Year 12 Physics Module 6 IQ1 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 6 of 21

Electric vs Magnetic Fields and Velocity Selection

Compare how electric and magnetic fields change charged-particle motion, then use vector force balance to derive the speed selected by crossed fields.

Today's hook: A beam passes straight through perpendicular electric and magnetic fields of $1.5 \times 10^4$ V/m and $3.4 \times 10^{-4}$ T. What is the particle speed, and does the answer depend on charge sign or mass?
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1
Retrieve the separate field rules and predict how crossed fields could balance.
Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A positively charged particle enters a region with both a uniform electric field (pointing downward) and a uniform magnetic field (pointing into the page). The particle is initially moving to the right.

Before reading on, answer:

  1. Sketch the path you predict if only the electric field is present.
  2. Sketch the path you predict if only the magnetic field is present.
  3. With both fields present, do you think the particle could travel in a straight line? If so, what condition would be required?

Warm-up, which type of charge does a magnetic field exert a force on?

Learning Intentions
goals

Know, Field Behaviours

  • E-fields act on stationary or moving charges; magnetic force requires motion across field lines
  • Electric forces can transfer energy; magnetic forces alone cannot
  • A uniform transverse E field can produce a parabola; perpendicular entry into a uniform B field produces a circle

Understand, Combined Fields

  • How crossed E and B fields create a velocity selector
  • Why $v = E/B$ is the condition for straight-line motion
  • When each field dominates the particle's trajectory

Can Do, Solve Crossover Problems

  • Compare and contrast E-field and B-field effects quantitatively
  • Derive and apply the velocity selector equation
  • Predict trajectories in combined field configurations
Scan these before reading
vocab
Velocity selectorRegion with crossed E and B fields where only particles with $v = E/B$ travel straight.
Crossed fieldsElectric and magnetic fields oriented perpendicular to each other and to the particle velocity.
Mass-spectrometer bridgeSome designs place a crossed-field selector before magnetic analysis. L18 owns the complete instrument and alternative known-potential design.
Cross-lesson links: L05 showed pure magnetic circular motion. L06 combines E and B fields: when the electric and magnetic force vectors oppose and have equal magnitude, $v=E/B$ passes undeflected. L18 owns the complete mass-analysis application.

Core Content

Before you continue

Commit to a force direction for the positive particle in the Think First prompt.

2
Compare force conditions, energy transfer, and the special trajectories produced by uniform fields.
1
Side-by-Side Comparison
+5 XP

How electric and magnetic fields treat charges differently

Place a stationary proton between two charged plates and it accelerates immediately, the electric field acts on it even at rest. Now place that same stationary proton in a region with only a magnetic field: nothing happens. Move the proton sideways and suddenly the magnetic force appears, pushing it perpendicular to its motion. Electric fields act on any charge regardless of motion; magnetic fields are invisible to stationary charges and only reveal themselves when a charge is moving.

Property Electric field Magnetic field
Acts on Stationary or moving charges Moving charges with a velocity component perpendicular to $\vec B$
Force $\vec F_E=q\vec E$ $\vec F_B=q(\vec v\times\vec B)$; $F_B=|q|vB\sin\theta$
Direction Along $\vec E$ for $q>0$ and opposite $\vec E$ for $q<0$ Perpendicular to $\vec v$ and $\vec B$, reversed when $q<0$
Energy Can change kinetic energy when the force has a component along the displacement A pure magnetic field does zero work because $\vec F_B\perp\vec v$
Speed Can increase, decrease, or initially remain unchanged, depending on geometry Unchanged in a pure B field
Special path Parabola for uniform transverse $\vec E$ and constant initial longitudinal velocity Circle for uniform $\vec B$ and $\vec v\perp\vec B$
Mass role Acceleration depends on $m$ through $\vec a=q\vec E/m$ Radius and period depend on $m$

Electric field lines

Point away from positive charge and toward negative charge, or extend to infinity. They never cross because the field has one direction at each point.

