OCR A A-level Physics · H556 · 6.3.1–6.3.2
Magnetic fields and charged-particle motion
Calculate magnetic forces, explain circular particle paths and show why crossed fields select one speed. Follow an instrument from its wire actuator to its ion analyser.
Start with a direction and energy check
A positive ion enters a uniform magnetic field into the page, moving to the right. There is no electric field and no appreciable energy loss. What happens?
By the end, you should be able to
- Map magnetic fields and distinguish the right-hand grip rule from Fleming’s left-hand force rule.
- Calculate forces on wires and individual charges, including the angle and charge-sign conditions.
- Combine magnetic force with circular motion and electrical energy to analyse particle beams.
- Explain velocity selection and evaluate a wire-and-balance investigation using graphs and uncertainty.
These notes cover OCR A 6.3.1 and 6.3.2. Induced e.m.f., Faraday’s law and transformers belong to the separate 6.3.3 resource. Cyclotrons appear only as an optional extension in the existing simulator.
Magnetic fields, field lines and mapping
Magnetic fields arise from moving charges, including electric currents, and permanent magnets. A field line’s tangent gives the local field direction: the direction a north magnetic pole would be pushed. Outside a magnet the arrows run from north to south; the lines continue through the magnet to form closed loops.
Closer field-line spacing represents a larger flux density in a given diagram. Lines never cross because the field cannot have two directions at the same point. A uniform field is represented by straight, parallel, equally spaced lines.
To map a field, put a plotting compass at several positions and mark the direction of its north-seeking end. Move it along the indicated direction to trace a line and repeat from other starting positions. Iron filings reveal a pattern but do not, on their own, identify the arrow direction.
Quick retrieval check: can field lines start or end at a magnetic pole?
No. They form continuous closed loops. The familiar north-to-south convention describes the part outside the magnet, not a beginning or an end of a line.
Field patterns: straight wire, flat coil and long solenoid
Long straight current-carrying wire
In a plane perpendicular to the wire, field lines are concentric circles centred on it. Use the right-hand grip rule: thumb in conventional-current direction, fingers curl with the field. Field strength decreases with distance from the wire.
Flat coil
The fields from different parts of the coil reinforce through its centre. The central field is perpendicular to the plane of the coil. Away from the coil, the field returns around the outside, giving a dipole-like pattern.
Long solenoid
Several coil turns reinforce the field inside the solenoid. Well inside a long solenoid, away from its ends, the field is nearly uniform and directed along its axis. The outside field is weaker and loops back from the north end to the south end.
Quick retrieval check: what changes if a solenoid’s current reverses?
The field direction reverses and north and south ends exchange. With the same current magnitude and geometry, the field-strength magnitude is unchanged.
Magnetic flux density B and the tesla
For a straight current-carrying wire placed perpendicular to a uniform field, magnetic flux density is force per unit current per unit active length.
1 T = 1 N A−1 m−1
A field of 1 tesla produces a force of 1 newton on 1 metre of perpendicular wire carrying 1 ampere. Here I is current and L is wire length inside the field; a metre of wire outside the magnet’s gap does not contribute to this force calculation.
Fleming’s left-hand rule: force, field and current
Hold the three digits of your left hand mutually perpendicular:
- First finger: magnetic field, from north to south across a gap.
- Second finger: conventional current, the direction positive charge moves.
- Thumb: magnetic force, perpendicular to both.
Use the right-hand grip rule to find the field produced by a current. Use Fleming’s left-hand rule to find the force on a current or a moving charge in a field. They answer different questions.
| Object | Direction to use as current | Initial magnetic force |
|---|---|---|
| Wire with current to the right | Right | Up |
| Positive ion moving right | Right | Up |
| Electron moving right | Left, opposite electron velocity | Down |
Reverse current or field alone: force reverses. Reverse both: the force returns to its original direction. For an electron, you can first find the force on a positive charge moving the same way, then reverse it.
Quick retrieval check: does a negative particle necessarily curve downwards?
No. Curvature depends on the charge sign, field direction and velocity direction together. “Downwards” is correct only for the specific rightward-motion/into-page-field arrangement above.
