DP Physics SL D.3 Motion in Electromagnetic Fields

DP Physics SL D.3 Motion in Electromagnetic Fields - Full notes

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DP Physics SL | D.3 Motion in Electromagnetic Fields

DP PHYSICS · THEME D: FIELDS

D.3

Motion in Electromagnetic Fields

Full notes for SL · IB-style

Contents

1	Charges in electric fields
2	Magnetic force on moving charges
3	Combined fields and particle instruments
4	Forces on currents
5	Energy and fields
6	Exam technique and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
analyse the motion of charges in uniform electric fields
use F = qvB sin θ and direction rules
derive and use r = mv/qB and T = 2πm/qB
explain velocity selectors, mass spectrometers and cyclotrons
use F = BIL sin θ and the force between parallel wires
describe motors, loudspeakers and detectors

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do charged particles move in electric and magnetic fields? How do the forces on currents in magnetic fields lead to useful devices? How can fields be used to measure properties of particles?

How this topic is examined

D.3 appears in Papers 1 and 2. Expect calculations of deflection in E fields, circular motion in B fields, F = BIL, parallel wires, and explanations of devices.

State the rule you use for directions and draw clear diagrams with field directions marked (dots and crosses).

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Charges in electric fields

Accelerating and deflecting charges.

Think about it

How does an X-ray tube give electrons enough energy to produce X-rays?

Electric and magnetic fields exert forces on charged particles. Controlling these forces lets us accelerate particles, steer beams, weigh atoms and build motors. This section starts with electric fields.

1.1 Acceleration through a potential difference

A charge q moving through a potential difference V gains kinetic energy qV. For an electron starting from rest, eV = ½mv². In a uniform field the force qE is constant, so the acceleration a = qE/m is uniform.

Figure 1. An electron gun accelerates electrons from the cathode to the anode.

WORKED EXAMPLE 1 Electron gun (standard)

An electron is accelerated from rest through 500 V. Find its kinetic energy and speed.

Solution

Ek = eV = 8.0 × 10⁻¹⁷ J (500 eV)A1

v = √(2Ek/m) = 1.33 × 10⁷ m s⁻¹M1 A1

WORKED EXAMPLE 2 Proton (standard)

Find the speed of a proton accelerated from rest through 2.0 kV.

Solution

½mv² = eV = 3.2 × 10⁻¹⁶ JM1

v = 6.2 × 10⁵ m s⁻¹A1

Quick check

1. Express 500 eV in joules.

Answers: 1. 8.0 × 10⁻¹⁷ J

1.2 The electronvolt

The electronvolt (eV) is the energy gained by an electron accelerated through 1 V: 1 eV = 1.60 × 10⁻¹⁹ J. It is a convenient unit for atomic and nuclear energies (keV, MeV).

WORKED EXAMPLE 3 Conversion (basic)

Convert 5.0 MeV to joules.

Solution

5.0 × 10⁶ × 1.6 × 10⁻¹⁹M1

= 8.0 × 10⁻¹³ JA1

1.3 Motion perpendicular to a uniform field

A charge entering a uniform field at right angles moves like a projectile: constant velocity parallel to the plates and uniform acceleration perpendicular to them. Its path between the plates is a parabola; after leaving, it travels in a straight line.

Figure 2. An electron deflected by the field between parallel plates.

Figure 3. Deflection of an electron entering at 2.0 × 10⁷ m s⁻¹ (E = 10⁴ V m⁻¹).

Key point

Treat the two directions separately, exactly as for projectiles: no force along the plates, constant force across them.

WORKED EXAMPLE 4 Deflection (exam level)

An electron enters a field E = 1.0 × 10⁴ V m⁻¹ at 2.0 × 10⁷ m s⁻¹ and travels 6.0 cm between the plates. Find its vertical deflection.

Solution

t = 0.060/2.0 × 10⁷ = 3.0 × 10⁻⁹ s; a = eE/m = 1.76 × 10¹⁵ m s⁻²M1

y = ½at² = 7.9 mmA1

WORKED EXAMPLE 5 Exit angle (exam level)

Find the angle at which the electron leaves the plates.

