Specification: AQA A-level Physics (7408) 3.12.1.3 Turning points option | Paper: 3 Section B | Data and formulae booklet: F = Bev = mv²/r and ½mv² = eV are given in the exam; e/m = v/(Br), e/m = 2V/(B²r²) and v = E/B are not, so learn how to derive them
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Specific charge: the charge per unit mass of a particle, e ÷ m, measured in C kg⁻¹.
Circular path in a magnetic field: in a fine-beam tube the magnetic force Bev is perpendicular to the velocity and provides the centripetal force, so Bev = mv² ÷ r, giving r = mv ÷ Be.
Thomson’s crossed fields: when the electric and magnetic forces on the beam are equal and opposite the beam is undeflected, so eE = Bev and v = E ÷ B = V ÷ Bd. Thomson found e ÷ m for the electron was about 1800 times larger than for the hydrogen ion, the largest then known, so the electron is a particle much less massive than an atom.
The equation
specific charge = speed ÷ (magnetic flux density × radius of path) e/m = v/(Br) combined with ½mv² = eV: e/m = 2V/(B²r²) crossed fields: v = E/B = V/(Bd)
| Quantity | Symbol | Unit |
|---|---|---|
| specific charge of the electron | e/m | coulombs per kilogram (C kg⁻¹) |
| speed of the electrons | v | metres per second (m s⁻¹) |
| magnetic flux density | B | teslas (T) |
| radius of the circular path | r | metres (m) |
| anode (accelerating) pd | V | volts (V) |
| electric field strength between the plates | E | volts per metre (V m⁻¹), equivalent to N C⁻¹ |
| plate separation | d | metres (m) |
Rearranged: r = mv ÷ Be | v = (e/m)Br | B = mv ÷ er | r = (1/B)√(2mV ÷ e) | Derive e/m = 2V/(B²r²) from Bev = mv²/r and ½mv² = eV, convert mT to T (× 10−3) and cm to m, and halve a measured diameter to get r.
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) In a fine-beam tube, electrons travelling at 9.4 × 106 m s⁻¹ move in a circle of radius 0.049 m in a uniform magnetic field of flux density 1.1 × 10−3 T. Calculate the specific charge of the electron, to 2 significant figures.
Show solution
F: e/m = v/(Br)
I: e/m = 9.4 × 106 ÷ (1.1 × 10−3 × 0.049)
F: e/m is already the subject
A: 1.743… × 1011 C kg⁻¹ = 1.7 × 1011 C kg⁻¹ (2 s.f.)
Q2 (Core) An electron travels at 8.0 × 106 m s⁻¹ perpendicular to a uniform magnetic field of flux density 9.0 × 10−4 T. Calculate the radius of its circular path, to 3 significant figures. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
F: r = mv ÷ Be
I: r = (9.11 × 10−31 × 8.0 × 106) ÷ (9.0 × 10−4 × 1.60 × 10−19)
F: r is already the subject
A: 0.05061… m = 0.0506 m (5.06 cm, 3 s.f.)
Q3 (Core) An electron beam passes undeflected between two parallel plates 5.0 cm apart with a pd of 2.0 kV between them. A uniform magnetic field of flux density 1.5 mT acts perpendicular to both the beam and the electric field. Calculate the speed of the electrons, to 3 significant figures.
Show solution
Convert first: 5.0 cm = 0.050 m; 2.0 kV = 2000 V; 1.5 mT = 1.5 × 10−3 T
F: eE = Bev, so v = E/B = V/(Bd)
I: v = 2000 ÷ (1.5 × 10−3 × 0.050) = 40 000 ÷ 1.5 × 10−3
F: v is already the subject
A: 2.666… × 107 m s⁻¹ = 2.67 × 107 m s⁻¹ (3 s.f.)
Watch out: v = E/B needs the field strength E = V/d in V m⁻¹, not the pd. Dividing 2000 V by B gives a speed 20 times too small.
Q4 (Core) Electrons are accelerated from rest through 250 V in a fine-beam tube. Calculate the magnetic flux density needed to make the beam form a circle of diameter 10.0 cm, to 3 significant figures. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: diameter 10.0 cm, so r = 5.00 cm = 0.0500 m
Step 1 – speed from ½mv² = eV
v = √(2 × 1.60 × 10−19 × 250 ÷ 9.11 × 10−31) = 9.370… × 106 m s⁻¹
Step 2 – flux density
F: r = mv ÷ Be
I: 0.0500 = (9.11 × 10−31 × 9.370… × 106) ÷ (B × 1.60 × 10−19)
F: B = mv ÷ er = (9.11 × 10−31 × 9.370… × 106) ÷ (1.60 × 10−19 × 0.0500)
A: 1.067… × 10−3 T = 1.07 mT (3 s.f.)