Force on a current-carrying wire: F = BIL sin θ
A laboratory sample-transfer instrument uses a short wire actuator to move a shutter. A straight section of its wire sits in a uniform magnetic gap.
F is force magnitude in newtons, B is flux density in teslas, I is current in amperes and L is active length in metres. θ is the angle between conventional-current direction and B. The equation gives magnitude; use the direction rule separately.
- Perpendicular: θ = 90°, sin θ = 1, so F = BIL is maximum.
- Parallel or antiparallel: θ = 0° or 180°, giving zero magnetic force.
- Oblique: use sin θ. If the diagram gives the complementary angle, first identify the angle to B.
Worked example 1 · an angled actuator wire
The active length is 65 mm, current 2.8 A and flux density 0.45 T. The wire makes 40° with the field. Calculate its magnetic force.
L = 0.065 m. F = 0.45 × 2.8 × 0.065 × sin 40° = 0.05264 N ≈ 0.053 N. Using F = BIL would give the perpendicular maximum, 0.0819 N, not the force at 40°.
Quick retrieval check: what happens if this wire is turned parallel to B?
The magnetic force becomes zero, even though current still flows. This is the magnetic force from the applied field; other mechanical forces may remain.
Force on a moving charge: F = BQv
The same instrument sends ions into an analyser. A single charge entering perpendicular to a uniform magnetic field experiences force magnitude
OCR writes F = BQv. When calculating a magnitude, use |Q|; determine direction from the sign separately. A singly charged ion has |Q| = e, an alpha particle has |Q| = 2e, and an electron has charge −e.
A stationary charge experiences no magnetic force from B alone. Motion parallel to B also gives zero magnetic force. For general oblique motion only the perpendicular velocity component produces magnetic force; perpendicular entry is the circular-orbit model used below.
Worked example 2 · force on an alpha particle
An alpha particle enters at right angles with v = 2.4 × 106 m s−1 and B = 0.28 T. Take e = 1.60 × 10−19 C.
|Q| = 2e = 3.20 × 10−19 C. F = 0.28 × 3.20 × 10−19 × 2.4 × 106 = 2.1504 × 10−13 N ≈ 2.2 × 10−13 N. If it enters moving right with B into the page, the force initially acts upwards.
Charged-particle circular motion and orbit radius
In a uniform magnetic field, with velocity perpendicular to B, the magnetic force is always perpendicular to the instantaneous velocity. It changes velocity direction and supplies the centripetal resultant force.
r = mv/(B|Q|)
The force does no work because it has no component along the motion. Kinetic energy and speed remain constant if other forces do no work and losses are negligible. The particle accelerates towards the centre, but its speed does not increase. Its velocity changes because its direction changes.
At fixed m and |Q|, a larger speed gives a larger radius. At fixed speed and particle type, a larger B gives a smaller radius. A negative charge curves the opposite way from a positive charge under otherwise identical conditions.
A complete circle requires the particle to stay inside the uniform field. On leaving a finite field region with no other force, it travels straight along the tangent at the exit.
Acceleration through a potential difference first
If an ion starts with negligible kinetic energy and gains energy through a p.d. of magnitude V, the non-relativistic model gives
v = √(2|Q|V/m)
r = √(2mV/|Q|)/B
The sign of the accelerating p.d. must accelerate that charge; V above denotes the positive magnitude of the energy-gaining p.d. If the particle already has appreciable kinetic energy, add the gained energy to the initial kinetic energy. The simple kinetic-energy formula is unsuitable when relativistic effects become significant.
Worked example 3 · proton beam accelerated before bending
Protons start from negligible speed, gain energy through 2.40 kV and enter a 0.420 T analyser at right angles. Take m = 1.67 × 10−27 kg and e = 1.60 × 10−19 C.
Energy gain = eV = 3.84 × 10−16 J. v = √(2 × 3.84 × 10−16 / 1.67 × 10−27) = 6.781 × 105 m s−1.
r = mv/(Be) = 0.01685 m ≈ 16.9 mm. If the beam follows a semicircle before detection, the entry-to-detector separation is its diameter, about 33.7 mm, not its radius.
