Generated ERQ

✓ passed D.2 Electric and magnetic fields × B.5 Current and circuits 12 marks HL 3 passes 134.17s $0.8349
## ERQ · 12 marks · Topics: D.2 Electric and magnetic fields + B.5 Current and circuits · Archetype: data_response **Integration:** primary=D.2 Electric and magnetic fields, secondary=B.5 Current and circuits (strength: supporting) **Stem.** A Wien velocity selector is used at the entrance of an ion-beam line to select singly-ionised neon-20 ions (mass 3.32 × 10⁻²⁶ kg, charge +1.60 × 10⁻¹⁹ C). Ions enter horizontally through a slit and pass between two parallel plates separated by d = 12.0 mm. A uniform magnetic field of magnitude B = 42.0 mT is directed horizontally and perpendicular to the ion velocity. The electric field between the plates is produced by a high-voltage supply of emf ε = 1.50 kV and internal resistance r = 250 kΩ, connected to the plates through a series protection resistor R = 750 kΩ. Only ions that travel in a straight line through the 80 mm long region reach the exit slit. A small leakage current flows continuously between the plates. The table below shows the relative number of ions detected at the exit slit when the plate voltage is varied: | Plate voltage V_p / V | 480 | 520 | 560 | 600 | 640 | 680 | 720 | |---|---|---|---|---|---|---|---| | Relative count (arb.) | 0.05 | 0.18 | 0.62 | 1.00 | 0.65 | 0.20 | 0.04 | ### Part (a) State [2 marks] · AO1 · Topic: D.2 Electric and magnetic fields State the direction of the electric force on a positive ion and the direction of the magnetic force on the same ion (relative to each other) such that the ion passes through the selector undeflected. Justify each direction with one sentence. ### Part (b)(i) Show that [3 marks] · AO2 · Topic: D.2 Electric and magnetic fields The plate voltage that produces the peak count is V_p = 600 V. Show that the selected ion speed is approximately 1.19 × 10⁶ m s⁻¹. ### Part (b)(ii) Determine [3 marks] · AO2 · Topic: D.2 Electric and magnetic fields Using the Show-that speed v = 1.19 × 10⁶ m s⁻¹, determine the kinetic energy of the selected ions in eV, and comment on whether the table is consistent with a beam whose mean energy lies near this value. ### Part (c) Determine [2 marks] · AO3 · Topic: D.2 Electric and magnetic fields Using the data in the table, determine the approximate fractional width Δv/v of the selected ion speeds (use full-width-at-half-maximum of the count distribution). ### Part (d) Explain [2 marks] · AO3 · Topic: D.2 Electric and magnetic fields + B.5 Current and circuits · ASSUMPTIONS DISCRIMINATOR The designer assumes the plate voltage exactly equals the supply emf ε = 1.50 kV when the dial is set to that value. Explain, using the circuit data, why the actual plate voltage may differ from the dial setting and how this would shift the peak of the count distribution in the table. --- ## Mark Scheme ### Part (a) [2 marks] - M1: Electric force and magnetic force are **antiparallel** (opposite directions, perpendicular to v) ✓ with justification that the net force must be zero for undeflected motion [no ECF] - M2: Correctly identifies that, for a given B direction perpendicular to v, the magnetic force qv×B is fixed in direction, **therefore** E must be oriented so that qE points opposite to qv×B (e.g. if B is into the page and v is right, qv×B is up, so E must point downward on the positive ion, i.e. upper plate positive) [no ECF] ### Part (b)(i) [3 marks] — Show that - M1: States balance condition qE = qvB, hence v = E/B, with E = V_p/d [no ECF] - M2: Substitutes E = 600 / 0.0120 = 5.00 × 10⁴ V m⁻¹ [no ECF] - M3: v = 5.00 × 10⁴ / 0.0420 = 1.190 × 10⁶ m s⁻¹ ✓ (target given to 3 sf; student-derived 1.18–1.20 × 10⁶ acceptable) [no ECF] ### Part (b)(ii) [3 marks] — Determine - M1: KE = ½mv² with substitution: ½ × (3.32 × 10⁻²⁶) × (1.19 × 10⁶)² = 2.35 × 10⁻¹⁴ J [no ECF; uses Show-that value per §4.5] - M2: Converts: KE = 2.35 × 10⁻¹⁴ / 1.60 × 10⁻¹⁹ ≈ 1.47 × 10⁵ eV ≈ 147 keV [ECF from M1] - M3: Comment tied to data: the peak count at V_p = 600 V corresponds to ions selected at this energy; the symmetric fall-off about V_p = 600 V in the table is consistent with a mean ion energy ≈ 147 keV [ECF from M2] ### Part (c) [2 marks] — Determine - M1: Identifies half-maximum count = 0.50; from table, this occurs between V_p = 520 V (0.18) and 560 V (0.62) on the low side, and between 640 V (0.65) and 680 V (0.20) on the high side, giving FWHM in V_p ≈ 640 − 555 ≈ 85 V (accept 80–95 V); since v ∝ V_p, Δv/v ≈ ΔV_p/V_p [no ECF] - M2: Δv/v ≈ 85 / 600 ≈ 0.14 (accept 0.13–0.16) [ECF from M1] ### Part (d) [2 marks] — Explain (causal chain per §4.4.1) - M1: A leakage current I flows through the series resistor R and internal resistance r, **therefore** the voltage across the plates V_p = ε − I(R + r) is less than the emf, **so** the dial reading systematically overestimates the actual plate voltage [no ECF] - M2: A reduced V_p means E = V_p/d is smaller than designed, **therefore** the balance speed v = E/B is lower than expected, **so** the measured peak of the count distribution shifts to a lower V_p (or, equivalently, the dial value at peak overstates the true selection voltage) [no ECF] ### Marker notes - Alternative for (a): a labelled diagram showing E, B, v, qE and qv×B with correct relative orientations earns both marks. - Alternative for (b)(i): using E = vB rearranged with V_p = vBd → 1.19 × 10⁶ × 0.0420 × 0.0120 = 600 V is acceptable working. - Alternative for (c): Gaussian-fit FWHM estimate giving 70–100 V on V_p axis acceptable; Δv/v in range 0.12–0.17 accepted. - (d) accept any one of the following causal chains for M1 in place of the leakage-current version: (i) **ripple variant** — supply ripple causes V_p to oscillate about the nominal value, **therefore** the instantaneous E fluctuates, **so** only ions arriving during the phase matching qE = qvB pass, broadening and shifting the peak; (ii) **fringing variant** — at the edges of finite plates E is non-uniform and weaker than V_p/d, **therefore** the effective field experienced by ions is below the design value, **so** the peak in the table shifts to a higher V_p than predicted by the uniform-field model. Each variant must show two explicit "therefore/so" links to score both marks. - Show-that target in (b)(i): v = 1.19 × 10⁶ m s⁻¹ given to 3 sf; subsequent parts use this value per §4.5.