Magnetic fields and charged-particle motion: worked solutions

Answer: Clockwise concentric field circles, with a decreasing field magnitude away from the wire. A flat coil has a field along its axis near the centre and returning outside; a long solenoid has nearly parallel, evenly spaced internal lines and an external return field. Electron drift is opposite to conventional current.

1. Draw circles centred on the actual wire cross-section, rather than on a label or leader line. Add clockwise arrows for current into the page.
2. Use increasing spacing between the circles to communicate weakening field, and state that the magnitude decreases with distance. This is a qualitative map, not a quantitative spacing rule.
3. Near the flat coil’s centre the field runs through the coil along its axis; outside it curves back, forming closed loops. The whole coil field is not uniform.
4. In the central interior of a long solenoid, lines are approximately parallel and evenly spaced, so the field is approximately uniform. The field returns outside and becomes less uniform near the ends.
5. Electrons have negative charge. Electrons drifting out of the page correspond to conventional current into the page. Magnetic fields arise from moving charge or permanent magnets.

Check: Field lines form closed loops and do not cross. “Approximately uniform” applies to the central long-solenoid region, not every point around the coil.

Mark checklist · 6 marks

• 1 mark — Shows concentric circles centred on the wire.
• 1 mark — Adds clockwise arrows for conventional current into the page.
• 1 mark — Clearly shows/describes weakening with distance, using increasing line spacing.
• 1 mark — Describes the flat coil’s axial central field and curved external return.
• 1 mark — Describes the solenoid’s approximately uniform central field and external return.
• 1 mark — Explains electron drift and conventional current have opposite directions.

Common mistake: The right-hand grip rule gives the field around a current. Fleming’s left-hand rule gives force once the field and conventional current are known; using one in place of the other gives the wrong construction.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, November 2021 — Q19(a); report p. 9

Answer: F = 0.0336 N; maximum = 0.0672 N at 90°. One tesla gives 1 N on a perpendicular 1 m wire carrying 1 A. Parallel current produces zero magnetic force even when B is nonzero.

1. F = BIL sinθ = 0.350 × 2.40 × 0.0800 × sin30.0° = 0.0336 N.
2. Fmax = BIL = 0.0672 N, at θ = 90°.
3. For a straight wire perpendicular to a uniform field, B = F/(IL). Therefore 1 T = 1 N A⁻¹ m⁻¹; state the perpendicular condition when defining the tesla.
4. At θ = 0° or 180°, sinθ = 0. The moving charge is parallel to B, so this arrangement gives no magnetic force; a zero reading alone cannot determine whether a field is absent.

Check: At 30°, the force is half the maximum because sin30° = 0.5.

Mark checklist · 6 marks

• 1 mark — Uses BIL sinθ with the stated active length.
• 1 mark — Obtains 0.0336 N.
• 1 mark — Obtains maximum 0.0672 N.
• 1 mark — States 90° to the field.
• 1 mark — Defines tesla with unit current, length, force and perpendicular condition.
• 1 mark — Explains parallel current gives zero force without requiring zero B.

Common mistake: Using cos30° would use the component parallel to the field. Resolve the current/active length perpendicular to B, or apply the supplied sine formula directly.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Answer: Force = 0.015696 N; B = (0.131 ± 0.009) T using the conservative estimate. Additional magnetic force on the magnet is downwards; force on the wire is upwards.

1. Δm = 1.60 g = 1.60 × 10⁻³ kg;
   F = Δmg = 0.015696 N.
2. B = F/(IL) = 0.015696/(3.00 × 0.0400) = 0.1308 T.
3. δ(Δm) = 0.02 + 0.02 = 0.04 g;
   δB/B ≈ 0.04/1.60 + 0.06/3.00 + 0.0008/0.0400
   = 0.025 + 0.020 + 0.020 = 0.065 = 6.5%.
   δB ≈ 0.1308 × 0.065 = 0.008502 T ≈ 0.009 T.
   Requested quote: (0.131 ± 0.009) T.
4. An increased reading means the magnetic force on the magnet is downwards, increasing the support force measured by the balance. The magnet’s magnetic force and the wire’s magnetic force are an equal-and-opposite interaction pair on different objects; the wire force is upwards.

Check: The magnetic change is only 1.60 g compared with the 112.40 g baseline. Using the whole 114.00 g would greatly overestimate B.

