Specification: AQA A-level Physics (7408) 3.7.5.2 | Paper: 2 | Data and formulae booklet: given in the exam
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Force on a moving charge: a charge Q moving with velocity v at right angles to a magnetic field of flux density B experiences a force F = BQv. The direction is given by Fleming’s left-hand rule, with the current in the direction of motion for a positive charge and opposite to the motion for a negative charge (such as an electron).
Circular motion in a field: the magnetic force is always perpendicular to the velocity, so it does no work: the speed and kinetic energy stay constant and the force provides the centripetal force, BQv = mv² ÷ r, giving r = mv ÷ BQ.
Period of the orbit: T = 2πm ÷ BQ, which is independent of the speed and of the radius. This is the principle of the cyclotron, where the alternating p.d. has a constant frequency f = BQ ÷ 2πm.
The equation
force = magnetic flux density × charge × velocity F = BQv (velocity at right angles to the field) and for a circular path r = mv ÷ BQ
| Quantity | Symbol | Unit |
|---|---|---|
| force on the charge | F | newtons (N) |
| magnetic flux density | B | teslas (T) |
| charge | Q | coulombs (C) |
| velocity (perpendicular to the field) | v | metres per second (m s⁻¹) |
| mass of the particle | m | kilograms (kg) |
| radius of the circular path | r | metres (m) |
Rearranged: B = F ÷ Qv | v = F ÷ BQ | from BQv = mv² ÷ r: r = mv ÷ BQ, v = BQr ÷ m, m = BQr ÷ v | The data booklet gives F = BQv and the centripetal force F = mv² ÷ r; you must combine them yourself. For an electron or proton, Q = e = 1.60 × 10−19 C.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An electron travels east at 4.0 × 106 m s⁻¹ through a uniform magnetic field of flux density 0.30 T directed vertically downwards. (a) Calculate the magnitude of the force on the electron, to 2 significant figures. (b) State the direction of the force. (e = 1.60 × 10−19 C)
Show solution
F: F = BQv
I: F = 0.30 × 1.60 × 10−19 × 4.0 × 106
F: F is already the subject
A: (a) 1.92 × 10−13 N = 1.9 × 10−13 N (2 s.f.)
(b) The electron is negative, so the conventional current is to the west. First finger down (field), second finger west (current): the thumb points south.
Watch out: for electrons the second finger must point opposite to their motion. This is the most common direction error in exams.
Q2 (Core) A proton moves at 2.0 × 106 m s⁻¹ at right angles to a magnetic field of flux density 0.80 T. Calculate the force on the proton, to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
F: F = BQv
I: F = 0.80 × 1.60 × 10−19 × 2.0 × 106
F: F is already the subject
A: 2.56 × 10−13 N = 2.6 × 10−13 N (2 s.f.)
Q3 (Core) A proton moving at 6.0 × 106 m s⁻¹ at right angles to a magnetic field experiences a force of 4.8 × 10−15 N. Calculate the magnetic flux density in mT. (e = 1.60 × 10−19 C)
Show solution
F: F = BQv
I: 4.8 × 10−15 = B × 1.60 × 10−19 × 6.0 × 106
F: B = 4.8 × 10−15 ÷ (1.60 × 10−19 × 6.0 × 106) = 5.0 × 10−3 T
A: 5.0 mT
Q4 (Core) A proton travels at 3.0 × 106 m s⁻¹ at right angles to a uniform magnetic field of flux density 0.50 T. Calculate the radius of its circular path in cm, to 2 significant figures. (mₚ = 1.67 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
F: BQv = mv² ÷ r, so r = mv ÷ BQ
I: r = (1.67 × 10−27 × 3.0 × 106) ÷ (0.50 × 1.60 × 10−19)
F: r is already the subject
A: 0.062625 m = 6.3 cm (2 s.f.)
Q5 (Exam) Electrons are accelerated from rest through a p.d. of 250 V and then enter a uniform magnetic field of flux density 1.2 mT at right angles to the field. Calculate (a) the speed of the electrons and (b) the radius of their path, both to 2 significant figures. (mₑ = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 1.2 mT = 1.2 × 10−3 T
Step 1 – (a) speed
eV = ½mv², so v = √(2eV ÷ m) = √(2 × 1.60 × 10−19 × 250 ÷ 9.11 × 10−31) = 9.370… × 106 m s⁻¹
A: 9.4 × 106 m s⁻¹ (2 s.f.)
Step 2 – (b) radius
F: r = mv ÷ BQ
I: r = (9.11 × 10−31 × 9.370… × 106) ÷ (1.2 × 10−3 × 1.60 × 10−19)
A: 0.04446… m = 4.4 cm (2 s.f.)
Watch out: keep the unrounded speed for step 2, and remember mT. Using 1.2 T gives a radius 1000 times too small.
Q6 (Exam) Protons circle in a uniform magnetic field of flux density 0.20 T. (a) Calculate the radius of the path of a proton moving at 2.0 × 106 m s⁻¹, to 2 significant figures. (b) Show that the time for one orbit is the same for a proton moving at 4.0 × 106 m s⁻¹, and calculate it to 2 significant figures. (mₚ = 1.67 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) radius
F: r = mv ÷ BQ
I: r = (1.67 × 10−27 × 2.0 × 106) ÷ (0.20 × 1.60 × 10−19)
A: 0.104375 m = 0.10 m (2 s.f.)
