Physics • Year 12 • Module 6 • Lesson 8
Forces Between Parallel Conductors
Build HSC Band 5–6 extended-response technique on multi-step calculations, three-wire analysis, and evaluating the physical basis of the SI definition of current.
1. Multi-step calculation: the three-wire problem (Band 4–5)
8 marks Band 4–5
Scenario. Three long parallel wires, P, Q, and R, run side by side in a vertical plane. Their positions and currents are shown below. All separations are measured between adjacent wires.
- Wire P: carries 15 A upward
- Wire Q: carries 10 A upward; positioned 4.0 cm to the right of P
- Wire R: carries 8.0 A downward; positioned 4.0 cm to the right of Q (and 8.0 cm from P)
Use μ0 = 4π × 10−7 T m A−1.
(a) Calculate the force per unit length on wire Q due to wire P. State the direction of this force. 2 marks
(b) Calculate the force per unit length on wire Q due to wire R. State the direction of this force. 2 marks
(c) Determine the magnitude and direction of the net force per unit length on wire Q due to the combined effect of wires P and R. Show your vector addition. 2 marks
(d) Interpret the result. A student argues: “Because both P and R exert forces on Q, wire Q will always be in unstable equilibrium and will move.” Evaluate this claim and suggest one condition under which wire Q could be in equilibrium. 2 marks
2. Data + scenario: current balance at the National Measurement Institute (Band 5–6)
8 marks Band 5–6
Scenario. A current balance is a precision instrument used to measure electrical current by exploiting the force between two parallel conductors. One conductor is fixed; the other is suspended on a balance arm. The downward gravitational force is balanced against the upward magnetic force. The table below shows five measurements recorded at NMI Australia during calibration of a precision ammeter.
| Run | d (m) | Length l (m) | Measured force F (N) | Calculated I (A) |
|---|---|---|---|---|
| 1 | 0.010 | 0.200 | 1.60 × 10−4 | |
| 2 | 0.010 | 0.200 | 1.58 × 10−4 | |
| 3 | 0.010 | 0.200 | 1.61 × 10−4 | |
| 4 | 0.010 | 0.200 | 1.59 × 10−4 | |
| 5 | 0.010 | 0.200 | 1.60 × 10−4 |
Both conductors carry the same current I. Geometry: conductors are parallel, horizontal, separated by d = 0.010 m. Illustrative data based on current-balance methodology. Use μ0 = 4π × 10−7 T m A−1.
Q2. Analyse and evaluate the current-balance data above to determine the current passing through both conductors and assess the reliability of the measurement method. In your response you must:
- Rearrange F/l = μ0I2 / (2πd) to make I the subject, showing each algebraic step.
- Calculate I for each run from the measured force F (using F/l = F/length l with l = 0.200 m).
- Calculate the mean current and the range (max − min) of your five values as a measure of spread.
- Identify one systematic error and one random error that could affect the current-balance measurement, and explain how each would affect the calculated current.
- Evaluate whether this method is appropriate for defining the ampere, using evidence from your calculations and the spread of results.
3. Experimental design — investigating the force law (Band 5–6)
8 marks Band 5–6
Research question: How does the force per unit length between two parallel wires depend on the separation d, when both currents are held constant at 5.0 A?
Constraints: You have two straight conducting rods of length 0.50 m each, a variable DC power supply (0–20 A), a precision force sensor (measures up to 0.1 N, precision ±0.001 N), and the ability to adjust the separation from 1.0 cm to 20.0 cm in steps.
3.1 Identify the independent variable, dependent variable, and at least two controlled variables for this experiment. 2 marks
3.2 Describe the method you would use to verify whether the force law is F/l ∝ 1/d. State what graph you would plot to linearise the data and what the gradient of that graph would represent. 3 marks
3.3 State two limitations of this experimental design and suggest a specific improvement for each. 2 marks
3.4 State what result would falsify the inverse-proportionality hypothesis and describe how you would detect it from your graph. 1 mark
Q1(a) — Force on Q due to P
P and Q carry currents in the same direction (both upward): force is attractive (Q is pulled toward P, i.e., to the left).
F/l = μ0IPIQ / (2πd) = (4π×10−7 × 15 × 10) / (2π × 0.040)
= (4π×10−7 × 150) / (0.080π) = (600×10−7) / 0.080 = 7.5 × 10−4 N/m toward P (leftward).
Q1(b) — Force on Q due to R
Q carries 10 A upward; R carries 8.0 A downward: opposite directions — force is repulsive (Q is pushed away from R, i.e., to the left).
F/l = (4π×10−7 × 10 × 8.0) / (2π × 0.040) = (4π×10−7 × 80) / (0.080π) = (320×10−7) / 0.080 = 4.0 × 10−4 N/m away from R (leftward).
