⁻⁵ T. Determine the emf induced between the wingtips.
Show answer
The wings act as a conducting rod of length L = 60 m moving perpendicular to the VERTICAL component of the field, so ε = BvL = 4.0 × 10⁻⁵ T × 250 m s⁻¹ × 60 m. Step: 4.0 × 10⁻⁵ × 250 = 1.0 × 10⁻². Then 1.0 × 10⁻² × 60 = 0.60 V (2 s.f.). Check/Trap: only the component of B perpendicular to the plane swept by the wings contributes — using the total field or the horizontal component is the standard error. No current flows round a circuit, so no wingtip "power" is generated and there is no retarding force; a pd simply exists across the wings. Units: T m s⁻¹ m = V.
- Worked problemHL onlyData booklet: Yes
A rod of length 0.25 m slides at 3.0 m s⁻¹ along frictionless rails in a 0.40 T field perpendicular to the plane of the rails. The circuit resistance is 0.50 Ω. Find the emf, current, force needed and power dissipated.
Show answer
emf: ε = BvL = 0.40 × 3.0 × 0.25 = 0.30 V. Current: I = ε/R = 0.30 / 0.50 = 0.60 A. Force on the rod from the induced current: F = BIL = 0.40 × 0.60 × 0.25 = 0.060 N, directed opposite to the motion (Lenz). For constant velocity the applied force equals 0.060 N. Mechanical power P = Fv = 0.060 × 3.0 = 0.18 W. Electrical power P = εI = 0.30 × 0.60 = 0.18 W — they agree, as energy is conserved (2 s.f.). Check/Trap: the magnetic force opposes motion, it does not drive the rod; and P = I²R = 0.60² × 0.50 = 0.18 W confirms the answer.
- Worked problemHL onlyData booklet: Yes
A graph of flux linkage against time for a coil rises linearly from 0 to 0.40 Wb turns in 20 ms, then stays constant. Determine the emf during each stage.
Show answer
Faraday's law: |ε| = Δ(NΦ)/Δt = gradient of the flux-linkage–time graph. Stage 1: |ε| = 0.40 Wb turns / (20 × 10⁻³ s) = 20 V, constant, because the gradient is constant. Stage 2: the flux linkage is constant so the gradient is zero and |ε| = 0 V, even though the flux itself is large. The emf–time graph is therefore a 20 V rectangular pulse of duration 20 ms followed by zero. Check/Trap: converting ms to s is where marks are lost — 0.40/20 = 0.020 V is wrong by 10³. A constant non-zero flux always gives zero emf; only the RATE of change matters.
- Worked problemHL onlyData booklet: No – derive
A generator coil of 500 turns and area 2.0 × 10⁻² m² rotates at 3000 revolutions per minute in a uniform field of 0.080 T. Determine the peak and rms emf.
Show answer
Frequency f = 3000/60 = 50 Hz, so ω = 2πf = 2π × 50 = 314 rad s⁻¹. Peak emf ε₀ = NBAω = 500 × 0.080 × 2.0 × 10⁻² × 314. Step: 500 × 0.080 = 40; 40 × 2.0 × 10⁻² = 0.80; 0.80 × 314 = 2.5 × 10² V (251 V). rms emf: ε_rms = ε₀/√2 = 251/1.414 = 1.8 × 10² V (178 V). Answers to 2 s.f. Check/Trap: convert rpm to Hz and then to ω — using 3000 or 50 directly as ω is the standard error, giving answers wrong by 2π or by 60. The output frequency equals the rotation frequency, 50 Hz.
- Worked problemHL onlyData booklet: Yes
An electric heater of mean power 2.0 kW operates from a 230 V rms mains supply. Determine the peak voltage, the resistance, the rms and peak currents, and the maximum instantaneous power.
