Physics • Year 12 • Module 6 • Lesson 19

Eddy Currents and Induction Applications

Build HSC Band 5–6 extended-response technique on eddy current physics, evaluating evidence, and designing investigations into electromagnetic braking.

Master · Extended Response

1. Data + scenario: investigating magnetic braking in a copper tube (Band 5–6)

8 marks   Band 5–6

Scenario. A student drops a strong neodymium magnet (mass = 25 g, diameter = 18 mm) down four vertical tubes, each 1.0 m long, and measures the time for the magnet to travel from top to bottom. The four tubes have the same internal diameter (20 mm) but different materials and designs. Results are shown in the table below. The free-fall time for the same 25 g object through 1.0 m (ignoring air resistance) is 0.45 s.

TubeMaterial and designFall time (s)
1PVC plastic (non-conductor)0.47
2Copper, 2 mm wall thickness, solid cylinder4.3
3Copper, 2 mm wall thickness, cut lengthwise (open seam)0.52
4Aluminium, 2 mm wall thickness, solid cylinder2.1

Illustrative data. Free-fall time through 1.0 m ≈ 0.45 s (g = 9.8 m s⁻²).

Q1. Analyse and evaluate the experimental data to explain the pattern of fall times across the four tubes. In your response you must:

  • Explain why Tube 1 gives a fall time close to the free-fall value, with reference to eddy currents.
  • Explain why the magnet falls significantly slower in Tube 2 than in Tube 1, using Faraday’s Law and Lenz’s Law.
  • Explain why Tube 3 (open-seam copper) gives a fall time close to Tube 1, referring to the conducting loop.
  • Compare Tubes 2 and 4 and explain why copper produces a longer fall time than aluminium of the same dimensions.
  • State one source of systematic error in this experiment and suggest how it could be reduced.
Plan: Tube 1 (no conductor → no eddy currents) → Tube 2 (closed loop, high conductivity, large emf → large braking) → Tube 3 (seam breaks loop → no closed circuit → tiny braking) → Tubes 2 vs 4 (copper has lower resistivity ρ than aluminium → same emf drives larger current in copper → stronger force) → systematic error (dropping from inconsistent heights; friction of magnet touching wall).

2. Experimental design — testing the effect of lamination on transformer core losses (Band 5–6)

7 marks   Band 5–6

Research question. A Year 12 student claims that using more laminations in a transformer core always increases efficiency because it reduces eddy current losses. Design a scientific investigation to test whether the number of laminations in a core affects the power lost as heat in the core during AC operation.

Constraints: You have access to: a signal generator (1–100 Hz AC), primary and secondary coils with 200 turns each, iron cores of the same total cross-section that can be assembled from 1, 4, 8, 16, or 32 laminations (each insulated), a power meter to measure input power, an infrared thermometer to measure core surface temperature, and a multimeter. Safety: low-voltage AC only (max 6 V rms).

Q2. Design the investigation and present it in the format below.

  • State your hypothesis as a testable prediction including the independent and dependent variables.
  • Identify the independent variable, dependent variable, and at least two controlled variables.
  • Describe the procedure in at least four numbered steps, including how you will measure and compare core losses.
  • Explain what result would falsify your hypothesis.
  • State two limitations of your design and one way to improve reliability.
Consider: IV = number of laminations (1, 4, 8, 16, 32); DV = input power at same output voltage (power loss = input − useful output), or core temperature rise; control = frequency, turns, applied voltage, core material. Falsification: if increasing laminations does not reduce core temperature or input power loss → hypothesis falsified. Limitation: contact resistance between laminations varies; infrared thermometer may underestimate internal temperature.
Answers — Do not peek before attempting

Q1 — Sample Band 6 response (8 marks), annotated

Tube 1 — plastic (fall time ≈ free fall): PVC is a non-conductor, so no charge carriers exist to form eddy currents when the magnetic flux through the tube wall changes as the magnet passes. Without eddy currents there is no opposing force beyond air resistance, and the fall time of 0.47 s is close to the theoretical free-fall value of 0.45 s (small difference due to air drag) [1].

Tube 2 — copper solid (fall time = 4.3 s): As the magnet falls, the magnetic flux through the copper walls changes continuously. By Faraday’s Law, this changing flux induces an emf in the copper, which drives eddy currents in large closed loops around the circumference of the tube [1]. By Lenz’s Law, these eddy currents produce a magnetic field that opposes the flux change — i.e., opposes the falling magnet with an upward force [1]. Because copper has very low electrical resistivity (ρ = 1.7 × 10⁻&sup8; Ω m), the eddy currents are large, producing a strong upward force that rapidly decelerates the magnet to a low terminal velocity (fall time ≈ 4.3 s vs 0.47 s in plastic) [1].