Magnetic field lines

Form closed loops because isolated magnetic poles have not been observed. They also never cross. Line spacing represents relative field strength in both models.

Electron motion in separate uniform electric and magnetic fields At left, a right-moving electron curves progressively upward because the downward electric field produces an upward force. At right, the same electron follows a clockwise circular path in a magnetic field directed out of the page; velocity is tangent and magnetic force points toward the centre. Uniform E field E down v − F up Uniform B field B out − F inward v tangent

Figure 1. The parabolic and circular paths are special cases requiring uniform fields and perpendicular entry geometry.

Stop & Check

For fixed $\vec E$, $\vec B$, and $\vec v$, explain why replacing a proton with an electron reverses both force vectors but does not make their magnitudes equal.

Electric field: $\vec F_E=q\vec E$ and it can change kinetic energy. A uniform transverse $E$ field gives a parabola under the stated component assumptions. Magnetic field: $\vec F_B=q(\vec v\times\vec B)$; its magnitude is $|q|vB\sin\theta$, it does no work, and perpendicular entry into a uniform field gives a circle. Oblique magnetic entry gives a helix.

Pause, copy the highlighted comparison table summary into your book before moving on.

A magnetic field can do work on a moving charged particle and increase its kinetic energy.

A stationary charge in a magnetic field experiences zero magnetic force.

Explain the contrast

State why a pure magnetic field can turn a particle without changing its kinetic energy.

3
Arrange opposing electric and magnetic forces and derive the selected speed.
Beyond the syllabus. Comparing how electric and magnetic fields affect a moving charge is core. The full crossed-field velocity selector and routine $v = E/B$ calculations are a useful application but are not named syllabus content — treat the selector section as extension, and expect the exam to assess the field comparison itself. The same crossed-fields idea returns as core in Module 8, where it is Thomson's charge-to-mass experiment.
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The Velocity Selector
+5 XP

When crossed E and B fields produce straight-line motion

We just saw that E-fields and B-fields exert forces in very different ways. That raises a question: what happens when both fields act on the same particle simultaneously? This card answers it → at exactly one speed ($v = E/B$) the forces cancel and the particle travels straight.

When both electric and magnetic fields are present and perpendicular to each other and to the particle's velocity, the particle experiences two forces. By arranging the fields so that these forces act in opposite directions, we can find a specific velocity where they exactly cancel.

  • Combined vector law: $\vec F=q(\vec E+\vec v\times\vec B)$
  • Cancellation: the fields must be arranged so $\vec E$ and $\vec v\times\vec B$ point in opposite directions.
Velocity Selector Condition

$q(\vec E+\vec v\times\vec B)=\vec 0$

$|q|E=|q|vB$ for mutually perpendicular vectors

$$v = \dfrac{E}{B}$$

Only particles at this speed are undeflected in the stated crossed-field geometry.

Notice that $q$ cancels out. The balance condition is independent of mass and charge magnitude: provided the two forces oppose, a charged particle with $v=E/B$ travels straight. A crossed-field selector can precede magnetic analysis, but it is not required in every mass-spectrometer design.

Velocity selector forces for positive and negative particles Particles move right through an electric field directed down and a magnetic field directed into the page. For a positive particle electric force is down and magnetic force is up. For a negative particle both forces reverse. Equal arrow lengths indicate balance at speed E over B. E downward; B into page, shown by crosses + F_B up F_E down − F_E up F_B down v right v right

Figure 2. Both force arrows reverse with charge sign, so the balance speed remains $v=E/B$.

Stop & Check

A velocity selector has E = 2000 V/m and B = 0.040 T. A proton and an alpha particle both enter at $5.0 \times 10^4$ m/s. Which particles (if any) pass through undeflected? Explain.

Velocity selector condition: first show $\vec E$ and $\vec v\times\vec B$ oppose, then equate magnitudes $|q|E=|q|vB$, giving $v=E/B$. The condition is independent of mass and charge sign. Above this speed magnetic force dominates; below it electric force dominates.

Add the highlighted velocity selector equation and key points to your notes before the check below.