Mark checklist · 7 marks

• 1 mark — Subtracts baseline to obtain 1.60 g and converts to kg.
• 1 mark — Obtains magnetic force 0.015696 N.
• 1 mark — Uses B = F/(IL) and obtains 0.1308 T.
• 1 mark — Adds the reading uncertainties to obtain 0.04 g.
• 1 mark — Adds the three fractional uncertainties to obtain 6.5%.
• 1 mark — Obtains about 0.009 T and quotes requested significant figures.
• 1 mark — Explains downwards force on magnet and upwards force on wire.

Common mistake: The balance measures its support force, not a direct wire-force reading. Newton’s third-law forces act on different objects and do not cancel each other on either free-body diagram.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, November 2021 — Q19(b); report p. 9 · OCR H556/02, November 2021 — Q19(b); report p. 9

Answer: Support the perpendicular wire independently, measure its active length, subtract/tare the no-current baseline, and convert display changes to force. Measure and repeat at several currents while fixing geometry. B is fitted gradient/L. Control heating, misalignment, offsets and balance uncertainty.

1. Place the magnet on the balance and fix the wire on independent supports. Measure only its length L in the approximately uniform gap and arrange it perpendicular to B; prevent contact with the pan or magnet.
2. Record the no-current reading or tare the balance. Measure I with a series ammeter. Convert the change in display from g to kg and calculate F = |Δm|g for one current direction.
3. Take several well-spaced currents over a suitable range and repeat readings. Keep active length, wire angle and position fixed. Check the measured current at each point; a supply setting alone does not guarantee it.
4. Fit F against I and calculate a large-triangle gradient G. For a perpendicular wire, G = BL, so B = G/L. Check the intercept against the baseline correction rather than automatically forcing the fit through the origin.
5. For example, switch off between readings and limit current to reduce heating and movement, or realign the wire to remove angle bias. Repeats reduce random spread but do not fix a systematic angle error.

Check: A magnet and wire sharing the same balance can put both internal magnetic reaction forces on the measured system, defeating this arrangement.

Mark checklist · 6 marks

• 1 mark — Independent support, perpendicular wire and measured active length.
• 1 mark — Baseline/tare, actual current measurement and Δm-to-force conversion.
• 1 mark — Uses several currents and repeats to judge random variation.
• 1 mark — Keeps length, angle and position fixed and checks actual current.
• 1 mark — Fits force against current and finds B = gradient/L.
• 1 mark — States a concrete limitation with a physically linked improvement.

Common mistake: “Repeat to remove all errors” is insufficient. Say which random uncertainty is reduced and how a systematic geometric error is addressed.

Exam technique: Make the apparatus, baseline, graph and uncertainty explicit. A straight trend does not prove the intercept is zero or every systematic error is absent.

Related examiner evidence: OCR H556/02, November 2021 — Q19(b); report p. 9

Answer: G = 0.0180 N A⁻¹; B = (0.30 ± 0.02) T with interval about 0.280–0.320 T. The 0.310 T value is consistent. Fraw/I is biased by the nonzero offset.

1. G = (49.0 − 4.0)/(2.50 − 0)
   = 18.0 mN A⁻¹ = 0.0180 N A⁻¹.
   Fraw = Foffset + BLI, so G = BL.
2. B = G/L = 0.0180/0.0600 = 0.300 T.
3. δB/B ≈ 0.9/18.0 + 0.0010/0.0600 = 0.0667;
   δB = 0.0200 T.
   Quote (0.30 ± 0.02) T.
4. 0.310 T is inside the estimated interval, so it is consistent under this uncertainty model. This does not prove every systematic effect is absent.
5. Because Fraw has a nonzero intercept, Fraw/I = BL + Foffset/I. At 2.00 A, 40.1 mN/2.00 A = 20.05 mN A⁻¹, above the fitted 18.0. Use the fitted slope or subtract the independently measured offset; do not force the raw fit through zero.

Check: The fitted intercept is roughly 4 mN, well outside the zero-current point’s error interval around zero. The near-linear data do not justify discarding this offset.

Mark checklist · 7 marks

• 1 mark — Calculates gradient 18.0 mN A⁻¹ and converts to 0.0180 N A⁻¹.
• 1 mark — Explains gradient = BL despite a constant intercept.
• 1 mark — Obtains B = 0.300 T.
• 1 mark — Obtains total fractional uncertainty about 6.67%.
• 1 mark — Obtains about 0.020 T and quotes a sensible value/uncertainty interval.
• 1 mark — Compares 0.310 T with the interval and judges consistency.
• 1 mark — Explains Fraw/I includes offset/I and is not the fitted slope.

Common mistake: A small triangle on a fitted line and F/I at a single measured point are different procedures. Use well-separated fitted-line coordinates and keep axis prefixes throughout the calculation.

