HSC Exam Practice

Sample response. The force per unit length is the mutual force between two infinitely long parallel current-carrying conductors divided by their length. It is given by F/l = μ₀I₁I₂ / (2πd), where F/l is the force per unit length in N/m; μ₀ is the permeability of free space (4π × 10⁻⁷ T m A⁻¹); I₁ and I₂ are the currents in each wire in amperes; and d is the perpendicular separation between the wires in metres.

Marking notes. 1 mark for a correct definition of F/l (force divided by length, or force per unit length between two parallel conductors). 1 mark for stating the correct formula F/l = μ₀I₁I₂ / (2πd). 1 mark for correctly defining all four symbols with correct SI units.

Sample response. Wire 1 carries a current and therefore produces a magnetic field circling around it; by the right-hand grip rule, if the current flows upward, the field at a point to the right of wire 1 (i.e. at the location of wire 2) is directed into the page [1]. Wire 2 also carries current in the same upward direction and is in this field, so by the motor effect (F = BIl) it experiences a force [1]. Applying the right-hand grip rule to the combination of current direction (upward) and field direction (into page), the force on wire 2 is directed toward wire 1, i.e., attractive [1]. By Newton’s third law, wire 1 simultaneously experiences an equal and opposite force also directed inward toward wire 2 [1].

Marking notes. 1 mark for applying the right-hand grip rule to state that wire 1 creates a magnetic field at wire 2’s location. 1 mark for stating the field direction at wire 2 (into the page). 1 mark for applying the motor effect (F = BIl) to wire 2 in that field. 1 mark for correctly deducing the force is directed toward wire 1 (attraction) using the right-hand grip rule or equivalent reasoning.

Sample response. 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 a vacuum, would produce between those conductors a force equal to 2 × 10⁻⁷ newtons per metre of length.

Marking notes. 1 mark for stating the current (1 A in each conductor). 1 mark for stating the separation (1 m, in vacuum / free space). 1 mark for stating the resulting force per unit length (2 × 10⁻⁷ N/m). Accept paraphrase that captures all three quantities.

Sample response. When currents flow in the same direction the force is attractive (wires drawn together); when in opposite directions the force is repulsive (wires pushed apart). Reversing only one current changes the relative current direction from same to opposite, so the force changes from attractive to repulsive — the direction of the force on each wire reverses. The magnitude of the force per unit length is unchanged because the formula F/l = μ₀I₁I₂/(2πd) depends on the magnitudes of the currents, which are unaltered by reversing direction.

Marking notes. 1 mark for correctly distinguishing same-direction (attract) from opposite-direction (repel). 1 mark for stating that reversing one current changes attraction to repulsion. 1 mark for stating that the magnitude of F/l is unchanged (since it depends on |I₁||I₂|).

Sample response. F/l = μ₀I₁I₂ / (2πd) = (4π×10⁻⁷ × 20 × 15) / (2π × 0.060) = (4π×10⁻⁷ × 300) / (0.12π) = (1200 × 10⁻⁷) / 0.12 = 1.0 × 10⁻³ N/m. The force is attractive because the currents flow in the same direction.

Marking notes. 1 mark for correctly substituting all values (including converting 6.0 cm to 0.060 m). 1 mark for a correct calculation arriving at 1.0 × 10⁻³ N/m. 1 mark for stating the force is attractive.

Sample response. In the formula F/l = μ₀I₁I₂/(2πd), the force depends on the product I₁I₂. If both I₁ and I₂ are reversed, their product (negative × negative) remains positive and equal in value to the original product, so the magnitude of F/l is unchanged. Furthermore, both wires are still flowing in the same direction relative to each other, so the force remains attractive. Newton’s third law guarantees that forces on each wire are equal and opposite regardless of how the currents are directed.

Marking notes. 1 mark for noting that I₁I₂ (product of magnitudes) is unchanged when both currents reverse. 1 mark for explaining that both currents are still in the same relative direction (both reversed = same relative sense), so attraction vs repulsion does not change. 1 mark for correctly invoking Newton’s third law to explain that each wire’s force remains equal and opposite to the other.