A velocity selector has $E = 3000$ V/m and $B = 0.015$ T. What speed must a particle have to pass through undeflected?

Check the vectors

For the electron diagram, verify the positive-charge $\vec v\times\vec B$ direction first, then reverse it.

4
Use the shared field model only to test force directions and the selector condition.
Beyond the syllabus. Comparing how an electric field and a magnetic field each push a moving charge is core, and so is the fact that a magnetic force does no work — both are what the One field at a time mode of the simulation below shows, and those are the parts to expect in an exam. The full crossed-field velocity selector and routine $v = E/B$ calculations are not named syllabus content here, so treat the Crossed fields mode as extension: drive it for the insight, but you will be assessed on the field comparison itself. The same apparatus returns as core in Module 8, where it is Thomson’s charge-to-mass experiment.
4
Drive It: Find the One Speed That Goes Straight
+XP for exploring

The balance condition measured by experiment instead of asserted, and the shape of the path when the balance fails

Interactive · Crossed fields velocity selector

Try this: in One field at a time , fire with E only and then with B only, and watch what the speed readout does in each case. Then switch to Crossed fields , set E = 4.0 kV m−1 and B = 0.020 T, and hunt for the entry speed that goes straight through. Fire one speed either side of it. Finally run the two gold probes: Charge flip and Same speed, different particle .

The single-field mode is the core field comparison. The crossed-field mode is the extension application flagged above; Lesson 18 takes the same apparatus further into mass spectrometry.

Use the simulation. In Crossed fields , set E = 4.0 kV m⁻¹ and B = 0.020 T, then fire protons at 1.5, 2.0 and 2.5 × 10⁵ m s⁻¹. Read the deflection after each shot. Which row matches what you see?

Use the simulation's single-field mode. Fire once with B only and once with E only , watching the speed readout and the exit speed each time. Then answer:

With B only, the particle speeds up as it curves, because the magnetic force is doing work on it.

With E only the exit speed is higher than the entry speed, but with B only the two readouts agree to every digit shown.

A charged particle moving parallel to a magnetic field (angle = 0°) experiences zero magnetic force.

In a velocity selector, an electron at $v = E/B$ and a proton at $v = E/B$ both travel in a straight line.

Record one observation

Describe what changes when the particle speed moves above or below $E/B$.

5
Solve a crossed-field problem by separating direction from magnitude.
3
Worked Example: Combined Fields Problem
+5 XP

Determining trajectory in crossed E and B fields

Problem

An electron enters a region with a uniform electric field of $E = 4000\,\text{V/m}$ directed downward and a uniform magnetic field of $B = 0.020\,\text{T}$ directed into the page. The electron enters moving horizontally to the right at $v = 2.0 \times 10^5\,\text{m/s}$.

  1. Calculate the electric force magnitude and direction on the electron.
  2. Calculate the magnetic force magnitude and direction on the electron.
  3. Determine whether the electron travels in a straight line or is deflected.
  4. What speed would be required for the electron to travel straight through?
Solution
  1. Electric force:
    $F_E = |q|E = (1.60 \times 10^{-19})(4000) = 6.40 \times 10^{-16}\,\text{N}$
    Direction: E-field points downward, so the negative electron experiences force upward.
  2. Magnetic force:
    $F_B = |q|vB = (1.60 \times 10^{-19})(2.0 \times 10^5)(0.020) = 6.40 \times 10^{-16}\,\text{N}$
    Direction: for a positive charge, rightward $\vec v$ crossed with $\vec B$ into the page points upward. Reverse for the electron, so its magnetic force is downward.
  3. Net force and trajectory:
    $F_E$ upward = $F_B$ downward = $6.40 \times 10^{-16}\,\text{N}$.
    Net force = 0. The electron travels in a straight line.
  4. Velocity selector speed:
    $v = \dfrac{E}{B} = \dfrac{4000}{0.020} = 2.0 \times 10^5\,\text{m/s}$
    This matches the given speed, confirming the result.
Stop & Check

If the electron's speed were increased to $4.0 \times 10^5$ m/s with the same E and B fields, in which direction would it be deflected? Justify your answer with a calculation.