Q1(c) — Net force on Q
Both forces on Q act in the same direction (leftward, toward P). They therefore add:
Net F/l = 7.5 × 10−4 + 4.0 × 10−4 = 1.15 × 10−3 N/m, directed toward P (to the left).
Q1(d) — Evaluation of equilibrium claim
The student’s claim is partially correct: with these currents, wire Q experiences a net force and is not in equilibrium [1 mark]. However, the claim that Q will “always” move is incorrect — wire Q could be in equilibrium if the forces from P and R were equal in magnitude but opposite in direction. This could be achieved, for example, by adjusting the current in R so that it is upward (same as Q) and the magnitudes of the two attractive forces are equal, or by adjusting the separations and currents so that the net force is zero [1 mark].
Q2 — Current-balance analysis (8 marks)
Rearrangement (1 mark): F/l = μ0I2/(2πd) ⇒ I2 = 2πd(F/l)/μ0 ⇒ I = √[2πd(F/l)/μ0].
Calculations (1 mark for method, 1 mark for correct values):
For each run, F/l = F/0.200 m. Factor: 2πd/μ0 = 2π×0.010/(4π×10−7) = 0.020/(4×10−7) = 5.00×104.
Run 1: F/l = 1.60×10−4/0.200 = 8.00×10−4 N/m; I = √(5.00×104 × 8.00×10−4) = √40.0 = 6.32 A.
Run 2: F/l = 7.90×10−4; I = √(5.00×104×7.90×10−4) = √39.5 = 6.28 A.
Run 3: F/l = 8.05×10−4; I = √40.25 = 6.34 A.
Run 4: F/l = 7.95×10−4; I = √39.75 = 6.30 A.
Run 5: F/l = 8.00×10−4; I = √40.00 = 6.32 A.
Mean and range (1 mark): Mean I = (6.32 + 6.28 + 6.34 + 6.30 + 6.32) / 5 = 31.56/5 = 6.31 A. Range = 6.34 − 6.28 = 0.06 A.
Systematic error (1 mark): e.g. The wires are not truly infinitely long or have finite cross-section, so the theoretical formula slightly overestimates the field, causing I to be systematically underestimated. Or: residual magnetism in the apparatus produces an additive background force, biasing all measured F values upward and so I would be slightly overestimated.
Random error (1 mark): e.g. Mechanical vibration of the balance arm causes fluctuating force readings, leading to run-to-run variation in F and therefore in the calculated I. Or: temperature changes in the conductor alter its resistance and the actual current, even if the supply setting is fixed.
Evaluation (2 marks): The spread of 0.06 A on a mean of 6.31 A represents a relative uncertainty of about 1%, which was typical of current-balance precision [1 mark]. This level of precision was sufficient to define the pre-2019 ampere to about 1 part in 107 under ideal laboratory conditions, but it is ultimately limited by practical constraints (non-ideal wire geometry, background fields), which motivated the 2019 redefinition based on the fundamental constant e [1 mark].
Q3.1 — Variables
Independent variable: separation d between the two wires (varied from 1.0 cm to 20.0 cm). Dependent variable: force per unit length F/l (measured by dividing force sensor reading by effective conductor length 0.50 m). Controlled variables (any 2): current in each wire (held at 5.0 A using the power supply); wire length inside the sensing region (0.50 m); temperature of conductors; ambient magnetic field (test performed far from ferromagnetic materials or other current sources).
Q3.2 — Method and linearisation
Method (1 mark): Set both currents to 5.0 A. Record the force at each separation d (at least 8 values from 1.0 cm to 20.0 cm). Calculate F/l = Fmeasured/0.50 m at each d.
Graph to linearise (1 mark): Plot F/l on the y-axis versus 1/d on the x-axis. If F/l ∝ 1/d, the data should lie on a straight line through the origin.
Gradient (1 mark): gradient = μ0I1I2/(2π) = (4π×10−7 × 5.0 × 5.0)/(2π) = 5.0 × 10−6 N (this can be verified by multiplying the measured gradient by the known constant).
Q3.3 — Limitations and improvements
Limitation 1 (1 mark): The conducting rods have finite length (0.50 m), so end effects mean the force is not purely that of infinite wires. Improvement: use longer rods (e.g. 2.0 m) to minimise the proportional contribution of end effects.
Limitation 2 (1 mark): At very small separations (d < 2 cm), mechanical mounting becomes difficult and the wires may not remain perfectly parallel, introducing geometric error. Improvement: use a precision-machined jig with micrometer-controlled wire positioning to maintain parallelism across all separations.
Q3.4 — Falsification criterion
The inverse-proportionality hypothesis would be falsified if the F/l vs 1/d graph did not produce a straight line through the origin — for example, if it curved, had a non-zero y-intercept, or if the gradient changed with 1/d. Any systematic departure from linearity on the linearised graph would indicate that F/l does not follow a simple 1/d law.