Worked example 4 · separating isotope ions
Singly positive ions of masses 20 u and 22 u enter at the same speed, 1.80 × 105 m s−1, in B = 0.350 T. Use 1 u = 1.66 × 10−27 kg and e = 1.60 × 10−19 C.
For 20 u, r = (20 × 1.66 × 10−27 × 1.80 × 105)/(0.350 × 1.60 × 10−19) = 0.1067 m. For 22 u, r = 0.1174 m. If both complete semicircles from a common entry point in the same direction, detector positions differ by 2(r22 − r20) = 0.02134 m ≈ 21.3 mm.
At the same accelerating p.d., equally charged ions instead have equal kinetic energies and different speeds, giving r22/r20 = √(22/20) = 1.049. Never assume equal speed just because the same p.d. was used.
Try the comparison: what is held constant?
Quick retrieval check: does a tighter circular path mean the particle is slowing down?
Not in the ideal magnetic-only model. Increasing B can tighten the path without changing speed or kinetic energy. A different radius may also arise from a different mass, charge or entry speed: state the controlled quantities.
Velocity selector: crossed electric and magnetic fields
A velocity selector can prepare an ion beam with one chosen speed before the magnetic analyser. The electric and magnetic fields are mutually perpendicular and are arranged to give opposing forces for the incident beam.
v = E/B and E = V/d
v = V/(dB)
The charge magnitude cancels. With the appropriate direction arrangement and negligible other forces, particles moving at E/B are undeflected. It is the forces that have equal magnitudes, not E and B: their units are different.
| Entry speed | Force comparison | Initial motion |
|---|---|---|
| v < E/B | Electric force larger | Deflects downwards |
| v = E/B | Forces equal and opposite | Passes straight |
| v > E/B | Magnetic force larger | Deflects upwards |
For a negative ion entering the same way, both force directions reverse. The selected speed is still E/B, but the initial deflections of faster and slower particles reverse. The table describes initial deflection; an unselected particle’s later motion in crossed fields is not simply the magnetic-only circle.
Worked example 5 · setting a selected speed
The plates have p.d. 1800 V and separation 0.040 m. B = 0.300 T. What speed passes undeflected?
E = 1800/0.040 = 4.50 × 104 V m−1. v = E/B = 45000/0.300 = 1.50 × 105 m s−1. A positive ion entering faster initially curves upwards in the arrangement shown because magnetic force exceeds electric force.
Quick retrieval check: is putting E perpendicular to B enough to make every particle pass straight?
No. The forces must oppose for the chosen beam direction, and their magnitudes balance only at v = E/B. Other entry speeds are deflected.
Velocity-selector explorer: balance the forces
The beam travels right, B points into the page and E points down. Change the entry speed and charge sign, then explain the initial force direction. This activity calculates the forces at entry; the diagram above illustrates the positive-ion arrangement.
| Quantity | Value |
|---|---|
| Undeflected speed E/B | 150 km s⁻¹ |
| Electric force | 7.20 × 10⁻¹⁵ N downwards ↓ |
| Magnetic force | 7.20 × 10⁻¹⁵ N upwards ↑ |
| Resultant force | 0 N |
The forces balance: this entry speed gives a straight path.
- At the starting settings, raise entry speed from 150 to 200 km s⁻¹. Which force changes, and which way does the positive ion initially deflect?
- Keep 200 km s⁻¹ and change the charge to −e. Explain why the initial deflection reverses while E/B stays the same.
- Reset. Increase B to 0.40 T. Calculate the new selected speed before reading the output. What happens initially to the positive ion still entering at 150 km s⁻¹?
Check the selector tasks
At 200 km s⁻¹ the magnetic force increases to 9.60 × 10⁻¹⁵ N; the electric force stays 7.20 × 10⁻¹⁵ N. The positive ion initially deflects upwards. With negative charge both forces reverse, so the initial resultant is downwards. Raising B to 0.40 T gives E/B = 112.5 km s⁻¹; the 150 km s⁻¹ positive ion is too fast and initially deflects upwards.