Exam technique: Make the apparatus, baseline, graph and uncertainty explicit. A straight trend does not prove the intercept is zero or every systematic error is absent.

Answer: Initial force is downwards. F = 4.61 × 10⁻¹⁶ N and a = 5.06 × 10¹⁴ m s⁻². The magnetic force changes velocity direction, does no work and preserves speed/kinetic energy.

1. For rightward positive charge and field into the page, force is upwards. The electron has negative charge, so its force is downwards; use conventional current opposite to electron velocity in Fleming’s rule.
2. F = B|Q|v = 1.20 × 10⁻³ × 1.60 × 10⁻¹⁹ × 2.40 × 10⁶ = 4.608 × 10⁻¹⁶ N.
3. a = F/me = (4.608 × 10⁻¹⁶)/(9.11 × 10⁻³¹) = 5.06 × 10¹⁴ m s⁻².
4. The magnetic force is always perpendicular to the instantaneous velocity and acts towards the orbit centre. The particle accelerates because its velocity direction changes.
5. There is no displacement component along the magnetic force, so it does no work. With no other work or losses, kinetic energy and speed stay constant. The path is circular in the ideal uniform-field, perpendicular-entry model.

Check: The speed is 0.008c, so the stated non-relativistic equations are appropriate. A large acceleration can occur in a very small orbit.

Mark checklist · 6 marks

• 1 mark — Gives downwards initial force and explains negative charge reversal.
• 1 mark — Uses B|Q|v with mT converted to T.
• 1 mark — Obtains force about 4.61 × 10⁻¹⁶ N.
• 1 mark — Uses electron mass to obtain a about 5.06 × 10¹⁴ m s⁻².
• 1 mark — Explains perpendicular force changes velocity direction/centripetal acceleration.
• 1 mark — Explains no work and constant kinetic energy/speed.

Common mistake: Do not add a separate “centripetal force” to the magnetic force. In this model the magnetic force is the centripetal resultant.

Exam technique: Make the apparatus, baseline, graph and uncertainty explicit. A straight trend does not prove the intercept is zero or every systematic error is absent.

Related examiner evidence: OCR H556/02, June 2019 — Q20(a); report p. 33

Answer: v = 3.10 × 10⁵ m s⁻¹; r = 0.0404 m; semicircle time = 4.10 × 10⁻⁷ s.

1. K = ½mpv² ⇒ v = √(2K/mp)
   = √[(2 × 8.00 × 10⁻¹⁷)/(1.67 × 10⁻²⁷)]
   = 3.095 × 10⁵ m s⁻¹.
2. Bev = mpv²/r ⇒ r = mpv/(Be)
   = 0.0403839… m ≈ 0.0404 m.
3. t = πr/v = πmp/(Be) = 4.10 × 10⁻⁷ s.

Check: At fixed proton mass, charge and B, the ideal semicircle time is independent of speed: the larger distance and larger speed cancel.

Mark checklist · 5 marks

• 1 mark — Uses K = ½mpv² and appropriate proton mass.
• 1 mark — Obtains speed about 3.10 × 10⁵ m s⁻¹.
• 1 mark — Equates magnetic and centripetal force to derive r.
• 1 mark — Obtains radius about 0.0404 m.
• 1 mark — Uses πr/v or πmp/(Be) to obtain about 4.10 × 10⁻⁷ s.

Common mistake: Kinetic energy is not speed; doubling energy multiplies speed by √2. A magnetic field does not supply the given entry energy.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, June 2019 — Q20(a); report p. 33

Answer: Equal speed: r24 = 0.1195 m, r26 = 0.1295 m, ratio 1.083. Equal p.d./equal charge: ratio √(26/24) = 1.041. A +2e ion of the same mass has radius 1/√2 of the +e ion at equal p.d.

1. m24 = 24 × 1.66 × 10⁻²⁷ = 3.984 × 10⁻²⁶ kg;
   r24 = mv/(Be) = 0.11952 m.
   r26 = 0.12948 m;
   r26/r24 = 26/24 = 1.0833….
2. ½mv² = |Q|V and r = mv/(B|Q|)
   ⇒ r² = 2mV/(B²|Q|).
3. At equal V, |Q| and B: r26/r24 = √(26/24) = 1.0408….
   The heavier ion is slower: v ∝ 1/√m, so radius is proportional to √m rather than m.
4. At fixed m, V and B: r ∝ 1/√|Q|.
   r(+2e)/r(+e) = 1/√2 ≈ 0.707.

Check: The heavier isotope has the larger radius in both comparisons, but the separation is smaller at equal p.d. than at equal speed.