Tube 3 — open-seam copper (fall time ≈ 0.52 s): The lengthwise seam breaks the continuous conducting path around the tube wall. Eddy currents require a closed loop; without a complete circuit, currents cannot flow circumferentially. Only very small, localised currents exist along the short axial strips beside the seam [1]. The braking force is negligible, so the fall time (0.52 s) is almost identical to the plastic result, confirming that a closed conducting path is essential for magnetic braking.

Tubes 2 vs 4 — copper vs aluminium: Aluminium has a higher electrical resistivity (ρ ≈ 2.7 × 10⁻&sup8; Ω m) than copper (ρ ≈ 1.7 × 10⁻&sup8; Ω m). The same changing flux induces the same emf in both tubes (same geometry and magnet), but aluminium’s higher resistance means smaller eddy currents flow by Ohm’s Law (I = ε/R) [1]. Smaller eddy currents produce a weaker opposing magnetic force, so the magnet falls faster in aluminium (2.1 s) than in copper (4.3 s) [1].

Systematic error: One source is the magnet touching the inside wall of the tube, causing friction that adds a non-electromagnetic braking force and artificially increases fall times, particularly for narrower clearances. This could be reduced by using a tube with a larger internal diameter (greater clearance) or by verifying the magnet does not contact the wall using a transparent acrylic tube [1].

Marking criteria summary (8 marks): 1 = Tube 1 correctly explained (no conductor, no eddy currents); 1 = Faraday’s Law applied to Tube 2 (emf due to changing flux); 1 = Lenz’s Law applied to Tube 2 (opposing upward force); 1 = Tube 2 copper conductivity link (low resistivity, large current, strong force); 1 = Tube 3 correctly explained (open seam breaks loop, no closed circuit); 1 = quantitative comparison of copper and aluminium (different resistivities → different current magnitudes); 1 = conclusion from comparison (higher resistivity → weaker braking → shorter fall time); 1 = valid systematic error named and a feasible improvement proposed.

Q2 — Sample Band 6 response (7 marks), annotated

Hypothesis: If increasing the number of laminations in a transformer core reduces eddy current losses, then as the number of laminations increases from 1 to 32 (independent variable), the power lost in the core (measured as input power minus useful output power, or core temperature rise) will decrease (dependent variable) [1].

Variables: IV = number of laminations (1, 4, 8, 16, 32). DV = power lost in core = input power − (Vs × Is), or maximum core surface temperature rise above ambient after 5 min of operation. Controlled variables: total core cross-sectional area (same for all trials), AC frequency (50 Hz), primary voltage (6 V rms), number of turns on primary and secondary (200 each), ambient temperature, duration of each trial (5 min). [1 — at least two controlled variables]

Procedure: (1) Set up the transformer with 200-turn primary and secondary coils wound on a single-lamination solid iron core. Connect the primary to the signal generator (50 Hz, 6 V rms); connect the secondary to a fixed load resistor (100 Ω) and measure secondary voltage and current with the multimeter. Use the power meter on the primary side to record input power. Record core surface temperature with the IR thermometer after 5 min. (2) Replace the core with a 4-lamination assembly of the same total cross-section area. Repeat all measurements under identical conditions. (3) Repeat for 8, 16, and 32 laminations. (4) Calculate power loss for each trial: Ploss = Pin − (Vs × Is). Plot Ploss and core temperature vs number of laminations. [1 — four clear steps with measurement method]

Falsification: If increasing the number of laminations produces no significant decrease in measured core power loss or temperature rise (within measurement uncertainty), the hypothesis would be falsified — there is no evidence that lamination number affects eddy current losses in this configuration [1].

Limitations: (1) Contact resistance between individual laminations varies depending on assembly pressure, introducing uncontrolled variation in eddy current paths [1]. (2) The IR thermometer measures surface temperature, which may differ from the internal core temperature where most eddy current heating occurs, leading to underestimation of losses at high lamination counts [1].

Improvement: Repeat each lamination configuration three times and average the results to improve reliability; use a calibrated calorimeter to measure heat produced in the core directly, rather than relying on surface temperature [1].

Marking criteria summary (7 marks): 1 = testable hypothesis naming IV, DV, and predicted direction; 1 = four numbered steps with a clear method for measuring power loss or temperature; 1 = correct identification of at least two controlled variables; 1 = falsification condition clearly stated; 1 = first valid limitation; 1 = second valid limitation; 1 = one specific, practical improvement to reliability.