Watch out: using the diameter as r. The tube shows you the diameter of the circle, so halve it first.
Q5 (Exam) In a fine-beam tube the anode pd is 300 V and the magnetic flux density is 0.95 mT. The beam forms a circle of radius 6.2 cm. (a) Show that e/m = 2V/(B²r²). (b) Calculate the specific charge of the electron, to 2 significant figures.
Show solution
Step 1 – (a) derivation
Bev = mv²/r gives v = Ber/m; substitute into ½mv² = eV: ½m(Ber/m)² = eV, so B²e²r²/(2m) = eV and e/m = 2V/(B²r²)
Convert first: 0.95 mT = 9.5 × 10−4 T; 6.2 cm = 0.062 m
Step 2 – (b) specific charge
F: e/m = 2V/(B²r²)
I: e/m = (2 × 300) ÷ ((9.5 × 10−4)² × 0.062²) = 600 ÷ 3.469… × 10−9
F: e/m is already the subject
A: 1.729… × 1011 C kg⁻¹ = 1.7 × 1011 C kg⁻¹ (2 s.f.)
Watch out: B and r are both squared. Squaring only one of them is the most common slip in this calculation.
Q6 (Exam) In a Thomson-style experiment an electron beam passes undeflected through crossed fields: the plates are 4.0 cm apart with a pd of 1500 V, and the magnetic flux density is 1.2 mT. The electric field is then switched off, and the beam moves in a circular arc of radius 15 cm in the magnetic field alone. Calculate (a) the speed of the electrons and (b) the specific charge of the electron, to 2 significant figures.
Show solution
Convert first: 4.0 cm = 0.040 m; 1.2 mT = 1.2 × 10−3 T; 15 cm = 0.15 m
Step 1 – (a) speed from the balanced fields
E = V/d = 1500 ÷ 0.040 = 37 500 V m⁻¹; v = E/B = 37 500 ÷ 1.2 × 10−3 = 3.125 × 107 m s⁻¹ = 3.1 × 107 m s⁻¹ (2 s.f.)
Step 2 – (b) specific charge from the arc
F: e/m = v/(Br)
I: e/m = 3.125 × 107 ÷ (1.2 × 10−3 × 0.15)
F: e/m is already the subject
A: 1.736… × 1011 C kg⁻¹ = 1.7 × 1011 C kg⁻¹ (2 s.f.)
Watch out: in the crossed fields the beam is straight, so there is no radius yet; r only applies once the electric field is switched off. Combining the steps gives e/m = E/(B²r).
Q7 (Exam) (a) Calculate the specific charge of the electron and of the hydrogen ion (a proton), each to 3 significant figures. (b) Calculate the ratio of the two values, to 3 significant figures, and explain what Thomson concluded from his result. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg, mₚ = 1.673 × 10−27 kg)
Show solution
Step 1 – (a) electron
F: specific charge = e ÷ m
I: e ÷ mₑ = 1.60 × 10−19 ÷ 9.11 × 10−31
A: 1.76 × 1011 C kg⁻¹ (3 s.f.)
Step 2 – (a) hydrogen ion
e ÷ mₚ = 1.60 × 10−19 ÷ 1.673 × 10−27 = 9.56 × 107 C kg⁻¹ (3 s.f.)
Step 3 – (b) ratio
ratio = mₚ ÷ mₑ = 1836.4… = 1840 (3 s.f.)
Thomson’s value of e/m was about 1800 times the largest specific charge then known (the hydrogen ion). If the electron carries the same size charge as the hydrogen ion, its mass must be much smaller, so it is a sub-atomic particle, a constituent of all atoms.
Watch out: Thomson measured e/m, not e or m separately. His result alone could also mean a much larger charge; Millikan’s later measurement of e settled it.
Q8 (Exam) In a fine-beam tube the beam forms a circle of radius 4.5 cm when the anode pd is 220 V. The anode pd is increased to 330 V with the magnetic field unchanged. (a) Calculate the new radius, to 2 significant figures. (b) Calculate the factor by which B must be changed to bring the radius back to 4.5 cm, to 3 significant figures.