Mark checklist · 7 marks

• 1 mark — Uses masses in kg and r = mv/(B|Q|).
• 1 mark — Obtains both equal-speed radii about 0.1195 and 0.1295 m.
• 1 mark — Obtains equal-speed ratio 1.083.
• 1 mark — Combines kinetic energy from accelerating p.d. with magnetic radius.
• 1 mark — Derives r² = 2mV/(B²|Q|) and ratio 1.041.
• 1 mark — Explains heavier equal-charge ions are slower at equal p.d.
• 1 mark — Obtains charge-state radius ratio 1/√2 at equal p.d.

Common mistake: r ∝ m requires fixed v, B and charge. At equal p.d. speed changes, and a +2e ion receives twice as much kinetic energy as a +e ion.

Exam technique: Write down which of speed, accelerating p.d., mass, charge and B stay constant before simplifying a radius ratio.

Answer: r = 0.150 m and p = 4.32 × 10⁻²¹ kg m s⁻¹. D is above H. A neutral particle has no magnetic force and continues straight in this ideal model.

1. B|Q|v = mv²/r ⇒ p = mv = B|Q|r.
2. H–D = 2r ⇒ r = 0.150 m;
   p = 0.180 × 1.60 × 10⁻¹⁹ × 0.150
   = 4.32 × 10⁻²¹ kg m s⁻¹.
3. A right-moving positive ion in B into the page initially feels an upward force. The semicircle returns to the entry boundary above H.
4. For Q = 0, magnetic force is zero. With no other force, a neutral particle continues in a straight line; it does not reach D along this curved path.

Check: B|Q|r has units T C m = kg m s⁻¹. Using 0.300 m as r would double the inferred momentum.

Mark checklist · 6 marks

• 1 mark — Equates magnetic force to mv²/r.
• 1 mark — Obtains p = B|Q|r without needing mass separately.
• 1 mark — Uses r = half the straight detector separation.
• 1 mark — Obtains p = 4.32 × 10⁻²¹ kg m s⁻¹ with units.
• 1 mark — Gives D above H using positive-charge force direction.
• 1 mark — Explains zero charge means zero magnetic force and straight motion.

Common mistake: The semicircle’s path length is πr, but the boundary separation is 2r. Neither is automatically the radius.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, June 2022 — Q7; report p. 10

Answer: E = 3.00 × 10⁴ V m⁻¹ downwards; selected speed = 1.50 × 10⁵ m s⁻¹. At 2.00 × 10⁵ m s⁻¹, +e feels 4.80 × 10⁻¹⁵ N down and 6.40 × 10⁻¹⁵ N up, giving 1.60 × 10⁻¹⁵ N up. Negative charge reverses the resultant; selected speed stays the same.

1. E = V/d = 600/0.0200 = 3.00 × 10⁴ V m⁻¹, downwards.
2. For undeflected motion, |Q|E = B|Q|v ⇒ v = E/B
   = 1.50 × 10⁵ m s⁻¹. The forces cancel in magnitude and direction; E and B do not have the same units.
3. For +e at 2.00 × 10⁵ m s⁻¹:
   FE = eE = 4.80 × 10⁻¹⁵ N downwards;
   FB = Bev = 6.40 × 10⁻¹⁵ N upwards;
   Fnet = 1.60 × 10⁻¹⁵ N upwards.
4. For −e both forces reverse, so the initial resultant is downwards at this speed. The same E/B balances them because charge magnitude cancels.
5. These are initial directions. An unselected particle’s later motion in crossed fields is not generally a magnetic-only circle, because the electric force can do work.

Check: At the selected speed the net force is zero for either sign of charge, even though each individual force is nonzero.

Mark checklist · 7 marks

• 1 mark — Finds E = 3.00 × 10⁴ V m⁻¹ with downward direction.
• 1 mark — Balances opposing electric and magnetic forces to derive v = E/B.
• 1 mark — Obtains selected speed 1.50 × 10⁵ m s⁻¹.
• 1 mark — Obtains electric/magnetic magnitudes 4.80 and 6.40 × 10⁻¹⁵ N.
• 1 mark — Subtracts to obtain net 1.60 × 10⁻¹⁵ N upwards for +e.
• 1 mark — Reverses both forces and obtains downwards net for −e.
• 1 mark — Explains selected speed is unchanged because charge cancels.

Common mistake: Equal numerical field strengths are not the condition. Balance forces and arrange their directions to oppose; perpendicular fields alone do not guarantee the right polarity.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, November 2021 — Q24(c); report p. 13

Answer: Selected speed = 2.00 × 10⁵ m s⁻¹. The 0.415 m peak corresponds to 4.98 × 10⁻²⁶ kg = 30.0u. The 28u peak is at 0.3873 m, giving separation 27.7 mm. Peak positions identify m/|Q| under the model, not relative numbers.