Magnetic-field explorer: predict, test and explain
Use the current two-tab explorer. It models perpendicular particle entry into a uniform field, with no energy losses. It does not simulate the wire-and-balance practical or a crossed-field velocity selector.
Open the magnetic-field explorer
1. Direction of the path
- Select positive charge and field into the page. Predict Up, then use Launch particle.
- Watch the force arrow point inward and the velocity arrow remain tangent. Describe what happens to speed, velocity direction and kinetic energy.
- Change only charge sign, then only field direction. Predict before each launch.
- Reverse both from the starting case. Explain why the original curvature returns.
2. Radius of the path
- Use Reset the activity, then select Radius of the path. Baseline: proton, B = 0.60 T, v = 3.0 × 106 m s−1; r ≈ 52.2 mm.
- Select Flux density, B. Change 0.60 T to 1.20 T. Predict r halves to about 26.1 mm.
- Select Speed, v. Its baseline controls reset; change speed from 3.0 to 6.0 × 106 m s−1. Predict r doubles to about 104.4 mm.
- Select m / |Q|. Compare proton, deuteron and alpha. At equal speed and B, deuteron and alpha radii are about twice the proton radius because their m/|Q| ratios are about twice as large.
Optional extension: the explorer’s cyclotron explanation separates electric acceleration across a gap from magnetic bending inside the dees. Detailed cyclotron operation is not a separately listed outcome in OCR 6.3.1–6.3.2.
Digital-balance practical: determining magnetic flux density
OCR includes measuring a uniform magnetic field using a current-carrying wire and a digital balance. A convenient arrangement puts the magnet on the balance and supports the wire independently, so magnetic forces transfer between the two objects without the wire pressing on the balance.
- Place the magnet on the balance and arrange a straight active wire segment perpendicular to B in the approximately uniform gap. Measure its length L within the field.
- Connect a suitable low-voltage supply, current control and ammeter in series with the wire. Record the no-current balance reading, or tare it. Ensure the wire and supports do not touch the magnet or balance pan.
- Pass a measured current I. After the reading settles, record its change Δm from the baseline. A display in grams must be converted to kilograms.
- Calculate force magnitude F = |Δm|g. The force on the magnet is opposite to that on the wire, by Newton’s third law.
- Vary current over a suitable range with several points and repeats. Keep L, wire angle, magnet position and wire position fixed.
- Plot F against I for one current direction. A best-fit gradient G = BL gives B = G/L. Check reversal of current produces an opposite signed change.
Worked example 6 · interpreting a mass-reading change
For I = 2.0 A and perpendicular L = 0.050 m, a balance reading rises by 1.83 g. Find B.
Δm = 1.83 × 10−3 kg. F = Δmg = 1.83 × 10−3 × 9.81 = 0.01795 N. B = F/(IL) = 0.0179523/(2.0 × 0.050) = 0.1795 T ≈ 0.18 T.
An increased reading means an additional downward force on the magnet, so the force on the wire is upwards. Using 1.83 directly as a force would be a unit error.
| I / A | F / N |
|---|---|
| 0.50 | 0.0120 |
| 1.00 | 0.0241 |
| 1.50 | 0.0358 |
| 2.00 | 0.0482 |
| 2.50 | 0.0599 |
Worked example 7 · gradient and uncertainty
A fit gives G = 0.0240 ± 0.0012 N A−1; L = 0.0600 ± 0.0010 m. Estimate B and its uncertainty.
B = G/L = 0.400 T. Using a conservative addition of fractional uncertainties for division, ΔB/B ≈ ΔG/G + ΔL/L = 0.0500 + 0.0167 = 0.0667. So ΔB ≈ 0.0267 T and a suitable result is B = 0.40 ± 0.03 T.
If a manufacturer quotes 0.42 T, it falls within this interval, so the values are consistent at this uncertainty. Mere numerical closeness without uncertainty is not enough to establish agreement.
Evaluation that identifies the cause
- Nonzero baseline: subtract or tare the no-current reading. A residual offset can give a nonzero intercept; do not force a line through the origin without justification.
