HSC Exam Practice

Sample response. A velocity selector is a region where a uniform electric field and a uniform magnetic field are oriented perpendicular to each other and to the initial velocity of a charged particle, so that the electric force and the magnetic force on the particle act in opposite directions. The condition for undeflected (straight-line) motion is v = E/B: only particles with exactly this speed experience equal and opposite forces that cancel, producing zero net force.

Marking notes. 1 mark for defining a velocity selector as a crossed-fields device with E and B perpendicular to each other and to v. 1 mark for stating that forces on the particle are in opposite directions. 1 mark for stating the condition v = E/B (or equivalently qE = qvB).

Sample response. Work done by a force is W = F · d cosθ, where θ is the angle between the force and the displacement. The magnetic force on a moving charged particle is always perpendicular to the particle’s velocity (θ = 90°) because F = qv × B is a cross product. Since cos 90° = 0, the work done is zero at all times. Consequently, the magnetic force changes the direction of the particle’s velocity but never its magnitude (speed), so kinetic energy remains constant.

Marking notes. 1 mark for stating the magnetic force is always perpendicular to velocity. 1 mark for linking this to W = Fd cosθ = 0 at θ = 90°. 1 mark for concluding that speed (and therefore kinetic energy) is unchanged.

Sample response. In a uniform electric field (perpendicular to v): the particle follows a parabolic path, analogous to a projectile in gravity. The constant electric force has a component along the direction of motion at some point, so it does work and the particle’s speed changes — it accelerates. In a uniform magnetic field (perpendicular to v): the particle follows a circular path at constant speed. The magnetic force is always perpendicular to the velocity, so it provides centripetal acceleration without doing work; the speed remains constant throughout.

Marking notes. 1 mark for identifying the E-field path as parabolic. 1 mark for stating that speed (or kinetic energy) changes in the E-field. 1 mark for identifying the B-field path as circular. 1 mark for stating that speed remains constant in the B-field.

Sample response. v = E/B = 8000 / 0.040 = 2.0 × 10⁵ m/s. This speed is the same for both the proton and the doubly charged oxygen ion. When setting qE = qvB, the charge q cancels from both sides, giving v = E/B — a result that is independent of the sign and magnitude of the charge and the mass of the particle. Changing the type of charge does not change the balance speed.

Marking notes. 1 mark for correct calculation of v = 2.0 × 10⁵ m/s with working. 1 mark for stating the speed is the same for both particles. 1 mark for explaining why (q cancels from qE = qvB, leaving v = E/B independent of q and m).

Sample response. Stage 1 — Acceleration: An electric field (no magnetic field) accelerates ions through a known voltage, giving them kinetic energy qV = ½mv². This ensures all ions start with defined kinetic energies but different speeds (depending on mass). Stage 2 — Velocity selection: Crossed electric and magnetic fields allow only ions with speed v = E/B to pass through a narrow slit; all others are deflected away. This ensures all ions entering the next stage have the same known speed. Stage 3 — Mass analysis: A pure magnetic field (no electric field) causes ions to follow circular paths of radius r = mv/(qB). Since v, q, and B are fixed, r varies only with m, and the radius difference is measured on a detector to identify the ion masses.

Marking notes. 1 mark per stage correctly described with both the field type present and the role in mass separation.

Sample response (a) — 4 marks. Electric force: F_E = qE = (1.60 × 10⁻¹⁹)(3000) = 4.80 × 10⁻¹⁶ N, directed upward (same direction as E for a positive charge) [1 mark]. Magnetic force: F_B = qvB = (1.60 × 10⁻¹⁹)(1.5 × 10⁵)(0.015) = 3.60 × 10⁻¹⁶ N, directed downward (by right-hand rule: v to the right × B out of page → force upward on positive; BUT since E is upward and both forces must be opposite, verify: v right × B out gives F upward on positive — so F_E is upward and F_B is downward means there is a configuration error; let us re-examine: if E is upward F_E = qE is upward on proton; for F_B to oppose, the magnetic force must be downward; v right × B out of page: using right-hand rule, fingers point right (v), curl toward out-of-page (B), result points — using v×B: right(î) × out-of-page(k̂) = î×k̂ = −ĵ̂ = downward. So F_B is downward on the positive proton.) [1 mark]. Since F_E = 4.80 × 10⁻¹⁶ N upward > F_B = 3.60 × 10⁻¹⁶ N downward, the net force is upward and the proton follows path X (curves upward) [1 mark for correct path; 1 mark for comparing magnitudes and justifying].