Show answer
Peak voltage: V₀ = √2 V_rms = 1.414 × 230 = 3.3 × 10² V (325 V). Resistance: P̄ = V_rms²/R so R = 230²/2000 = 52900/2000 = 26 Ω. rms current: I_rms = P̄/V_rms = 2000/230 = 8.7 A. Peak current: I₀ = √2 × 8.7 = 12 A. Maximum instantaneous power: P_max = I₀V₀ = 12.3 × 325 = 4.0 × 10³ W, which is exactly 2P̄ as expected since P̄ = ½I₀V₀. All to 2 s.f. Check/Trap: mains voltages are always quoted as rms; using 325 V in P = V²/R would give a power twice too large. R = V₀/I₀ = V_rms/I_rms — resistance is the same either way.
- Worked problemHL onlyData booklet: No – derive
A 400-turn coil of area 1.5 × 10⁻³ m² sits perpendicular to a 0.60 T field. The field is switched off. If the circuit resistance is 8.0 Ω, determine the charge that flows.
Show answer
Charge q = IΔt and I = NΔΦ/(RΔt), so q = NΔΦ/R — independent of how quickly the field is switched off. ΔΦ = BA = 0.60 × 1.5 × 10⁻³ = 9.0 × 10⁻⁴ Wb. NΔΦ = 400 × 9.0 × 10⁻⁴ = 0.36 Wb turns. q = 0.36 / 8.0 = 4.5 × 10⁻² C (2 s.f.). Check/Trap: students try to find the time; it cancels. Physically, switching off faster gives a larger emf and current for a shorter time, so the same charge flows — this is exactly how a search coil and ballistic galvanometer/charge meter measures flux density B.
- Worked problemHL onlyData booklet: Yes
An ideal transformer has 1200 primary turns and 60 secondary turns and is connected to a 240 V rms supply. The secondary supplies 5.0 A. Determine the secondary voltage and the primary current.
Show answer
Voltage ratio: ε_p/ε_s = N_p/N_s, so ε_s = 240 × 60/1200 = 240/20 = 12 V rms (step-down). Ideal transformer: input power = output power. Output P = ε_sI_s = 12 × 5.0 = 60 W. Primary current I_p = P/ε_p = 60/240 = 0.25 A. Equivalently N_p/N_s = I_s/I_p gives I_p = 5.0 × 60/1200 = 0.25 A. Answers to 2 s.f. Check/Trap: the current ratio is the INVERSE of the turns ratio — a step-down transformer reduces voltage but increases current. Real transformers are not 100% efficient, so the actual primary current would be slightly larger than 0.25 A.
- Worked problemHL onlyData booklet: Yes
A power station transmits 2.0 MW through cables of total resistance 4.0 Ω. Compare the power lost at a transmission voltage of 10 kV and at 200 kV.
Show answer
At 10 kV: I = P/V = 2.0 × 10⁶ / 1.0 × 10⁴ = 200 A. Loss = I²R = 200² × 4.0 = 4.0 × 10⁴ × 4.0 = 1.6 × 10⁵ W = 160 kW, which is 8.0% of the transmitted power. At 200 kV: I = 2.0 × 10⁶ / 2.0 × 10⁵ = 10 A. Loss = 10² × 4.0 = 4.0 × 10² W = 0.40 kW, which is 0.020%. Raising the voltage by a factor of 20 reduces the loss by 20² = 400. Check/Trap: use P_loss = I²R with the CABLE resistance; using P = V²/R with the transmission voltage gives an absurd 25 MW at 10 kV. The pd across the cable at 10 kV is only IR = 800 V.
- Worked problemHL onlyData booklet: Yes
A solenoid with 2000 turns per metre carries a current that rises uniformly from 0 to 3.0 A in 0.50 s. A small coil of 50 turns and area 4.0 × 10⁻⁴ m² lies inside, coaxially. Determine the emf induced in the small coil.