Show solution
Step 1 – how r depends on V and B
from e/m = 2V/(B²r²): r = (1/B)√(2mV ÷ e), so r ∝ √V at fixed B and r ∝ 1/B at fixed V
Step 2 – (a) new radius
I: r = 4.5 × √(330 ÷ 220) = 4.5 × √1.5
A: 5.51… cm = 5.5 cm (2 s.f.)
Step 3 – (b) restoring the radius
to cancel the √1.5 increase, B must be multiplied by √1.5
A: B must increase by a factor of 1.22 (3 s.f.)
Watch out: faster electrons are harder to bend, so a higher anode pd gives a larger circle, not a smaller one.
Q9 (Stretch) Electrons are accelerated from rest through 2.50 kV and enter crossed fields. The plates are 3.0 cm apart with a pd of 900 V, and the beam is undeflected when the magnetic flux density is 1.01 mT. (a) Calculate the speed of the electrons, to 3 significant figures. (b) Use this speed and the anode pd to find e/m, to 3 significant figures, and compare it with the value of e/mₑ calculated from the data below. (c) The electric field is switched off. Calculate the radius of the beam’s path, to 3 significant figures. (e = 1.60 × 10−19 C, mₑ = 9.11 × 10−31 kg)
Show solution
Convert first: 2.50 kV = 2500 V; 3.0 cm = 0.030 m; 1.01 mT = 1.01 × 10−3 T
Step 1 – (a) speed
E = 900 ÷ 0.030 = 30 000 V m⁻¹; v = E/B = 30 000 ÷ 1.01 × 10−3 = 2.970… × 107 m s⁻¹
A: 2.97 × 107 m s⁻¹ (3 s.f.)
Step 2 – (b) specific charge
F: ½mv² = eV, so e/m = v² ÷ 2V
I: e/m = (2.970… × 107)² ÷ (2 × 2500)
A: 1.764… × 1011 C kg⁻¹ = 1.76 × 1011 C kg⁻¹ (3 s.f.)
from the data: 1.60 × 10−19 ÷ 9.11 × 10−31 = 1.756… × 1011 C kg⁻¹, so the result is about 0.5% too high: good agreement
Step 3 – (c) radius in B alone
F: e/m = v/(Br), so r = v ÷ (B × e/m)
I: r = 2.970… × 107 ÷ (1.01 × 10−3 × 1.764… × 1011)
A: 0.167 m (3 s.f.)
Watch out: v = E/B uses the plate pd (900 V) to find E, while ½mv² = eV uses the anode pd (2500 V). Mixing up the two pds is a common error.
Q10 (Stretch) A fine-beam tube is operated at a fixed anode pd of 250 V. The flux density B is varied and the radius r of the beam is measured. A graph of r (y-axis) against 1/B (x-axis) is a straight line through the origin, passing through the point (1/B = 800 T⁻¹, r = 4.32 cm). (a) Show that the gradient of the graph is √(2V ÷ (e/m)). (b) Use the graph to find e/m, to 2 significant figures. (c) Find the radius when B = 1.20 mT, to 2 significant figures.
Show solution
Step 1 – (a) gradient
from e/m = 2V/(B²r²): r² = 2V ÷ ((e/m)B²), so r = √(2V ÷ (e/m)) × (1/B). This is y = kx with k = √(2V ÷ (e/m))
Convert first: 4.32 cm = 0.0432 m
Step 2 – (b) e/m from the gradient
gradient = 0.0432 ÷ 800 = 5.4 × 10−5 T m
F: gradient² = 2V ÷ (e/m), so e/m = 2V ÷ gradient²
I: e/m = (2 × 250) ÷ (5.4 × 10−5)²
A: 1.714… × 1011 C kg⁻¹ = 1.7 × 1011 C kg⁻¹ (2 s.f.)
Step 3 – (c) radius at 1.20 mT
1/B = 1 ÷ 1.20 × 10−3 = 833.3… T⁻¹; r = 5.4 × 10−5 × 833.3…
A: 0.045 m = 4.5 cm (2 s.f.)
Watch out: plotting r against B gives a curve, not a straight line. Choose the axes so the equation matches y = mx (or y = mx + c) before finding a gradient.
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