- Balance resolution and random variation: repeat readings and estimate spread. Include error bars and use a large gradient triangle on the best-fit line. Repeats do not remove a systematic misalignment.
- Length and angle: measure only the active segment. A wire not perpendicular gives G = BL sin θ, so assuming 90° underestimates B when sin θ < 1.
- Nonuniform field: keep the segment in the uniform gap and fixed in position; moving it changes the field sampled.
- Heating and movement: use suitable current limiting and switch off between readings. Heating can change current or wire position; re-read the ammeter, rather than assuming the supply setting fixes I.
- Uncertainty in the result: use acceptable steepest/shallowest gradients based on error bars and include L uncertainty. If current uncertainty is significant, represent it too.
Quick retrieval check: why must the wire support be separate from the balance?
In this arrangement the balance must respond to the magnetic reaction force on the magnet, not to a variable mechanical load from the wire or its support touching the pan.
Magnetic-field graph skills
Worked example 8 · momentum from magnetic curvature
A singly charged ion follows a circular arc of radius 0.075 m in a perpendicular 0.180 T field. Its mass is not given. Find its momentum, using e = 1.60 × 10−19 C.
B|Q|v = mv²/r gives p = mv = B|Q|r. So p = 0.180 × 1.60 × 10−19 × 0.075 = 2.16 × 10−21 kg m s−1. Mass is unnecessary when field, charge and radius are known.
| Graph | Held constant | Gradient or shape |
|---|---|---|
| F against I for a wire | B, active L, θ = 90° | Gradient BL; force intercept ideally zero after baseline correction |
| F against v for a charge | B and |Q|, perpendicular entry | Gradient B|Q|; proportional relationship |
| r against 1/B | m, |Q| and speed | Gradient mv/|Q|; straight line through origin in the ideal model |
| r against v | m, |Q| and B | Gradient m/(B|Q|) |
| r² against accelerating V | m, |Q|, B; negligible initial speed | Gradient 2m/(B²|Q|) |
Read axis prefixes before calculating a gradient. If F is plotted in mN, convert the resulting mN A−1 gradient to N A−1 before finding B in teslas. A radius–field graph is not linear in B; it is linear in 1/B for a fixed-speed beam.
Common magnetic-field exam mistakes
- Using electron velocity as conventional current: reverse it for a negative charge.
- Missing the perpendicular condition: F = B|Q|v applies directly when v is perpendicular to B; F = BIL needs a perpendicular wire.
- Calling the magnetic force an extra centripetal force: magnetic force is the centripetal resultant in the simple model. Do not add a separate mv²/r force.
- Saying “the velocity accelerates”: the particle accelerates; velocity direction changes; speed stays constant.
- Claiming the magnetic field increases kinetic energy: its force is perpendicular to motion and does no work.
- Equating field strengths in a selector: balance electric and magnetic forces, |Q|E = B|Q|v.
- Using mass-proportional radii at equal p.d.: equal energy is not equal speed; equally charged ions then have r ∝ √m.
- Confusing radius and diameter: a semicircular entry-to-detector separation is 2r.
- Using a balance display as newtons: convert grams to kilograms, then multiply by g.
- Claiming an experiment agrees without an uncertainty: calculate a justified interval and compare values within it.
Original OCR A magnetic-fields exam practice · 47 marks
Apply the ideas to the sample-transfer instrument, actuator magnet and ion analyser. Attempt the questions before revealing the answers. State equations, convert to SI and keep intermediate values unrounded. Mark points here support self-checking; they are not copied OCR questions or an official OCR mark scheme.
1. [2 marks]
A sample-transfer instrument uses a long solenoid. Describe the field pattern well inside it, and explain how reversing the current affects that pattern.
Show worked answer and mark points
- The field is approximately uniform, with straight, parallel, equally spaced field lines along the solenoid axis, away from the ends. [1]
- Reversing the current reverses the field direction and exchanges the magnetic poles; for the same current magnitude the field strength is unchanged. [1]
2. [2 marks]
A technician describes an actuator magnet as having flux density 1.0 T. Define magnetic flux density using the force on a wire and explain the meaning of 1.0 T.