Show answer
Field inside the solenoid: B = μ₀NI/L = μ₀nI = 4π × 10⁻⁷ × 2000 × 3.0 = 7.5 × 10⁻³ T at the final instant, and 0 initially. Flux through the small coil: ΔΦ = ΔB × A = 7.5 × 10⁻³ × 4.0 × 10⁻⁴ = 3.0 × 10⁻⁶ Wb. Flux linkage change = 50 × 3.0 × 10⁻⁶ = 1.5 × 10⁻⁴ Wb turns. |ε| = 1.5 × 10⁻⁴ / 0.50 = 3.0 × 10⁻⁴ V = 0.30 mV (2 s.f.). Check/Trap: use the AREA OF THE SMALL COIL, not of the solenoid — the field exists over the solenoid's cross-section but only the small coil's area is linked. n is turns per metre, so N/L = 2000 m⁻¹.
- Worked problemHL onlyData booklet: No – derive
An AC generator produces a peak emf of 12 V at a frequency of 50 Hz. The number of turns is doubled and the rotation frequency is reduced to 25 Hz. Deduce the new peak emf and output frequency.
Show answer
Peak emf ε₀ = NBAω = NBA(2πf), so ε₀ ∝ Nf when B and A are unchanged. New peak emf = 12 × (2N/N) × (25/50) = 12 × 2 × 0.50 = 12 V — unchanged. The output frequency equals the rotation frequency, so it becomes 25 Hz. The emf–time graph therefore has the same amplitude but twice the period (0.040 s instead of 0.020 s). rms emf is also unchanged at 12/√2 = 8.5 V. Check/Trap: a Paper 1 favourite — the two changes cancel for amplitude but NOT for period. Do not assume that doubling the turns must double the output; always write the proportionality ε₀ ∝ NBAω first and check each factor.
- Exam technique/trapHL only
Explain the correct way to answer "Explain the direction of the induced current" questions, and the trap that costs most marks.
Show answer
Trap: writing that the induced current "opposes the flux" or "opposes the magnetic field". Lenz's law says it opposes the CHANGE in flux — so when flux is decreasing the induced field is in the SAME direction as the original field, to try to maintain it. Students fall for this because the standard example (magnet approaching) has opposition in both senses. Correct approach, for the command term "explain": (1) state whether flux through the coil is increasing or decreasing; (2) state the direction of the induced field needed to oppose that change; (3) use the right-hand grip rule to convert that to a current direction (clockwise/anticlockwise viewed from a stated side). Always specify the viewing direction — "clockwise" alone is ambiguous.
- Exam technique/trapHL onlyData booklet: Yes
Outline the symbol, unit and sign the most common errors in Faraday's law questions.
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Traps: (1) omitting N — the law is ε = −N ΔΦ/Δt, and the coil often has hundreds of turns; (2) quoting flux linkage in Wb instead of Wb turns (weber-turns); (3) using B rather than ΔB, or the final flux rather than the CHANGE in flux; (4) forgetting to convert ms, cm² or rpm; (5) treating the minus sign as something to compute with — it only encodes direction, and "magnitude of the emf" questions want a positive answer; (6) using θ from the plane rather than the normal in Φ = BA cosθ. Approach: write the equation symbolically, list N, ΔΦ and Δt with SI units before substituting, and quote the answer to the same s.f. as the least precise datum.
- Exam technique/trapHL only
Give the marking points expected when asked to "sketch" the emf–time graph for a magnet falling through a coil.
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"Sketch" means show the correct shape with key features labelled; no accurate scales are required, but axes must be labelled with quantity and unit. Marks: two pulses of OPPOSITE sign; emf momentarily zero between them (when the magnet's midpoint is level with the coil); second pulse of LARGER magnitude and SHORTER duration because the magnet is faster on leaving; approximately equal areas under the two pulses, since area = NΔΦ. Trap: drawing two identical symmetric peaks, or a continuous positive curve. Related "suggest" question: dropping the magnet from higher makes both pulses taller and narrower; a coil of more turns raises both emfs proportionally; the area under a pulse is unchanged by speed.
- Exam technique/trapHL onlyData booklet: Yes
Explain the peak-versus-rms traps in AC power calculations.