Combined-field method: (1) draw $\vec E$, $\vec B$, and $\vec v$; (2) find each force direction, including charge sign; (3) calculate magnitudes $F_E=|q|E$ and $F_B=|q|vB$ for perpendicular entry; (4) compare. At $v=E/B$ the net force is zero. Above it $F_B$ dominates; below it $F_E$ dominates.

Pause, write the highlighted problem-solving method into your book before moving on.

In a velocity selector, a proton travels straight through at speed $v = E/B$. If the proton is replaced by an electron with the same speed, what happens?

HSC Tip: Combined Fields Trap

In combined field problems, students often forget that direction depends on charge sign for both forces. The electric force reverses for negative charges (opposite to E), and the magnetic force also reverses (opposite to the right-hand rule prediction). If you reverse both forces, they still oppose each other, and the velocity selector condition $v = E/B$ remains the same regardless of charge sign. Always draw arrows and check signs before calculating.

Complete the worked check

Predict the deflection at twice the selected speed before reading your calculation.

6
Connect velocity selection to later mass analysis, then consolidate the Lesson 6 boundary.
4
Preview, Connection to Mass Analysis
+5 XP

A short bridge only; the complete instrument belongs to L18

We just saw how crossed E and B fields select a single speed. That raises a question: what could a later analysing region reveal when speed is known? This card sets the boundary: Lesson 6 owns selection; Lesson 18 owns the complete mass-analysis instrument.

A crossed-field selector can supply a beam with a known speed. A later device may then use magnetic curvature to investigate mass-to-charge ratio, but selectors are not present in every mass-spectrometer design.

Lesson 6 owns

Force-vector comparison, cancellation, $v=E/B$, and predicting rejection above or below the selected speed.

Lesson 18 owns

Ion acceleration, analyser radius, detector geometry, resolving isotopes, calculations, and model limitations.

Boundary: a velocity selector transmits particles at $v=E/B$ when its force vectors oppose. It can feed a later analyser, but the complete instrument and all mass-analysis calculations belong to Lesson 18.

Which task belongs to this Lesson 6 velocity-selector boundary?

Activity 1, Velocity Selector Analysis
ApplyBand 4

Calculate and compare field effects on different particles

  1. A selector sends particles right through $\vec E$ downward and $\vec B$ into the page. It has $E = 5000$ V/m and $B = 50$ mT. Calculate $v_{\text{selector}} = E/B$. Show all working.
  2. An electron enters at $v = 1.0 \times 10^5$ m/s. Calculate both the electric and magnetic forces on it and determine whether it travels straight or is deflected.
  3. A proton replaces the electron with the same $E$, $B$, and speed. Does the proton also travel straight? Explain why the velocity selector condition is independent of charge sign and mass.
  4. With the proton at the selector speed, the electric field is doubled. What speed would now be required for straight-line motion?
Activity 2, Vector-Balance Response
AnalyseBand 5

Explain selection and rejection without crossing into Lesson 18 analysis

A beam of positive ions moves right through a selector with $\vec E$ downward and $\vec B$ into the page. The selected speed is $v_s=E/B$.

  1. Draw and label $\vec F_E$ and $\vec F_B$ for an ion at $v_s$.
  2. Predict the deflection of an ion entering at $2v_s$, and justify it by comparing force magnitudes.
  3. Repeat the force-direction reasoning for a negative ion at $v_s$.
  4. Explain in one sentence what this selector can establish for a later analyser, without calculating any mass or detector separation.
Wrap-up, Synthesis & Summary

Key Contrasts

  • E-field: stationary or moving charges
  • B-force: motion with a perpendicular component
  • Electric force can do work; magnetic force alone cannot
  • Uniform transverse E: parabola; uniform perpendicular B: circle

Velocity Selector

  • $q(\vec E+\vec v\times\vec B)=0$ first
  • $|q|E=|q|vB \Rightarrow v=E/B$
  • Independent of $q$ and $m$

Lesson Boundary

  • L06: compare and select
  • L18: complete instrument
  • No detector or isotope calculations here

HSC Check Signs

  • Draw field and velocity vectors first
  • Apply RH rule then reverse for $-q$
  • Both forces reverse → $v = E/B$ holds

Respond

Use Activity 2 to explain the selector boundary before opening Practice.

