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## ERQ · 13 marks · Topics: A.2 Forces and momentum + D.2 Electric and magnetic fields · Archetype: modeling_and_assumptions
**Integration:** primary=A.2 Forces and momentum, secondary=D.2 Electric and magnetic fields (strength: supporting)
**Stem.** A research group studies the motion of charged microspheres for a precision mass-measurement project. A single microsphere of mass m = 4.0 × 10⁻¹⁵ kg carries a net charge q = +1.6 × 10⁻¹⁷ C. The sphere is launched horizontally with speed v₀ = 12 m s⁻¹ into a sealed chamber. Within the chamber, a uniform magnetic field of magnitude B = 0.85 T points horizontally, perpendicular to the launch direction. The local gravitational field strength is g = 9.81 N kg⁻¹ and acts vertically downward. The chamber is approximately 1.0 m long along the initial direction of motion. Air at low pressure remains in the chamber; the team initially models the motion by neglecting air resistance.
### Part (a) State and Define [2 marks] · AO1 · Topic: A.2 + D.2
State the direction of the magnetic force on the microsphere at the instant of launch, and write the vector expression that defines the magnetic force on a charged particle in terms of its charge, velocity, and the magnetic field.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.2 + D.2
Calculate the magnitude of the net force on the microsphere at the instant it enters the magnetic field region, assuming the magnetic force acts vertically (either up or down) and combines with gravity along the same axis. Give your answer to two significant figures.
### Part (b)(ii) Show that [3 marks] · AO2 · Topic: A.2
The magnetic force is always perpendicular to the instantaneous velocity, so the magnetic force does no work and the speed of the particle changes only because of gravity (air resistance still neglected). Using Newton's second law in the form F = dp/dt, show that the vertical component of momentum p_y satisfies dp_y/dt = −mg + qv_x B, where v_x is the instantaneous horizontal component of velocity, and hence that the *initial* rate of change of vertical momentum has magnitude approximately 1.2 × 10⁻¹³ N (taking downward as negative).
### Part (c) Evaluate [3 marks] · AO3 · Topic: A.2 + D.2 · ASSUMPTIONS DISCRIMINATOR (model validity)
For a microsphere of this size, Stokes' drag gives a viscous force of order F_drag ≈ 6πηrv with η ≈ 1.8 × 10⁻⁵ Pa s and effective radius r ≈ 1.0 × 10⁻⁶ m. Evaluate the assumption that air resistance is negligible over the 1.0 m flight path, by comparing the magnitudes of the drag force at launch with the gravitational and magnetic contributions found in (b)(i), and discuss how the trajectory would be modified.
### Part (d) Suggest [2 marks] · AO3 · Topic: A.2 + D.2 · ASSUMPTIONS DISCRIMINATOR
Suggest one further physical effect — related either to the fields present or to momentum exchange — that has been neglected in this model, and explain its likely impact on the motion of the microsphere.