Show worked answer and mark points
- B is magnetic force per unit current per unit wire length when the straight wire is perpendicular to a uniform field: B = F/(IL). [1]
- A 1.0 m length carrying 1.0 A perpendicular to a 1.0 T field experiences a magnetic force of 1.0 N. [1]
3. [3 marks]
An actuator contains an 80 mm straight wire segment carrying 4.2 A in a 0.38 T field. The wire makes 35° with the field. Calculate the magnetic force magnitude.
Show worked answer and mark points
- Use F = BIL sin θ with L = 0.080 m and θ between current direction and B. [1]
- F = 0.38 × 4.2 × 0.080 × sin 35°. [1]
- F = 0.0732 N, or 0.073 N to two significant figures. [1]
4. [2 marks]
In the same actuator the current is to the right and the field is into the page. State the force direction. Explain the effect of reversing both current and field.
Show worked answer and mark points
- The magnetic force on the wire is upwards on the page. [1]
- Reversing either alone reverses the force; reversing both leaves the force upwards (and unchanged in magnitude if I and B magnitudes are unchanged). [1]
5. [3 marks]
A magnet rests on a balance below an independently supported wire. A perpendicular 60 mm wire segment carries 1.8 A. The balance reading increases by 2.40 g when the current flows. Find B, using g = 9.81 m s⁻², and state the direction of force on the wire.
Show worked answer and mark points
- Convert mass change to force: F = Δmg = 2.40 × 10⁻³ × 9.81 = 0.023544 N. [1]
- B = F/(IL) = 0.023544/(1.8 × 0.060) = 0.218 T ≈ 0.22 T. [1]
- The increased reading means a downward extra force on the magnet, so the equal and opposite magnetic force on the wire is upwards. [1]
6. [6 marks]
Plan a digital-balance experiment to determine B for an actuator magnet. Explain (a) the arrangement and measurements, (b) controlled variables and safe data collection, and (c) the graph, calculation and uncertainty evaluation. This is scaffolded six-mark-style practice, not an official OCR level-of-response rubric.
Show worked answer and mark points
- Place the magnet on a balance and independently support a straight wire in its uniform gap, perpendicular to B; measure only the active length L in the field. [1]
- Measure current with an ammeter in series; record the no-current baseline and the change in balance reading for each current. Convert Δm from g to kg and use F = |Δm|g. [1]
- Vary I over a suitable range, with several readings and repeats; keep L, orientation and magnet/wire position fixed, allowing the balance to settle. [1]
- Use a suitable low-voltage supply with current limiting; switch off between readings to limit heating and check reversal of current reverses the reading change. [1]
- Plot force magnitude against current magnitude for one direction; gradient G = BL, so B = G/L. Use a best-fit line and a large gradient triangle, not just two nearby raw readings. [1]
- Use reading spread/resolution to estimate force uncertainty and acceptable steepest/shallowest lines for gradient uncertainty; include L uncertainty and compare the resulting B interval with any quoted value. [1]
7. [3 marks]
An alpha-particle beam enters a 0.20 T magnetic field at right angles at 3.0 × 10⁶ m s⁻¹. An alpha particle has mass 6.64 × 10⁻²⁷ kg and charge +3.2 × 10⁻¹⁹ C. Determine its magnetic force and acceleration magnitudes.
Show worked answer and mark points
- F = B|Q|v = 0.20 × 3.2 × 10⁻¹⁹ × 3.0 × 10⁶ = 1.92 × 10⁻¹³ N. [1]
- a = F/m = (1.92 × 10⁻¹³)/(6.64 × 10⁻²⁷). [1]
- a = 2.89 × 10¹³ m s⁻² ≈ 2.9 × 10¹³ m s⁻²; it is perpendicular to the instantaneous velocity. [1]
8. [3 marks]
An electron enters a uniform field into the page, moving to the right. State its initial force direction and explain why its speed stays constant even though it accelerates.