Solution

vy = at = 5.3 × 10⁶ m s⁻¹M1

tan θ = 5.3 × 10⁶/2.0 × 10⁷: θ = 15°A1

1.4 The cathode-ray tube

In a cathode-ray tube, an electron gun produces a narrow beam that passes between pairs of deflecting plates and strikes a fluorescent screen. Varying the pd across the plates moves the spot: this was the basis of oscilloscopes and older televisions.

Figure 4. A cathode-ray tube with vertical-deflection plates.

Quick check

1. What happens to the spot if the pd across the Y-plates is reversed?

Answers: 1. it is deflected the other way

1.5 Energy changes during deflection

As an electron is deflected towards the positive plate, the field does work on it, increasing its kinetic energy by eΔV, where ΔV is the pd between its entry and exit points.

WORKED EXAMPLE 6 Energy gain (exam level)

An electron entering midway between plates 2.0 cm apart with 400 V across them is deflected 4.0 mm towards the positive plate. Find its gain in kinetic energy.

Solution

E = 2.0 × 10⁴ V m⁻¹; ΔV = E × 0.0040 = 80 VM1

ΔEk = 80 eV (1.3 × 10⁻¹⁷ J)A1

1.6 Comparing particles in the same field

In the same electric field, the force depends on charge and the acceleration on charge-to-mass ratio. Electrons are deflected much more than protons; positive and negative particles are deflected in opposite directions.

WORKED EXAMPLE 7 Electron vs proton (standard)

Compare the accelerations of an electron and a proton in the same field.

Solution

a = eE/m: ratio = mp/meM1

Electron's is 1840 times larger (opposite direction)A1

2 Magnetic force on moving charges

F = qvB sin θ.

2.1 The magnetic force

A charge q moving with velocity v at angle θ to a magnetic field B experiences a force F = qvB sin θ, perpendicular to both v and B. A charge moving parallel to the field feels no force; a stationary charge feels no magnetic force.

WORKED EXAMPLE 8 Force (standard)

A proton moves at 3.0 × 10⁶ m s⁻¹ perpendicular to a 0.20 T field. Find the force.

Solution

F = 1.6 × 10⁻¹⁹ × 3.0 × 10⁶ × 0.20M1

= 9.6 × 10⁻¹⁴ NA1

WORKED EXAMPLE 9 At an angle (standard)

An electron moves at 5.0 × 10⁶ m s⁻¹ at 30° to a 0.10 T field. Find the force.

Solution

F = 1.6 × 10⁻¹⁹ × 5.0 × 10⁶ × 0.10 × sin 30°M1

= 4.0 × 10⁻¹⁴ NA1

Quick check

1. What magnetic force acts on an electron moving parallel to a field?

Answers: 1. none

2.2 Direction rules

Fleming's left-hand rule: first finger = field, second finger = conventional current (direction of positive charge motion), thumb = force. For negative charges, the current is opposite to their velocity, so the force is reversed. Diagrams use • for out of the page and × for into the page.

Figure 5. Fleming's left-hand rule: Field, Current, Motion (force) mutually perpendicular.

Common mistake

Applying the left-hand rule to electrons without reversing direction. Conventional current is opposite to electron flow.

Quick check

1. An electron moves east in a field pointing vertically upwards. Which way is the force?

Answers: 1. north (a positive charge would be pushed south)

2.3 Circular motion in a magnetic field

The magnetic force is always perpendicular to the velocity, so it does no work: the speed is constant and the charge moves in a circle. qvB = mv²/r gives r = mv/qB. The period T = 2πm/qB does not depend on speed.

Figure 6. A positive charge moving in a circle in a field into the page.

WORKED EXAMPLE 10 Electron circle (exam level)

An electron at 4.0 × 10⁶ m s⁻¹ enters a 2.0 mT field at right angles. Find the radius and period.