Show answer
Trap: substituting peak values into power or Ohm's-law expressions meant for rms values. Rules: P̄ = I_rms V_rms = ½I₀V₀; P_max = I₀V₀ = 2P̄; R = V₀/I₀ = V_rms/I_rms (both ratios give the same R). A quoted mains or appliance voltage is always rms unless stated otherwise. Students fall for it because "maximum" appears in both P_max and V₀, so they mix families. Approach: decide first whether the quantity asked for is an average or an instantaneous maximum, then keep to one family of values throughout. A useful check: any answer for mean power computed from peak values will be exactly twice too big. Also note the mean CURRENT over a cycle is zero, which is why rms exists.
- Exam technique/trapHL onlyData booklet: Yes
List the traps in transformer questions and the assumptions you must state.
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Traps: (1) inverting the current ratio — the correct relation is ε_p/ε_s = N_p/N_s = I_s/I_p, so a step-up transformer raises voltage and LOWERS current; (2) assuming a transformer works on DC — constant flux gives no induced emf; (3) claiming a transformer "increases power" — at best power is conserved; (4) forgetting that N_s/N_p is a ratio of turns, so the answer has no unit. Assumptions to state for an "ideal" transformer: 100% efficient (no eddy-current, hysteresis or resistive losses), all flux links both coils (no leakage). If asked to "suggest" why a real device is worse, name a specific loss and its remedy. Efficiency = output power/input power × 100%.
- Exam technique/trapHL onlyData booklet: No – derive
Describe how to process data from an experiment measuring peak emf ε₀ against angular frequency ω for a rotating generator coil.
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Theory: ε₀ = NBAω, so a graph of ε₀ (y) against ω (x) should be a straight line through the origin with gradient NBA — this is the linearisation. Plot with error bars (absolute uncertainty in ε₀ from repeated oscilloscope readings, half-range or half the smallest division). Draw the best-fit line plus maximum and minimum gradient lines through the error bars; uncertainty in gradient = (max gradient − min gradient)/2. Then B = gradient/(NA), and the fractional uncertainty in B is the sum of the fractional uncertainties in gradient, N and A. A non-zero y-intercept suggests a systematic error (e.g. oscilloscope zero offset); random errors show as scatter about the line. Quote the gradient to a sensible 2–3 s.f.
- Exam technique/trapHL only
Describe a practical investigation of how the emf induced in a coil depends on the number of turns, including variables and limitations.
Show answer
Apparatus: signal generator driving a solenoid (or an AC electromagnet) to produce an alternating field, a set of search coils of identical area with different numbers of turns, and a calibrated oscilloscope (or rms voltmeter) to measure the induced emf. Method: place each search coil at the same fixed position coaxially inside the solenoid, record the peak emf from the oscilloscope trace, repeat three times and mean. Independent variable N; dependent variable ε₀; controlled: coil area, position and orientation, driving frequency, driving current (monitor with an ammeter), temperature. Expect ε₀ ∝ N through the origin. Limitations: field not perfectly uniform, coil alignment error, stray mains pick-up, heating of the solenoid changing its current. Improvements: fix coils in a jig, use a Helmholtz pair, shield leads, use twisted-pair connections.
- Exam technique/trapHL only
Describe how to investigate the relationship between the emf induced in a coil and the speed of a magnet passing through it, and evaluate the method.
Show answer
Apparatus: fixed coil clamped vertically, bar magnet released from measured heights h above the coil, datalogger/oscilloscope recording emf against time, light gates or v = √(2gh) to find entry speed. Method: release the magnet from rest at several heights, record the peak emf of the first pulse, repeat at least three times per height and average. Independent variable: entry speed v; dependent: peak emf; controlled: same magnet and orientation, same coil and position, same release mechanism, no ferrous objects nearby. Plot ε₀ against v — expect a straight line through the origin since ε ∝ dΦ/dt ∝ v. Limitations: magnet may rotate or not fall centrally, air resistance and induced braking make v slightly less than √(2gh), datalogger sampling rate may miss the peak. Improvements: guide tube, higher sampling rate, electromagnetic release.
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