Physics • Year 12 • Module 6: Electromagnetism • Lesson 6
Electric vs Magnetic Fields — Integrated Comparison
Apply the velocity selector formula, force direction rules, and trajectory concepts to real data, scenario-based problems, and quantitative calculations.
1. Velocity selector — calculate and compare
A velocity selector has a uniform electric field E = 6000 V/m directed downward and a uniform magnetic field B = 0.030 T directed into the page. Answer the following. 10 marks
1.1 Show that the selector speed (the speed at which a particle travels straight through) is v = 2.0 × 105 m/s. 2 marks
1.2 A proton (m = 1.67 × 10−27 kg, q = +1.60 × 10−19 C) enters at the selector speed. Calculate the magnitudes of (a) the electric force and (b) the magnetic force on the proton, and confirm they are equal. 3 marks
1.3 An electron (same E, B, and selector speed) enters the region. State the direction of each force on the electron and explain why it also travels straight at v = E/B. 2 marks
1.4 A particle enters the same selector at v = 3.0 × 105 m/s (faster than the selector speed). State which force dominates and predict the direction of deflection for a positive charge. Justify your answer. 3 marks
2. Interpret field comparison data
A student records the following observations for three different particle scenarios. Complete the missing columns. 9 marks (1 each)
| # | Situation | Field type(s) present | Does the field do work on the particle? | Expected trajectory | Speed change? |
|---|---|---|---|---|---|
| 2.1 | Electron enters perpendicular to a uniform electric field between two plates | Electric only | |||
| 2.2 | Proton enters perpendicular to a uniform magnetic field directed out of the page | Magnetic only | |||
| 2.3 | Electron enters a velocity selector at exactly v = E/B | Both E and B (crossed) |
3. Interpret a radius vs mass graph from a mass spectrometer
In a mass spectrometer experiment, singly charged ions (q = +e) are accelerated and then passed through a velocity selector with E = 4000 V/m and B1 = 0.020 T. They then enter a second uniform magnetic field B2 = 0.50 T and follow a semicircular path. The radius of the path for six different ions is plotted below. 8 marks
Figure 3. Radius of semicircular ion path vs ion mass for singly charged ions in a mass spectrometer. E = 4000 V/m, B1 = 0.020 T (velocity selector), B2 = 0.50 T (analysis field). Illustrative data.
3.1 Calculate the selector speed v = E/B1. Show your working. 2 marks
3.2 Using r = mv/(qB2), calculate the expected radius for an ion of mass m = 12 u (1 u = 1.66 × 10−27 kg). Compare your result with the graph. 3 marks
3.3 Describe the relationship shown in the graph between ion mass and radius. Explain this relationship using the formula r = mv/(qB2). 3 marks
4. Predict and justify — the ANSTO SIMS instrument
The Australian Nuclear Science and Technology Organisation (ANSTO) at Lucas Heights, NSW, operates a Secondary Ion Mass Spectrometry (SIMS) instrument used to analyse materials by mass spectrometry. A simplified model uses E = 5000 V/m, Bselector = 0.025 T and Banalysis = 0.60 T. 7 marks
4.1 Calculate the selector speed for this instrument. 1 mark
4.2 An ion with mass 28 u and charge +e enters the analysis field at the selector speed. Calculate the radius of its circular path. (1 u = 1.66 × 10−27 kg, e = 1.60 × 10−19 C.) 3 marks
4.3 A second ion has the same charge (+e) but twice the mass (56 u). Predict its radius and explain whether it would land at the same detection point as the 28 u ion, without repeating the full calculation. 2 marks
4.4 The instrument operator doubles the electric field (to 10 000 V/m) but keeps Bselector unchanged. Predict what happens to (i) the selector speed, (ii) the radius of the 28 u ion in the analysis field. 1 mark
Q1.1 — Selector speed (2 marks)
v = E/B = 6000 / 0.030 = 2.0 × 105 m/s. [1 mark for correct substitution; 1 mark for correct answer with units]
Q1.2 — Electric and magnetic forces on proton (3 marks)
FE = qE = (1.60 × 10−19)(6000) = 9.6 × 10−16 N [1]. FB = qvB = (1.60 × 10−19)(2.0 × 105)(0.030) = 9.6 × 10−16 N [1]. FE = FB, confirming the proton travels straight [1].