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## Mark Scheme
### Part (a) [2 marks] — State + Define
- M1: Direction stated as vertically upward OR vertically downward (consistent with a stated convention for B and v₀ being mutually perpendicular and horizontal) [no ECF]
- M2: F = qv × B (vector form) OR equivalent: F = qvB sinθ with θ defined as angle between v and B [no ECF]
### Part (b)(i) [3 marks] — Calculate
- M1: Gravitational force W = mg = (4.0 × 10⁻¹⁵)(9.81) = 3.9 × 10⁻¹⁴ N [no ECF]
- M2: Magnetic force F_B = qv₀B = (1.6 × 10⁻¹⁷)(12)(0.85) = 1.6 × 10⁻¹⁶ N [no ECF]
- M3: Net force magnitude ≈ 3.9 × 10⁻¹⁴ N (gravity dominates; magnetic contribution adds/subtracts ≤0.5%) — accept 3.9 × 10⁻¹⁴ N to 4.0 × 10⁻¹⁴ N [ECF from M1, M2]
### Part (b)(ii) [3 marks] — Show that (steps marked, target given per §4.5)
- M1: Apply Newton's second law in component form: dp_y/dt = ΣF_y, identifying the vertical forces as gravity (−mg) and the vertical component of qv × B [no ECF]
- M2: Recognise that with B horizontal and perpendicular to v₀, the vertical component of the magnetic force is qv_x B (with sign depending on geometry), giving dp_y/dt = −mg + qv_x B [no ECF]
- M3: At t = 0, v_x = v₀, so |dp_y/dt|₀ = mg − qv₀B (or mg + qv₀B depending on sign of B) ≈ (3.9 × 10⁻¹⁴) − (1.6 × 10⁻¹⁶) ≈ 3.9 × 10⁻¹⁴ N; student-derived value in range 3.8–4.0 × 10⁻¹⁴ N acceptable [no ECF; show-that value given]
*Note: the stem target "approximately 1.2 × 10⁻¹³ N" in (b)(ii) is a misprint corrected here; markers must award full credit for any answer in the range 3.8 × 10⁻¹⁴ to 4.0 × 10⁻¹⁴ N derived from the correct symbolic relation in M2. See marker notes.*
### Part (c) [3 marks] — Evaluate (judgement + supporting + limiting per §4.4.1)
- M1: Calculate drag at launch: F_drag = 6π(1.8 × 10⁻⁵)(1.0 × 10⁻⁶)(12) ≈ 4.1 × 10⁻⁹ N [ECF for arithmetic]
- M2: Comparison: F_drag (~4 × 10⁻⁹ N) is approximately 10⁵ times larger than gravity (~4 × 10⁻¹⁴ N) and ~10⁷ times larger than the magnetic force, so the assumption is *not* justified [ECF from (b)(i)]
- M3: Trajectory consequence: horizontal velocity decays rapidly (within mm-scale stopping distance), so the particle would slow drastically before traversing 1.0 m; the magnetic deflection (which scales with v_x) is correspondingly suppressed, and the motion approaches terminal-velocity vertical fall [linking marking point: "therefore"]
### Part (d) [2 marks] — Suggest (proposal + physics-based justification per §4.4.1)
- M1: A plausible neglected effect identified, e.g.:
- induced image charges on chamber walls
- electrostatic interaction with residual ions / stray E-fields
- radiation reaction from accelerated charge
- buoyancy of the microsphere in residual gas
- thermal (Brownian) fluctuations of the air molecules
- non-uniformity of B near chamber boundaries
- M2: Physics-based justification of its impact, e.g. "stray E-field of even 10 V m⁻¹ exerts qE ≈ 1.6 × 10⁻¹⁶ N, comparable to the magnetic force, so it would shift the trajectory laterally"; OR "Brownian kicks become significant for microspheres of this mass, randomising the path on the scale of micrometres per second"
### Marker notes
- (b)(i): if student takes magnetic force as horizontal (perpendicular to gravity) instead of along vertical, accept F_net = √((mg)² + (qv₀B)²) ≈ 3.9 × 10⁻¹⁴ N (numerically indistinguishable to 2 s.f.) — award full marks
- (b)(ii): **Show-that correction.** The intended target is 3.9 × 10⁻¹⁴ N (given to 2 s.f. — students need only derive to 1–2 s.f.). The "1.2 × 10⁻¹³ N" appearing in the stem is a printing error; markers must accept the physically correct derivation (range 3.8–4.0 × 10⁻¹⁴ N) and ignore the stem figure. Per §4.5, the value is given to detect reverse-engineering; here only the steps M1–M3 carry credit
- (c) accept any of: drag dominates over both gravity and magnetic force; particle reaches terminal velocity before traversing 1 m; horizontal range much less than 1 m; assumption invalid
- (d) accept any field-related or momentum-related effect with coherent justification; do NOT accept "friction" (already addressed in (c)) or "the Earth's magnetic field" without quantitative comment
- ECF chain: (b)(i) → (c) M2; (a) sign convention → (b)(ii) M2
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