Physics • Year 12 • Module 6 • Lesson 8
Forces Between Parallel Conductors
Apply the force-per-unit-length formula and attraction/repulsion rule to real data, multi-step calculations, and the definition of the ampere.
1. Interpret experimental data — force between transmission-line conductors
A student measures the force per unit length between pairs of overhead transmission-line conductors at a substation. The table records the currents and separation for five scenarios. 8 marks
| Scenario | I1 (A) | I2 (A) | Separation d (m) | F/l (N/m) | Attract or repel? |
|---|---|---|---|---|---|
| A — same direction | 100 | 100 | 0.50 | ||
| B — same direction | 100 | 200 | 0.50 | ||
| C — same direction | 100 | 100 | 0.25 | ||
| D — opposite directions | 100 | 100 | 0.50 | ||
| E — same direction | 200 | 200 | 1.00 |
Use μ0 = 4π × 10−7 T m A−1. Illustrative scenario.
1.1 Calculate F/l for each scenario and state whether the force is attractive or repulsive. Show your working for Scenario A. 6 marks (1 per row)
1.2 Compare Scenarios A and C. Identify which variable changed and use the formula to explain why the force doubled. 2 marks
2. Interpret graph — force per unit length vs. separation
The graph below shows the force per unit length between two parallel wires each carrying I = 50 A, as the separation d is varied. 7 marks
Figure 2.1. Force per unit length between two parallel wires each carrying I = 50 A, plotted against separation d. Both currents in the same direction. Calculated from F/l = μ0I2/(2πd) with μ0 = 4π × 10−7 T m A−1.
2.1 Describe the relationship between F/l and d as shown in the graph. State the mathematical form of this relationship. 2 marks
2.2 Using the graph, estimate the force per unit length at d = 0.20 m and verify this by calculation. Show your working. 2 marks
2.3 The graph would look different if the two currents were in opposite directions. Describe what would change in the graph and what would stay the same. Explain your answer using the formula. 3 marks
3. Compare two conductor configurations across five features
Complete the two-column table below. For each feature, write a concise response that contrasts the same-direction and opposite-direction configurations of two parallel wires each carrying 10 A, separated by 5.0 cm. 10 marks (1 per cell)
| Feature | Same-direction currents | Opposite-direction currents |
|---|---|---|
| Force direction (attract/repel) | ||
| Magnitude of F/l (N/m) | ||
| Field direction at wire 2 due to wire 1 | ||
| Effect of doubling both currents on F/l | ||
| Real-world analogy or application |
4. Predict and justify — the Sydney–Newcastle high-voltage transmission line
Two parallel high-voltage AC transmission conductors on the eastern seaboard run side by side, separated by 1.2 m. During peak demand, each conductor carries 500 A. At any instant when the currents are in the same direction, the conductors experience a mutual force.
5 marks
4.1 Calculate the force per unit length between the conductors. State whether the force is attractive or repulsive. Show full working. 3 marks
4.2 In AC circuits the current direction reverses 100 times per second (at 50 Hz, both conductors reverse simultaneously). Predict whether the force direction between the conductors would change as the current reverses. Justify your answer with reference to the force formula. 2 marks
5. Case study — defining the ampere
Before 2019, the SI definition of the ampere was based on the force between two parallel conductors. The National Measurement Institute Australia (NMI) maintained current standards using the classical definition: “One ampere is the constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, placed one metre apart in vacuum, would produce between these conductors a force equal to 2 × 10−7 newtons per metre of length.”
In 2019, the SI was redefined so that the ampere is now defined by fixing the elementary charge e = 1.602 176 634 × 10−19 C exactly. 5 marks
5.1 Using the classical definition, verify that substituting I1 = I2 = 1 A and d = 1 m into F/l = μ0I1I2 / (2πd) gives exactly 2 × 10−7 N/m. Show your substitution clearly. 2 marks
5.2 Explain one advantage of defining the ampere using the force between parallel conductors and one limitation that motivated the 2019 redefinition. 3 marks
Q1.1 — Data table calculations
Formula: F/l = μ0I1I2 / (2πd) with μ0 = 4π × 10−7.
Scenario A (worked): F/l = (4π×10−7 × 100 × 100) / (2π × 0.50) = (4π×10−7 × 104) / (2π × 0.50) = (4 × 10−3) / 1 = 4.0 × 10−3 N/m, Attractive.
Scenario B: F/l = (4π×10−7 × 100 × 200) / (2π × 0.50) = 8.0 × 10−3 N/m, Attractive.