Show worked answer and mark points
- The force is downwards because electron charge is negative; it is opposite to the force on a positive charge moving the same way. [1]
- The magnetic force remains perpendicular to the instantaneous velocity and changes its direction, giving centripetal acceleration. [1]
- It does no work, so kinetic energy and hence speed stay constant, assuming no other work or energy losses. [1]
9. [3 marks]
Protons travel perpendicular to a 0.50 T field at 2.0 × 10⁶ m s⁻¹. Use m = 1.67 × 10⁻²⁷ kg and |Q| = 1.60 × 10⁻¹⁹ C to find the orbit radius and the distance between entry and detection points after a semicircle.
Show worked answer and mark points
- B|Q|v = mv²/r, giving r = mv/(B|Q|). [1]
- r = (1.67 × 10⁻²⁷ × 2.0 × 10⁶)/(0.50 × 1.60 × 10⁻¹⁹) = 0.04175 m ≈ 0.042 m. [1]
- Semicircle entry-to-detector separation is 2r = 0.0835 m ≈ 0.084 m, not r. [1]
10. [3 marks]
Two singly positive isotope ions have masses 20 u and 22 u. Compare their magnetic orbit radii (22 u / 20 u) when (a) their speeds are equal and (b) they start from negligible speed and are accelerated through the same potential difference. The analyser field is unchanged.
Show worked answer and mark points
- At equal speed and equal |Q|, r ∝ m, so the ratio is 22/20 = 1.10. [1]
- At equal accelerating p.d., ½mv² = |Q|V, so r² = 2mV/(B²|Q|), giving r ∝ √m for these equally charged ions. [1]
- The radius ratio is √(22/20) = 1.049 ≈ 1.05; the heavier ion is slower at equal kinetic energy. [1]
11. [4 marks]
An electron starts from negligible speed and is accelerated through 2.00 kV, then enters a 0.0120 T field at right angles. Use me = 9.11 × 10⁻³¹ kg and e = 1.60 × 10⁻¹⁹ C. Calculate its non-relativistic speed and orbit radius.
Show worked answer and mark points
- Energy gain = eV = 1.60 × 10⁻¹⁹ × 2000 = 3.20 × 10⁻¹⁶ J. [1]
- Use ½mev² = eV, so v = √(2eV/me). [1]
- v = 2.65 × 10⁷ m s⁻¹. [1]
- r = mev/(Be) = 0.0126 m = 12.6 mm using the unrounded speed. [1]
12. [4 marks]
A positive-ion velocity selector has an upper positive plate and lower negative plate, 0.030 m apart, with p.d. 1200 V. The ions enter to the right and B = 0.160 T into the page. Determine the selected speed and explain what happens initially to faster ions.
Show worked answer and mark points
- E = V/d = 1200/0.030 = 4.00 × 10⁴ V m⁻¹, directed downwards. [1]
- For undeflected motion the upward magnetic force and downward electric force balance: B|Q|v = |Q|E. [1]
- v = E/B = (4.00 × 10⁴)/0.160 = 2.50 × 10⁵ m s⁻¹. [1]
- Faster positive ions have B|Q|v > |Q|E, so their initial resultant force and deflection are upwards. [1]
13. [3 marks]
For a fixed-speed proton beam, a graph of orbit radius r in metres against 1/B in T⁻¹ is straight, through the origin, with gradient 0.0240 m T. Use m = 1.67 × 10⁻²⁷ kg and e = 1.60 × 10⁻¹⁹ C to find the speed.
Show worked answer and mark points
- r = (mv/e)(1/B), so the gradient is mv/e. [1]
- v = gradient × e/m = (0.0240 × 1.60 × 10⁻¹⁹)/(1.67 × 10⁻²⁷). [1]
- v = 2.30 × 10⁶ m s⁻¹. [1]
14. [3 marks]
The selector in question 12 is used for a negative-ion beam entering in the same direction. Explain whether its selected speed changes, and the initial force direction for a negative ion travelling faster than that speed.