Sample response (b) — 3 marks. v_sel = E/B = 3000 / 0.015 = 2.0 × 10⁵ m/s [1 mark]. At v = 2.0 × 10⁵ m/s, F_B = qvB = (1.60 × 10⁻¹⁹)(2.0 × 10⁵)(0.015) = 4.80 × 10⁻¹⁶ N downward, which exactly equals F_E = 4.80 × 10⁻¹⁶ N upward. Net force is zero, so the proton follows path Y (straight line) [1 mark for identifying path Y; 1 mark for showing the forces balance].

Sample response (c) — 2 marks. For an electron (q = −e) at v = 1.5 × 10⁵ m/s: F_E = qE is downward (opposite to E because charge is negative); F_B = qvB is upward (opposite to proton’s magnetic force). Magnitudes are the same as for the proton at this speed: F_E = 4.80 × 10⁻¹⁶ N downward, F_B = 3.60 × 10⁻¹⁶ N upward. Net force is downward (F_E > F_B), so the electron follows path Z (curves downward) [1 mark for correct path Z; 1 mark for correctly reversing both force directions and comparing magnitudes].

Sample Band 6 response. Electric fields and magnetic fields each exert forces on charged particles, but in fundamentally different ways. An electric field of strength E exerts a force F_E = qE on any charged particle, whether stationary or moving. Because the force can act parallel (or anti-parallel) to the particle’s displacement, it does work and changes the particle’s kinetic energy. In a uniform field perpendicular to the initial velocity, the particle follows a parabolic path, accelerating as it goes — analogous to projectile motion. A magnetic field of strength B exerts a force F_B = qvB sinθ only on moving charges. Since this force is always perpendicular to the velocity (a cross product: F = q(v × B)), it does no work, and the particle’s speed remains constant. In a uniform magnetic field perpendicular to the velocity, the particle follows a circular path. The claim that “only one field can change a particle’s energy” is correct: only the electric field can do work and change kinetic energy; the magnetic force is a centripetal force that redirects the particle without altering its energy. This fundamental difference arises because work requires a force component along the direction of motion, which the magnetic force never has. In the velocity selector, both fields act simultaneously. By arranging E and B perpendicular to each other and to the particle’s velocity, the electric force and magnetic force point in opposite directions. They balance when qE = qvB, giving the mass- and charge-independent condition v = E/B. Particles with exactly this speed experience zero net force and travel straight; all others are deflected and blocked by the selector slit. In Australia, mass spectrometers incorporating velocity selectors are used at the Australian Nuclear Science and Technology Organisation (ANSTO) at Lucas Heights, NSW, for Secondary Ion Mass Spectrometry (SIMS) analysis of materials, isotope dating of geological samples, and nuclear forensics. The selector ensures uniform ion speeds entering the analysis magnet, so differences in radius of curvature r = mv/(qB) reflect mass-to-charge ratio differences only, enabling accurate identification of elemental and isotopic compositions. In summary, while both fields influence charged particles, the electric field’s unique ability to do work means it has different physical consequences — changing speed and trajectory shape — compared to the magnetic field, which only steers. Their combination in a velocity selector elegantly exploits both properties to filter particles by speed, and this principle underpins a wide range of real-world Australian analytical technologies.

Marking criteria (8 marks). 1 = correctly describes electric force (F = qE, acts on all charges, does work) with reference to trajectory shape (parabola). 1 = correctly describes magnetic force (F = qvB sinθ, acts only on moving charges, does no work) with reference to trajectory shape (circle). 1 = explicitly evaluates the claim that only the electric field can change energy — correct, with explanation referencing the perpendicularity of F_B to v. 1 = derives or states the velocity selector condition v = E/B and explains why forces balance at this speed. 1 = explains why the condition is independent of charge and mass (q cancels). 1 = correctly describes the role of the velocity selector in a mass spectrometer (uniform speed for mass analysis via r = mv/(qB)). 1 = names a specific Australian scientific or industrial application with sufficient context (ANSTO SIMS, isotope dating, geological analysis, forensic chemistry). 1 = integrates the discussion with an evaluative conclusion and uses precise physics terminology throughout (centripetal, cross product or perpendicularity, work, kinetic energy, charge-to-mass ratio).