Q1.3 — Electron in same fields (2 marks)
For the electron: the electric force is upward (E-field is downward; electron is negative, so force reverses to upward) [½ mark]. The magnetic force is downward (apply RH rule for positive charge → upward on positive; reverse for negative electron → downward) [½ mark]. Both forces reverse compared to the proton, but they still oppose each other with the same magnitude at v = E/B, so the electron also travels straight [1 mark]. Award 1 mark for explaining that q cancels and 1 mark for correct force directions.
Q1.4 — Faster particle deflection (3 marks)
At v = 3.0 × 105 m/s the magnetic force is FB = qvB = q(3.0 × 105)(0.030) > FE = q(6000) [1 mark]. Specifically FB = 1.44 × 10−15 N compared to FE = 9.6 × 10−16 N, so the magnetic force dominates [1 mark]. For a positive charge with v to the right and B into the page, the magnetic force is upward (RH rule: v × B is upward). Since FB > FE (downward on positive charge), the net force is upward and the positive charge is deflected upward [1 mark].
Q2 — Data table (9 marks, 1 per cell)
2.1 (Electric only): Work done? Yes — electric force has a component along displacement, transferring kinetic energy to the particle. Trajectory: Parabola. Speed change: Yes — kinetic energy increases as the particle accelerates along the field direction.
2.2 (Magnetic only): Work done? No — magnetic force is always perpendicular to velocity, so force · displacement = 0. Trajectory: Circle. Speed change: No — constant speed, direction only changes.
2.3 (Both, at v = E/B): Work done? No net work — the net force is zero, so no net displacement in the force direction; the particle travels straight. Trajectory: Straight line. Speed change: No — no net force means constant velocity.
Q3.1 — Selector speed (2 marks)
v = E/B1 = 4000 / 0.020 = 2.0 × 105 m/s [1 mark substitution, 1 mark answer].
Q3.2 — Radius for m = 12 u (3 marks)
m = 12 × 1.66 × 10−27 = 1.99 × 10−26 kg [1]. r = mv/(qB2) = (1.99 × 10−26)(2.0 × 105) / (1.60 × 10−19 × 0.50) = 3.99 × 10−21 / 8.0 × 10−20 ≈ 0.050 m = 5.0 cm [1]. Reading the graph at m = 12 gives approximately 5.0 cm — consistent with calculation [1].
Q3.3 — Relationship mass vs radius (3 marks)
The graph shows a linear relationship: radius increases proportionally with ion mass [1]. From r = mv/(qB2), with v, q, and B2 all fixed, r ∝ m, predicting a straight line through the origin [1]. This linear proportionality means the mass spectrometer can directly distinguish ions of different masses by measuring where they land on the detector (larger radius = larger mass) [1].
Q4.1 — Selector speed (1 mark)
v = E/Bselector = 5000 / 0.025 = 2.0 × 105 m/s.
Q4.2 — Radius of 28 u ion (3 marks)
m = 28 × 1.66 × 10−27 = 4.648 × 10−26 kg [1]. r = mv/(qBanalysis) = (4.648 × 10−26)(2.0 × 105) / (1.60 × 10−19 × 0.60) [1]. Numerator: 9.296 × 10−21; denominator: 9.6 × 10−20; r ≈ 0.097 m ≈ 9.7 cm [1].
Q4.3 — 56 u ion (2 marks)
Since r ∝ m (with v, q, Banalysis all constant), doubling the mass doubles the radius [1]. The 56 u ion has radius 2 × 9.7 cm ≈ 19.4 cm. It lands at a different point on the detector, confirming the mass spectrometer separates ions by mass [1].
Q4.4 — Effect of doubling E (1 mark)
(i) Selector speed doubles: v = E/B → 2E/B = 2v; new selector speed = 4.0 × 105 m/s. (ii) Since r = mv/(qBanalysis) and v doubled, the radius doubles to approximately 19.4 cm for the 28 u ion.