Generated ERQ

✓ passed D.3 Motion in electromagnetic fields × A.1 Kinematics 12 marks HL 3 passes 112.61s $0.6405
## ERQ · 12 marks · Topics: D.3 Motion in electromagnetic fields + A.1 Kinematics **Stem.** A mass spectrometer's velocity selector ejects protons (charge +1.6 × 10⁻¹⁹ C, mass 1.67 × 10⁻²⁷ kg) into an analyser chamber containing a uniform magnetic field of magnitude **B = 0.50 T** directed along the +x-axis. A proton enters the field at the origin with speed **v = 2.0 × 10⁶ m s⁻¹**, with its velocity vector lying in the x–y plane at an angle of **30° to the +x-axis**. The resulting trajectory is helical: the velocity component parallel to **B** is unchanged, while the perpendicular component executes uniform circular motion in the y–z plane. A detector plate is mounted perpendicular to the x-axis at **x = 1.0 m** from the entry point. Edge effects and gravity are negligible inside the analyser. ### Part (a) State [2 marks] · AO1 · Topic: D.3 State the direction of the magnetic force on the proton at the instant it enters the field, and explain why this force does no work on the proton. ### Part (b)(i) Determine [4 marks] · AO2/AO3 · Topic: D.3 + A.1 By equating the magnetic force on the proton to the centripetal force required for circular motion in the y–z plane (a kinematic condition for uniform circular motion at radius r with tangential speed v_⊥), determine the radius r of the helix. ### Part (b)(ii) Calculate [2 marks] · AO2 · Topic: D.3 Calculate the period T of one revolution in the y–z plane. ### Part (c) Show that [2 marks] · AO2/AO3 · Topic: D.3 + A.1 Using the kinematic equation for uniform motion along the field direction, show that the time for the proton to reach the detector plate at x = 1.0 m is approximately 5.8 × 10⁻⁷ s, and hence determine how many complete revolutions the helix executes between entry and detection. ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR The pitch predicted by this model assumes the axial motion is at constant velocity (a A.1 kinematic idealisation). In a real solenoid analyser, the magnetic field weakens near the ends ("fringing"). Suggest how this fringing would cause the measured number of revolutions before impact to differ from your answer in (c). --- ## Mark Scheme ### Part (a) [2 marks] - M1: Force is in the +z direction (or "out of the x–y plane" / perpendicular to **v** and **B**, consistent with F = qv × B for a positive charge with v in x–y plane and B along +x). [ECF: no] - M2: Force is always perpendicular to velocity, therefore F · v = 0 and no work is done (accept: kinetic energy / speed is unchanged). [ECF: no] ### Part (b)(i) [4 marks] - M1: Resolves perpendicular component v_⊥ = v sin 30° = 1.0 × 10⁶ m s⁻¹ (accept v sin θ). [ECF: no] - M2: States force balance qv_⊥B = mv_⊥²/r (Lorentz force = centripetal kinematic requirement). [ECF: no] - M3: Rearranges to r = mv_⊥/(qB) and substitutes: r = (1.67 × 10⁻²⁷ × 1.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.50). [ECF: yes — accepts wrong v_⊥ from M1] - M4: r = 2.1 × 10⁻² m (accept 0.021 m, 2.09 × 10⁻² m). [ECF: yes] ### Part (b)(ii) [2 marks] - M1: Uses T = 2πm/(qB) or T = 2πr/v_⊥ with correct substitution. [ECF: yes from (b)(i)] - M2: T = 1.3 × 10⁻⁷ s (accept 1.31 × 10⁻⁷ s). [ECF: yes] ### Part (c) [2 marks] - M1: Identifies v_∥ = v cos 30° = 1.73 × 10⁶ m s⁻¹ and applies kinematic equation t = x/v_∥ = 1.0 / 1.73 × 10⁶ → t ≈ 5.78 × 10⁻⁷ s ≈ 5.8 × 10⁻⁷ s. [ECF: no — "show that"] - M2: Number of revolutions N = t/T = 5.78 × 10⁻⁷ / 1.31 × 10⁻⁷ ≈ 4.4 revolutions (accept 4 complete revolutions, or 4.4). [ECF: yes from (b)(ii)] ### Part (d) [2 marks] - M1: Identifies that in fringing regions B is smaller, so the cyclotron period T = 2πm/(qB) increases (equivalently, angular speed ω = qB/m decreases) near the ends. [ECF: no] - M2: Consequence: fewer revolutions are completed in the same axial transit time than predicted, OR the assumption of constant v_∥ also fails because the field has off-axis components that exert a force with an x-component, accelerating/decelerating the axial motion and invalidating the uniform-kinematics model used in (c). [ECF: no] ### Marker notes - (a) M1: accept clear diagram with labelled +z arrow; reject "upward" without axes. - (b)(i): full credit for an algebraic derivation r = mv sin θ/(qB) followed by correct numerical answer. If candidate uses v (not v_⊥) throughout, award M2 only; cap at 2/4. - (b)(ii): accept derivation from T = 2πr/v_⊥ using the candidate's r and v_⊥ (ECF route). - (c): "Show that" — working must be visible; bare answer scores 0 for M1. Accept N stated as "4 full + partial" or 4.4. - (d) discriminator: accept any of (i) reduced B → longer period near ends; (ii) radial B_r component in fringe field → axial Lorentz force, violating constant-v_∥ kinematic assumption; (iii) non-uniform B → r and pitch vary along trajectory, so the helix is not geometrically regular. Do NOT accept generic "experimental error" or "air resistance".