Sample response (a). F/l = (4π×10⁻⁷ × 5.0 × 5.0) / (2π × 0.25) = (4π×10⁻⁷ × 25) / (0.50π) = (100 × 10⁻⁷) / 0.50 = 2.0 × 10⁻⁵ N/m. This matches the graph value for Trial 1. [2 marks: 1 substitution, 1 correct answer]

Sample response (b). Between Trial 1 and Trial 3, I₁ changed from 5.0 A to 10 A and I₂ changed from 5.0 A to 10 A [1 mark]. Since F/l ∝ I₁×I₂, doubling both currents multiplies the product by 4 (2 × 2 = 4) [1 mark]. Therefore F/l increases from 2.0 × 10⁻⁵ N/m to 8.0 × 10⁻⁵ N/m, a factor of 4, consistent with the bar chart [1 mark].

Sample response (c). The student’s conclusion is partially correct but incomplete [1 mark]. The formula F/l = μ₀I₁I₂/(2πd) depends only on the magnitudes of the currents; reversing one current does not change the magnitudes, so the magnitude of F/l is indeed unchanged [1 mark]. However, what does change is the direction (nature) of the force: Trial 1 (same-direction currents) produces an attractive force, pulling the wires together, whereas Trial 4 (opposite-direction currents) produces a repulsive force, pushing the wires apart. The graph cannot show this difference because it only displays the magnitude [1 mark].

Sample response. The force between two parallel current-carrying conductors arises from a two-step electromagnetic process. First, each current-carrying wire generates a magnetic field around itself, described by B = μ₀I/(2πr); the direction of this field at any point is determined by the right-hand grip rule — wrapping the right hand around the wire with the thumb pointing in the direction of conventional current gives the sense of the field circles. Second, the second wire, carrying its own current and lying within this field, experiences a motor-effect force F = BIl (for a wire perpendicular to B), directed either toward or away from the first wire depending on the relative sense of the currents. When both currents are in the same direction, the field from wire 1 at wire 2’s location is directed so that the motor-effect force on wire 2 points toward wire 1 (attraction); when they are in opposite directions, the field direction at wire 2 reverses, so the force reverses (repulsion). Newton’s third law guarantees that wire 1 experiences an equal and opposite force due to wire 2, so the mutual force is symmetric in magnitude. Combining the two steps gives F/l = μ₀I₁I₂/(2πd). This force relationship served as the basis for the pre-2019 SI definition of the ampere. Its primary strength was precision: by setting I₁ = I₂ = 1 A and d = 1 m, the formula gives F/l = 4π×10⁻⁷/(2π) = 2×10⁻⁷ N/m exactly, providing a macroscopic, reproducible and directly measurable link between mechanical force and electrical current without reference to atomic constants. National metrology institutes could, in principle, realise the ampere by measuring forces with a mechanical balance. However, a significant limitation was that truly “infinite, negligibly thin” wires cannot be constructed; practical implementations used finite-length, finite-diameter conductors, and the separation had to be controlled to sub-millimetre precision. These constraints limited achievable uncertainty to roughly 1 part in 10⁷. The 2019 redefinition, which fixes the elementary charge e exactly, overcomes this by making the ampere independent of macroscopic geometrical measurements and reproducible in any laboratory using quantum electrical standards such as the quantum Hall resistance and Josephson voltage standards, achieving far smaller uncertainties. In summary, the force between parallel conductors provides an elegant and physically transparent foundation for the ampere, grounded in the right-hand grip rule, the motor effect, and Newton’s third law, but its realisation in the classical definition was ultimately limited by practical measurement constraints.

Marking criteria (7 marks). [1] Correctly explains that wire 1’s current produces a magnetic field at wire 2’s location, with reference to the right-hand grip rule and the formula B = μ₀I/(2πr). [1] Correctly explains that the motor effect (F = BIl) acts on wire 2 in this field, producing a force toward or away from wire 1 depending on relative current directions. [1] Correctly applies Newton’s third law to explain the symmetry of the force (each wire experiences an equal and opposite force). [1] Correctly states the formula F/l = μ₀I₁I₂/(2πd) and links it to the definition of the ampere, including the specific numerical example (I = 1 A, d = 1 m, F/l = 2 × 10⁻⁷ N/m). [1] Identifies a genuine strength of the parallel-conductor definition (e.g., macroscopically measurable, precise, reproducible from mechanical force). [1] Identifies a genuine limitation of the parallel-conductor definition with physical justification (e.g., finite wire length/diameter, geometric alignment uncertainty, limited achievable precision of ~10⁻⁷). [1] Reaches a clear, integrated evaluative judgement connecting the physical basis (grip rule + motor effect) to the suitability of the definition, going beyond mere description to assess overall adequacy.