P
Retrieve the comparison and apply vector balance to a determinate selector case.
Quick recall, Electric vs Magnetic Fields
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 8 marks
+5 XP

ApplyBand 5(4 marks) 1. Compare and contrast the effect of a uniform electric field and a uniform magnetic field on a moving charged particle. In your answer, address: (i) the conditions under which each field exerts a force, (ii) whether each field can change the particle's kinetic energy, and (iii) the shape of the trajectory when the field is perpendicular to the initial velocity.

1 mark: E-field acts on stationary or moving charges; B-force requires a perpendicular velocity component · 1 mark: electric force can change KE; magnetic force alone does not · 1 mark each: qualified parabolic E path and circular B path

AnalyseBand 6(4 marks) 2. A positive ion moves right into uniform crossed fields. $\vec E$ points downward and $\vec B$ points into the page. Its entry speed is $2E/B$. State the directions and relative magnitudes of the two forces, predict its initial deflection, and explain what can be concluded about its speed at entry and later in the motion.

1 mark: $F_E=|q|E$ downward and $F_B=2|q|E$ upward · 1 mark: net $|q|E$ upward, so rejected upward · 1 mark: instantaneous speed unchanged at entry because both forces are perpendicular to entry velocity · 1 mark: magnetic force never does work, while electric force can change KE after the path turns

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (4 marks): (i) An electric field exerts $\vec F_E=q\vec E$ on stationary or moving charges (1 mark). A magnetic field exerts $\vec F_B=q(\vec v\times\vec B)$ only when a charge has a velocity component perpendicular to $\vec B$ (1 mark). (ii) Electric force can change kinetic energy; magnetic force is perpendicular to velocity and does zero work (1 mark). (iii) With uniform fields and perpendicular entry, the electric case has constant transverse acceleration and a parabolic path, while the magnetic case is circular at constant speed (1 mark).

Q2 (4 marks): $\vec F_E$ is downward with magnitude $|q|E$. Rightward $\vec v$ crossed with $\vec B$ into the page points upward, and at $v=2E/B$ the magnetic-force magnitude is $|q|vB=2|q|E$ (1 mark). The net force is therefore $|q|E$ upward, so the ion is initially rejected upward (1 mark). At entry both forces are perpendicular to the rightward velocity, so instantaneous power is zero and the speed is momentarily unchanged (1 mark). The magnetic force remains perpendicular to velocity and never does work. As the trajectory turns, the electric force may gain a component along or against velocity and can then change kinetic energy; monotonic increase cannot be claimed without solving the later motion (1 mark).

R
Check retention, revisit the opening calculation, and connect forward to Lesson 18.
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quiz

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How did your thinking change?

At the start, a beam passed straight through perpendicular fields with $E=1.5 \times 10^4$ V/m and $B=3.4 \times 10^{-4}$ T. What speed did it have, and does the balance depend on charge sign or mass?

The answer is $v=E/B=(1.5 \times 10^4)/(3.4 \times 10^{-4})\approx4.4 \times 10^7$ m/s, to two significant figures. The result is independent of mass and charge sign because both force vectors reverse when the sign of $q$ reverses. This conclusion applies only when the field geometry makes the electric and magnetic forces oppose.

Review your Think First predictions from this lesson and earlier lessons:

  • L06: Was $v = E/B$ the condition you proposed for straight-line motion?
  • L01–05: How does the circular path in a B-field connect to what you learned about circular motion in a magnetic field?
← Previous: Circular Motion in Magnetic Fields Next: Mass Spectrometers →

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