Show worked answer and mark points
- The selected speed is unchanged: reversing Q reverses both forces, and |Q| cancels in |Q|E = B|Q|v. [1]
- For a negative ion the electric force is upwards and the magnetic force is downwards. [1]
- At a speed greater than E/B the magnetic force magnitude is larger, so the initial resultant force is downwards. [1]
15. [3 marks]
The actuator practical gives an F–I gradient of 0.026 ± 0.002 N A⁻¹, with active length 0.065 ± 0.002 m. Calculate B and a conservative uncertainty by adding fractional uncertainties. Decide whether a manufacturer’s 0.45 T value is consistent with this interval.
Show worked answer and mark points
- B = gradient/L = 0.026/0.065 = 0.400 T. [1]
- ΔB/B ≈ 0.002/0.026 + 0.002/0.065 = 0.1077, so ΔB ≈ 0.0431 T; report about 0.40 ± 0.04 T. [1]
- The unrounded interval is about 0.357–0.443 T. The quoted 0.45 T lies outside it, so it is not consistent at the stated uncertainty; this alone does not identify which value or assumption is at fault. [1]
OCR A H556 specification coverage
The ten learning outcomes in 6.3.1–6.3.2 are taught below; circle-force and energy calculations also draw on earlier mechanics and electricity. Induction and transformers are deliberately outside this page’s core scope.
- 6.3.1(a) Sources of magnetic fields
- 6.3.1(b) Field lines and mapping
- 6.3.1(c) Straight wire, flat coil and solenoid patterns
- 6.3.1(d) Fleming’s left-hand rule
- 6.3.1(e)(i) Force on a current-carrying conductor
- 6.3.1(e)(ii) Wire and digital-balance investigation
- 6.3.1(f) Flux density and the tesla
- 6.3.2(a) Force on a perpendicular moving charge
- 6.3.2(b) Uniform-field circular orbits
- 6.3.2(c) Crossed fields and velocity selection
Cross-checked against the official OCR A-level Physics A specification, version 3.0. The coil and solenoid figures are schematic cross-sectional field patterns, not quantitative 3D field maps.
Magnetic-fields revision questions
What is magnetic flux density and what does one tesla mean?
For a straight wire perpendicular to a uniform field, B = F/(IL). One tesla gives a force of one newton on a one-metre wire carrying one ampere at right angles to the field. The SI unit is T = N A⁻¹ m⁻¹.
When do I use F = BIL sin θ rather than F = BQv?
Use BIL sin θ for the magnetic force on a straight current-carrying wire of active length L. Use B|Q|v for the force magnitude on an individual charge moving perpendicular to B. The angle for the wire is between conventional current and field.
How do I find the magnetic force direction for an electron?
Use Fleming’s left-hand rule with current opposite to electron motion, or first find the force on a positive charge moving in that direction and reverse it. Thumb gives force, first finger field and second finger conventional current.
Why does a particle accelerate without speeding up in a magnetic field?
The force is perpendicular to velocity. It changes velocity direction, providing centripetal acceleration, but does no work. With no other work or energy losses, kinetic energy and speed stay constant.
How do you calculate the radius of a charged-particle orbit?
For perpendicular entry into a uniform magnetic field, set B|Q|v = mv²/r to obtain r = mv/(B|Q|). Use charge magnitude for radius and charge sign for the direction of curvature. A semicircular detector separation is 2r.
Why is the selected speed in crossed fields v = E/B?
The fields must be arranged so the electric and magnetic forces oppose. Undeflected particles have |Q|E = B|Q|v, giving v = E/B. The forces balance, not the field strengths, whose units differ.
How can a digital balance measure magnetic flux density?
Independently support a wire perpendicular to the uniform field of a magnet on a balance. Convert the change in mass reading into force using |Δm|g. Plot F against I: its gradient is BL, so B = gradient/L. Keep wire length, angle and position fixed.
Do heavier ions always have proportionally larger magnetic radii?
Only if speed, charge magnitude and field are unchanged. At equal accelerating potential difference with equal charge, r is proportional to the square root of mass because the heavier ion travels more slowly. State what is held constant before using a ratio.
Continue your OCR A revision
Return to the Module 6 revision hub or OCR Paper 2 hub. Compare forces using electric fields, revisit circular motion and strengthen the current convention with charge and current. Then try problem-solving practice or MCQ practice.