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## ERQ · 12 marks · Topics: D.2 Electric and magnetic fields + A.2 Forces and momentum · Archetype: theory_application
**Integration:** primary=D.2 Electric and magnetic fields, secondary=A.2 Forces and momentum (strength: supporting)
**Stem.** A laboratory proton accelerator produces a narrow beam for a materials-analysis experiment. Protons are released from rest near a heated filament and accelerated horizontally through a potential difference of 2.5 kV in an evacuated chamber. The accelerated protons then enter a velocity selector consisting of a region where a uniform electric field of magnitude 1200 V m⁻¹ points vertically downward, crossed with a uniform horizontal magnetic field perpendicular to the beam direction. Only protons travelling at the selected speed emerge undeflected through a narrow exit slit. The mass of a proton is 1.67 × 10⁻²⁷ kg and its charge is 1.60 × 10⁻¹⁹ C. Treat gravitational effects on the protons as negligible.
### Part (a) State [2 marks] · AO1 · Topic: D.2
State the condition on the electric force and magnetic force acting on a charged particle for it to pass undeflected through a velocity selector, and write the resulting relationship between the particle's speed v, the electric field strength E and the magnetic flux density B.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: D.2
Calculate, in joules, the kinetic energy gained by a proton accelerated from rest through the 2.5 kV potential difference.
### Part (b)(ii) Determine [2 marks] · AO2 · Topic: D.2
Determine the speed of the proton on entering the velocity selector.
### Part (c) Show that [3 marks] · AO2 · Topic: D.2
Using the speed obtained in (b)(ii) and the electric field strength of 1200 V m⁻¹, show that the magnetic flux density required for the proton to pass undeflected through the selector is approximately 1.73 × 10⁻³ T.
### Part (d) Discuss [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: D.2+A.2
The momentum of a proton entering the velocity selector at the selected speed appears unchanged on exit. Discuss whether the proton alone constitutes an isolated system for momentum conservation in this region, identifying one assumption made in the analysis above that may not hold for a real beam of many protons.
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## Mark Scheme
### Part (a) [2 marks] — State
- M1: Electric force on the particle is equal in magnitude and opposite in direction to the magnetic force (so the net force is zero) [no ECF]
- M2: Resulting condition qE = qvB ⇒ v = E/B [no ECF]
### Part (b)(i) [3 marks] — Calculate
- M1: Recognises KE gained = qV with correct substitution: KE = (1.60 × 10⁻¹⁹)(2.5 × 10³) [no ECF]
- M2: Correct evaluation step shown, e.g. KE = 1.60 × 2.5 × 10⁻¹⁶ J [ECF from substitution]
- M3: KE = 4.0 × 10⁻¹⁶ J [ECF from (b)(i) M1]
### Part (b)(ii) [2 marks] — Determine
- M1: Uses KE = ½mv² ⇒ v = √(2KE/m), substitutes v = √(2 × 4.0 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) [ECF from (b)(i)]
- M2: v ≈ 6.9 × 10⁵ m s⁻¹ (accept 6.92 × 10⁵ m s⁻¹) [ECF from (b)(i)]
### Part (c) [3 marks] — Show that
- M1: States/uses condition v = E/B ⇒ B = E/v [no ECF]
- M2: Correct substitution B = 1200 / 6.9 × 10⁵ [ECF from (b)(ii)]
- M3: B = 1.74 × 10⁻³ T, consistent with the stated 1.73 × 10⁻³ T (accept 1.7–1.8 × 10⁻³ T) [ECF from (b)(ii)]
### Part (d) [2 marks] — Discuss (balanced consideration per §4.4.1)
- M1 (perspective 1 — proton not isolated): The proton experiences external electric and magnetic forces from the apparatus, therefore the proton alone is NOT an isolated system; momentum is only conserved overall if the field sources (and apparatus producing them) are included in the system.
- M2 (perspective 2 — limitation of the single-proton analysis): Identifies one assumption that fails for a real beam, e.g. inter-proton Coulomb repulsion is neglected (beam protons exert forces on each other so individual momenta are not preserved), OR fringe/non-uniform field regions at entry/exit are neglected, OR gravitational force on the proton has been ignored.
### Marker notes
- (a): accept a clearly labelled diagram showing F_E and F_B equal and opposite as equivalent to M1.
- (b)(i): full marks if answer is given as 4.0 × 10⁻¹⁶ J with one line of working.
- (b)(ii): accept any speed in the range 6.9–6.93 × 10⁵ m s⁻¹.
- Show-that target in (c): 1.73 × 10⁻³ T given to 3 sig figs; student-derived 1.7–1.8 × 10⁻³ T acceptable for M3.
- (d) accept any one of: space-charge / Coulomb repulsion between protons in the beam; non-uniformity / fringing of E or B fields near the slits; finite beam width so not all protons travel at the selected speed; gravitational force neglected; relativistic correction neglected.
- ECF applies throughout (b)(ii)→(c) for numerical chain.
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