HSC Exam Practice

Sample response. Magnetic flux (Φ) is the total magnetic field passing through a given area, defined as Φ = BA cos θ, where B is the magnetic field strength (T), A is the area (m²), and θ is the angle between the magnetic field direction and the normal to the area. Flux is measured in webers (Wb). Flux through a fixed loop is zero when the magnetic field is parallel to the plane of the loop (θ = 90° between B and the area normal).

Marking notes. 1 mark for correct definition of magnetic flux (field through an area); 1 mark for stating the formula Φ = BA cos θ with correct identification of θ; 1 mark for correctly stating the condition for zero flux (θ = 90° between B and the normal, or equivalently B parallel to the plane).

Sample response. Lenz’s Law states that the direction of an induced current is always such that its magnetic effect opposes the change in flux that produced it. This is consistent with conservation of energy because if the induced current aided the flux change instead of opposing it, the induced current would increase the flux, which would induce a larger current, which would increase the flux further — creating a perpetual increase in energy with no input. This is impossible. The opposing direction ensures that energy must be supplied (e.g. by an external force pushing a magnet) to maintain the flux change, and this input work appears as electrical energy in the induced current.

Marking notes. 1 mark for a correct statement of Lenz’s Law (induced current opposes the flux change); 1 mark for identifying the conservation of energy link (work must be done to maintain the flux change); 1 mark for explaining why an aiding current would violate energy conservation (perpetual amplification argument or equivalent).

Sample response. r = m_ev/(eB) = (9.11 × 10⁻³¹ × 5.0 × 10⁵) ÷ (1.60 × 10⁻¹⁹ × 0.050) = 4.555 × 10⁻²⁵ ÷ 8.0 × 10⁻²¹ ≈ 5.7 × 10⁻⁵ m (0.057 mm).

Marking notes. 1 mark for correct formula r = mv/(qB); 1 mark for correct substitution of all values; 1 mark for correct answer with appropriate units and significant figures (accept 5.6–5.7 × 10⁻⁵ m).

Sample response. ε_back = V − IR = 120 − (6.0)(1.5) = 120 − 9.0 = 111 V. The back emf arises because the rotating armature coil sweeps through the stator magnetic field, changing the flux linkage through the coil. By Faraday’s Law, this changing flux induces an emf in the coil. By Lenz’s Law, this induced emf opposes the supply voltage, limiting the current and accounting for the energy converted to mechanical output.

Marking notes. 1 mark for correct formula and substitution (ε_back = V − IR); 1 mark for correct numerical answer (111 V); 1 mark for explaining the physical origin via Faraday’s Law (rotating coil, changing flux, induced emf opposing supply).

Sample response. V_s/V_p = N_s/N_p, so V_s = V_p × (N_s/N_p) = 240 × (50/500) = 240 × 0.10 = 24 V. Since V_s < V_p, this is a step-down transformer.

Marking notes. 1 mark for stating the transformer voltage ratio formula correctly; 1 mark for correct calculated secondary voltage (24 V); 1 mark for correctly identifying it as a step-down transformer with a brief justification (V_s < V_p or N_s < N_p).

Sample response. The rotor of an AC induction motor always runs below synchronous speed because induced rotor currents (and hence torque) depend on a relative motion between the rotor and the rotating stator magnetic field — this relative motion is called slip. If the rotor reached synchronous speed, there would be no relative motion between them, no changing flux through the rotor conductors, no induced rotor current by Faraday’s Law, and therefore no magnetic force or torque. Without torque, friction and mechanical load would slow the rotor below synchronous speed, restoring slip and therefore rotor current. The motor thus always operates with a small slip, typically 1–5%.

Marking notes. 1 mark for correctly identifying that slip (speed difference) is required for flux change and induced rotor current; 1 mark for explaining what would happen if speeds were equal (no flux change → no induced current → no torque); 1 mark for completing the logical argument (motor would slow, slip restored) or for correctly naming slip and its typical magnitude.

Sample response (a). For 100 kV: I = P/V = 500 × 10³ ÷ 100 × 10³ = 5 A. P_loss = I²R = 25 × 5.0 = 125 W = 0.125 kW; % loss = 0.125/500 × 100% = 0.025%. For 500 kV: I = 1 A; P_loss = 1² × 5 = 5 W = 0.005 kW; % loss = 0.001%.

Marking notes (a). 1 mark for correct P_loss and % for 100 kV row; 1 mark for correct P_loss and % for 500 kV row; 1 mark for showing full working for one row (I = P/V, then P_loss = I²R, then % = P_loss/P × 100%).

Sample response (b). The data show that when transmission voltage is doubled, the line current is halved (I ∝ 1/V at constant P), and power loss decreases to one quarter (P_loss = I²R ∝ 1/V²). General rule: “If transmission voltage is doubled, power loss is reduced to one quarter (decreases by a factor of four).”

Marking notes (b). 1 mark for correctly identifying the inverse-square relationship (P_loss ∝ 1/V² or equivalent); 1 mark for correctly stating the doubling rule (loss reduces to one quarter).