Step 2 – (b) period
at 4.0 × 106 m s⁻¹ the radius doubles to 0.20875 m. T = 2πr ÷ v, and both r and v have doubled, so T is unchanged
in general T = 2πr ÷ v = 2πm ÷ BQ = 2π × 1.67 × 10−27 ÷ (0.20 × 1.60 × 10−19)
A: 3.279… × 10−7 s = 3.3 × 10−7 s (2 s.f.) at either speed
Q7 (Exam) In a mass spectrometer, singly charged positive neon ions all move at 4.0 × 105 m s⁻¹ as they enter a uniform magnetic field of flux density 0.25 T at right angles to the field. They travel through a semicircle before hitting a detector. The masses of neon-20 and neon-22 ions are 3.32 × 10−26 kg and 3.65 × 10−26 kg. Calculate the separation of the two points where the ions hit the detector, to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
Step 1 – radius for neon-20
F: r = mv ÷ BQ
I: r = (3.32 × 10−26 × 4.0 × 105) ÷ (0.25 × 1.60 × 10−19) = 0.332 m
Step 2 – radius for neon-22
r = (3.65 × 10−26 × 4.0 × 105) ÷ (0.25 × 1.60 × 10−19) = 0.365 m
Step 3 – separation (each ion lands one diameter from the entry point)
separation = 2 × (0.365 − 0.332) = 2 × 0.033
A: 0.066 m = 6.6 cm (2 s.f.)
Watch out: after a semicircle the ions land a diameter (2r) from the entry slit, so the separation is twice the difference in radii.
Q8 (Exam) A positive ion with charge +e moves in a circle of radius 0.13 m in a magnetic field of flux density 0.40 T at a speed of 2.5 × 106 m s⁻¹. (a) Calculate the mass of the ion, to 2 significant figures. (b) Suggest what the ion is. (mₚ = 1.67 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
F: BQv = mv² ÷ r, so m = BQr ÷ v
I: m = (0.40 × 1.60 × 10−19 × 0.13) ÷ 2.5 × 106
F: m is already the subject
A: (a) 3.328 × 10−27 kg = 3.3 × 10−27 kg (2 s.f.)
(b) m ÷ mₚ = 3.328 × 10−27 ÷ 1.67 × 10−27 = 1.99…, about 2 nucleon masses with charge +e, so it is a deuteron (a hydrogen-2 nucleus)
Q9 (Stretch) A velocity selector has two parallel plates 2.0 cm apart with a p.d. of 400 V between them, and a uniform magnetic field of flux density 0.050 T at right angles to both the electric field and the beam. Singly charged positive ions that pass straight through then enter a second magnetic field of flux density 0.25 T, travel through a semicircle and hit a detector 0.73 m from the entry slit. (a) Calculate the speed of the ions that pass straight through the selector. (b) Calculate the mass of the ions, to 2 significant figures, and estimate their mass number. (mₚ = 1.67 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 2.0 cm = 0.020 m
Step 1 – electric field strength
E = V ÷ d = 400 ÷ 0.020 = 20 000 V m⁻¹
Step 2 – (a) undeflected speed
the ions go straight when the electric and magnetic forces balance: EQ = BQv, so v = E ÷ B = 20 000 ÷ 0.050
A: 4.0 × 105 m s⁻¹
Step 3 – (b) mass
r = 0.73 ÷ 2 = 0.365 m; m = BQr ÷ v = (0.25 × 1.60 × 10−19 × 0.365) ÷ 4.0 × 105
A: 3.65 × 10−26 kg = 3.7 × 10−26 kg (2 s.f.); m ÷ mₚ = 21.8…, so a mass number of about 22
Watch out: the charge cancels in v = E ÷ B, so the selector picks out one speed whatever the charge or mass. Faster ions feel a larger magnetic force and are deflected one way; slower ions are deflected the other way.
Q10 (Stretch) A cyclotron accelerates protons in a uniform magnetic field of flux density 1.5 T. (a) Show that the frequency of the alternating p.d. needed is independent of the speed of the protons, and calculate it, to 2 significant figures. (b) The radius of the dees is 0.50 m. Calculate the maximum kinetic energy of the protons in MeV, to 2 significant figures. (mₚ = 1.67 × 10−27 kg, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)
Show solution
Step 1 – (a) frequency
r = mv ÷ BQ, so the time for one orbit T = 2πr ÷ v = 2πm ÷ BQ, which does not contain v. The p.d. must reverse twice per orbit, so it has frequency f = 1 ÷ T = BQ ÷ 2πm
I: f = (1.5 × 1.60 × 10−19) ÷ (2π × 1.67 × 10−27)
A: 2.287… × 107 Hz = 23 MHz (2 s.f.)
Step 2 – (b) maximum speed
v = BQr ÷ m = (1.5 × 1.60 × 10−19 × 0.50) ÷ 1.67 × 10−27 = 7.185… × 107 m s⁻¹
Step 3 – maximum kinetic energy
Eₖ = ½mv² = ½ × 1.67 × 10−27 × (7.185… × 107)² = 4.311… × 10−12 J
A: 4.311… × 10−12 ÷ 1.60 × 10−13 = 26.9… = 27 MeV (2 s.f.)
Watch out: this ignores relativistic effects. At about 0.24c the proton’s mass effectively increases, so in a real cyclotron the orbits slowly fall out of step with a fixed-frequency p.d.
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