1. v = E/Bselector = (4.00 × 10⁴)/0.200 = 2.00 × 10⁵ m s⁻¹.
   |Q| cancels in the balance; mass does not appear.
2. r = 0.415/2 = 0.2075 m;
   m = Banalyser|Q|r/v
   = 0.300 × 1.60 × 10⁻¹⁹ × 0.2075/(2.00 × 10⁵)
   = 4.98 × 10⁻²⁶ kg = 30.0u.
3. m28 = 28 × 1.66 × 10⁻²⁷ = 4.648 × 10⁻²⁶ kg;
   D28 = 2m28v/(Banalyser e) = 0.38733… m;
   ΔD = 0.415 − 0.38733… = 0.02767 m = 27.7 mm.
4. Positions describe the momentum/charge and, after speed selection, mass/charge. Relative abundance needs measured counts/areas with appropriate equal-response or calibrated detector and collection efficiencies. Two positions contain no population information.

Check: At fixed selected speed and charge, D ∝ m. The 30u spot is farther from H than the 28u spot, with D30/D28 = 30/28.

Mark checklist · 8 marks

• 1 mark — Uses selector E/B and obtains speed 2.00 × 10⁵ m s⁻¹.
• 1 mark — Explains mass independence from opposing-force balance.
• 1 mark — Uses radius half the detector separation and the analyser field.
• 1 mark — Derives m = B|Q|r/v and obtains 4.98 × 10⁻²⁶ kg.
• 1 mark — Converts to 30.0u with the supplied u.
• 1 mark — Obtains 28u separation about 0.3873 m.
• 1 mark — Obtains peak spacing about 27.7 mm.
• 1 mark — Explains positions alone give no abundance; counts/response calibration are needed.

Common mistake: Bselector sets speed; Banalyser sets radius. Substituting 0.200 T into the analyser equation would infer the wrong mass. Unknown multiple charge states could also mimic a different m/|Q|.

Exam technique: State what is held constant and identify the correct force before calculating. These are original questions with PhysicsUK’s own mark checklists.

Related examiner evidence: OCR H556/02, June 2019 — Q20(a); report p. 33 · OCR H556/02, November 2021 — Q24(c); report p. 13 · OCR H556/02, June 2022 — Q7; report p. 10

Answer: SI gradient = 8.30 × 10⁻⁶ m² V⁻¹. Mass = (16.0 ± 0.7)u using the conservative estimate. The graph identifies m/|Q|; the stated charge is needed to infer m.

1. |Q|V = ½mv² and r = mv/(B|Q|)
   ⇒ r² = [2m/(B²|Q|)]V.
   SI gradient G = 2m/(B²|Q|).
2. G = 8.30 × (10⁻³ m²)/(10³ V)
   = 8.30 × 10⁻⁶ m² V⁻¹.
3. m = GB²e/2 = (8.30 × 10⁻⁶) × 0.200² × (1.60 × 10⁻¹⁹)/2
   = 2.656 × 10⁻²⁶ kg = 16.0u.
4. δm/m ≈ δG/G + 2δB/B
   = 0.20/8.30 + 2 × 0.002/0.200
   = 0.04410 ≈ 4.41%.
   δm = 16.0 × 0.04410 = 0.7055u ≈ 0.7u.
   Quote (16.0 ± 0.7)u.

Check: At 1.00 kV, r = √0.00830 ≈ 0.0911 m. Quadrupling V doubles r; it does not quadruple it.

Mark checklist · 7 marks

• 1 mark — Uses accelerating energy and magnetic-radius equations.
• 1 mark — Derives r² proportional to V with gradient 2m/(B²|Q|).
• 1 mark — Converts displayed gradient to 8.30 × 10⁻⁶ m² V⁻¹.
• 1 mark — Uses m = GB²e/2 to obtain 2.656 × 10⁻²⁶ kg.
• 1 mark — Obtains mass 16.0u.
• 1 mark — Includes twice the fractional B uncertainty to obtain about 4.41%.
• 1 mark — Obtains about 0.7u and quotes a suitable result.

Common mistake: The same graph would imply 32.0u for charge +2e. Magnetic curvature gives mass/charge after known acceleration, so charge state must be established separately. Do not count the radius uncertainty again after it is already included in δG.

Exam technique: Carry both axis prefixes into the SI gradient. The squared B dependence gives twice its fractional uncertainty.