Sample response (c). The claim is partially correct but overstated. Higher voltages do reduce I²R losses, but losses can never be completely eliminated — at 500 kV the loss is 0.001% rather than zero. Furthermore, transmitting at very high voltages introduces significant practical limitations: electrical insulation requirements increase dramatically, tower heights and electrical clearances must be larger (more costly infrastructure), safety risks of corona discharge and electromagnetic interference increase, and the transformers themselves have copper and iron losses. An engineer must balance lower line losses against higher capital costs; beyond a certain voltage, the marginal gain in efficiency does not justify the additional infrastructure expense.

Marking notes (c). 1 mark for correctly explaining that losses are reduced but cannot be eliminated (any finite I in a resistive line produces some I²R loss); 1 mark for identifying at least one valid practical constraint of very high voltage (insulation cost, safety clearances, corona discharge, transformer losses, infrastructure expense).

Sample response (a). In the velocity selector, an ion travels undeflected when the electric force equals the magnetic force: qE = qvB₁, so v = E/B₁ = 4.0 × 10⁴ ÷ 0.020 = 2.0 × 10⁶ m/s. [1 formula, 1 answer]

Sample response (b). Using r = mv/(qB₂), rearranging: m = qB₂r/v = (1.60 × 10⁻¹⁹ × 0.15 × 0.18) ÷ (2.0 × 10⁶). Numerator = 4.32 × 10⁻²¹. m = 4.32 × 10⁻²¹ ÷ 2.0 × 10⁶ = 2.16 × 10⁻²⁷ kg (approximately 1.3 atomic mass units; likely a light isotope). [1 formula, 1 substitution, 1 answer]

Marking notes. (a) 1 mark for v = E/B; 1 mark for correct answer 2.0 × 10⁶ m/s. (b) 1 mark for rearranging r = mv/(qB) to m = qBr/v; 1 mark for correct substitution; 1 mark for correct answer approximately 2.2 × 10⁻²⁷ kg (accept 2.1–2.2 × 10⁻²⁷ kg).

Sample response. Electromagnetic induction, described by Faraday’s Law (ε = −NΔΦ/Δt), is the foundational principle enabling modern electricity generation and distribution. In an AC generator, a coil rotates in a magnetic field provided by permanent magnets or an electromagnet. This rotation continuously changes the flux through the coil at a rate proportional to angular frequency (ω), producing an alternating emf with peak value ε₀ = NBAω. Lenz’s Law ensures that the induced current in the coil creates a magnetic force opposing the rotation, so mechanical energy from a turbine (driven by steam, wind, or water) is converted to electrical energy while energy is conserved. At Loy Yang A Power Station in Victoria, steam turbines drive 500 MW generators at 50 Hz, producing AC at around 22 kV at the generator terminals. Transmitting 500 MW at 22 kV would give I = P/V = 22 700 A; with line resistance of just 1 Ω, P_loss = I²R ≈ 520 MW — exceeding the transmitted power. Step-up transformers at the power station raise voltage to 500 kV (turns ratio N_s/N_p ≈ 23:1), reducing line current to 1 000 A. At 1 Ω resistance, losses fall to just 1 MW (0.2% of 500 MW). These transformers operate on Faraday’s Law: the alternating primary current creates a changing flux in the iron core, which induces an emf in the secondary coil in proportion to its turns. Lenz’s Law governs that the secondary induced emf will always act to oppose the changing primary flux, which is what drives the secondary current. At the receiving end, step-down transformers progressively reduce voltage (500 kV → 33 kV → 11 kV → 240 V) for distribution. The trade-off is that higher voltages require more expensive insulation, larger tower clearances (1.5–5 m for 500 kV), and greater safety infrastructure. Ultra-high voltage systems (>1 000 kV DC) are used in China for very long distances where the I²R savings justify the cost; in Australia the national grid uses 500 kV AC as the economic optimum. In summary, Faraday’s Law enables both generation (rotating coil) and transformation (coupled coils), while Lenz’s Law ensures energy conservation is maintained throughout, and the P_loss = I²R relationship drives the engineering choice to transmit at high voltage with transformers at each end.

Marking criteria (8 marks): 1 = correctly explains how Faraday’s Law (ε = −NΔΦ/Δt) enables AC generation in a rotating coil, with reference to changing flux; 1 = correctly explains the role of Lenz’s Law in ensuring energy conservation during induction (opposing force requires mechanical input); 1 = correctly explains how transformers use Faraday’s Law to step voltage up or down via turns ratio, with formula V_p/V_s = N_p/N_s; 1 = uses P_loss = I²R (NOT P = V²/R) correctly to explain why high voltage reduces losses; 1 = provides a valid quantitative comparison showing that lower line current at higher voltage dramatically reduces I²R losses; 1 = names a specific Australian example (Loy Yang, Eraring, Snowy Hydro, or any named NSW/Vic/Qld power station or grid component) and applies it correctly; 1 = discusses at least one valid trade-off or limitation of very high voltage transmission (cost, safety, insulation, corona discharge, transformer losses); 1 = reaches an explicit evaluative judgement integrating generation, transformation, transmission, and the voltage/cost trade-off into a coherent conclusion.