Scenario C: F/l = (4π×10−7 × 100 × 100) / (2π × 0.25) = 8.0 × 10−3 N/m, Attractive.
Scenario D: Same magnitude as A: 4.0 × 10−3 N/m, Repulsive (opposite-direction currents).
Scenario E: F/l = (4π×10−7 × 200 × 200) / (2π × 1.00) = 8.0 × 10−3 N/m, Attractive.
Q1.2 — Comparing Scenarios A and C
The separation d was halved from 0.50 m (A) to 0.25 m (C) [1 mark]. Since F/l ∝ 1/d, halving d doubles the force per unit length from 4.0 × 10−3 N/m to 8.0 × 10−3 N/m [1 mark].
Q2.1 — F/l vs d relationship
As d increases, F/l decreases in a hyperbolic (inverse) fashion. The mathematical relationship is F/l ∝ 1/d; it is an inverse proportionality [2 marks: 1 for hyperbolic/decreasing description, 1 for stating F/l ∝ 1/d].
Q2.2 — Estimate and verify at d = 0.20 m
Graph estimate: approximately 2.5 × 10−4 N/m [1 mark]. Calculation: F/l = (4π×10−7 × 50 × 50) / (2π × 0.20) = (4π×10−7 × 2500) / (0.40π) = (4 × 2500 × 10−7) / 0.40 = 2.5 × 10−4 N/m [1 mark]. Graph and calculation agree.
Q2.3 — Opposite-direction currents
The shape and magnitude of the graph would remain identical because the formula F/l = μ0I1I2 / (2πd) depends on the products of the magnitudes of the currents, which are unchanged [1 mark]. What would change is the direction of the force: the force would be repulsive rather than attractive [1 mark]. The graph would have the same hyperbolic curve but would represent a repulsive force per unit length [1 mark].
Q3 — Compare and contrast table
Force direction: Same = Attractive (wires drawn together). Opposite = Repulsive (wires pushed apart).
Magnitude F/l: Both configurations: F/l = (4π×10−7 × 10 × 10) / (2π × 0.05) = 4.0 × 10−4 N/m (same in both cases).
Field direction at wire 2 due to wire 1: Same direction = field at wire 2 is perpendicular to the plane of the wires, directed into the page (if currents flow upward, wire 2 is to the right). Opposite direction = same field direction from wire 1 (wire 1’s field does not depend on wire 2’s current); the field is still into the page at wire 2.
Effect of doubling both currents: Both configurations: F/l increases by a factor of 4 (F/l ∝ I1I2; doubling both multiplies by 2 × 2 = 4).
Real-world analogy/application: Same = Two parallel power transmission cables on the same phase attract; DC bus bars in a switchboard attract when current-loaded. Opposite = Two-wire power cables (live and neutral running close together in opposite directions) repel slightly; rails of a linear induction motor may repel.
Q4.1 — Transmission-line calculation
F/l = μ0I1I2 / (2πd) = (4π×10−7 × 500 × 500) / (2π × 1.2) [1 mark for correct substitution] = (4π×10−7 × 2.5×105) / (2.4π) = (4 × 2.5 × 10−2) / 2.4 = 0.10 / 2.4 = 4.17 × 10−2 N/m [1 mark for correct calculation]. The force is attractive (same-direction currents) [1 mark].
Q4.2 — AC current reversal and force direction
When both currents reverse simultaneously they remain flowing in the same direction relative to each other [1 mark]. In the formula F/l = μ0I1I2 / (2πd), both I1 and I2 change sign, but their product I1I2 remains positive. Therefore the force remains attractive and its magnitude is unchanged [1 mark].
Q5.1 — Verification of ampere definition
Substituting I1 = I2 = 1 A and d = 1 m: F/l = (4π×10−7 × 1 × 1) / (2π × 1) = 4π×10−7 / 2π = 2 × 10−7 N/m. This exactly matches the defined value, confirming the consistency of the definition [2 marks: 1 for correct substitution, 1 for arriving at 2 × 10−7 N/m and stating it matches].
Q5.2 — Advantage and limitation
Advantage (1 mark): The force between parallel conductors can be measured with extremely high precision using a mechanical balance, providing a direct and reproducible link between the unit of current and measurable mechanical quantities.
Limitation (2 marks): Practical implementation requires near-ideal conditions (infinitely long, thin wires, perfect vacuum) that cannot be exactly achieved [1 mark]. Measurement uncertainty was around 1 part in 107, which is insufficient for modern precision metrology; the 2019 redefinition fixes the elementary charge exactly, giving a more fundamental and universally reproducible standard independent of macroscopic measurements [1 mark].