Solution

r = 9.11 × 10⁻³¹ × 4.0 × 10⁶/(1.6 × 10⁻¹⁹ × 2.0 × 10⁻³)M1

= 1.1 cmA1

T = 1.8 × 10⁻⁸ sA1

WORKED EXAMPLE 11 Momentum from radius (exam level)

A particle of charge e moves in a circle of radius 0.25 m in a 0.80 T field. Find its momentum.

Solution

p = mv = qBrM1

= 1.6 × 10⁻¹⁹ × 0.80 × 0.25 = 3.2 × 10⁻²⁰ kg m s⁻¹A1

Quick check

1. Why does the magnetic force not change the kinetic energy?

Answers: 1. it is perpendicular to the velocity, so no work is done

2.4 Radius and speed relationship

Since r = mv/qB, the radius is proportional to speed. A graph of radius against speed for a given particle and field is a straight line through the origin; heavier particles give steeper lines.

Figure 7. Radius against speed in a 0.10 T field.

Quick check

1. Why is the alpha line steeper than the proton line?

Answers: 1. m/q is twice as large for the alpha particle

2.5 Different particles in the same magnetic field

At the same speed, r = mv/qB depends on m/q. Electrons curve tightly one way; protons and alpha particles curve the other way with much larger radii (alpha has twice the m/q of a proton).

Figure 8. Paths of different particles entering the same field at the same speed (schematic, not to scale).

WORKED EXAMPLE 12 Radius ratio (standard)

Find the ratio of radii for an alpha particle and a proton at the same speed in the same field.

Solution

r ∝ m/q: (4m/2e)/(m/e)M1

= 2A1

2.6 Entering and leaving a field region

A particle entering a region of uniform magnetic field follows a circular arc while inside, then leaves in a straight line in a new direction. The deflection angle depends on the arc length relative to the radius.

Figure 9. A charged particle follows an arc inside the field region, then a straight line outside.

Quick check

1. Does the particle leave the field faster, slower or at the same speed?

Answers: 1. at the same speed

2.7 Helical paths

If a charge enters a magnetic field at an angle, the velocity component parallel to the field is unchanged, while the perpendicular component produces circular motion. The result is a helix wound around the field lines.

Figure 10. A charged particle spiralling along a magnetic field line.

In the real world

Charged particles from the Sun spiral along Earth's field lines into the polar regions, where they excite gases and produce aurorae.

Quick check

1. What path does a charge take if v is at 60° to B?

Answers: 1. a helix

3 Combined fields and particle instruments

Selecting, weighing and accelerating particles.

3.1 Velocity selectors

When perpendicular electric and magnetic fields act on a beam, a particle passes straight through if qE = qvB, so v = E/B, independent of mass and charge. Faster particles are deflected one way, slower the other.

Figure 11. A velocity selector: only particles with v = E/B pass through the slit.

WORKED EXAMPLE 13 Selector (exam level)

E = 3.0 × 10⁴ V m⁻¹ and B = 0.15 T. Find the selected speed.

Solution

v = 3.0 × 10⁴/0.15M1

= 2.0 × 10⁵ m s⁻¹A1

3.2 The mass spectrometer

Ions are produced, accelerated, passed through a velocity selector, then enter a uniform magnetic field where they travel in semicircles of radius r = mv/qB. Heavier ions follow larger semicircles, so isotopes land at different positions on a detector.

Figure 12. A mass spectrometer: heavier ions travel in larger semicircles.

WORKED EXAMPLE 14 Isotopes (exam level)

Ions of ²⁰Ne⁺ and ²²Ne⁺ enter a 0.50 T field at 1.0 × 10⁵ m s⁻¹. Find the separation of their landing points (u = 1.66 × 10⁻²⁷ kg).

Solution

r = mv/qB: r20 = 4.15 cm, r22 = 4.57 cmM1 A1

Separation = 2(r22 − r20) = 8.3 mmA1

In the real world

Mass spectrometers detect drugs in athletes, date archaeological samples using isotope ratios and identify pollutants.

3.3 The cyclotron

In a cyclotron, charged particles move in semicircles inside two D-shaped electrodes ('dees') in a magnetic field and are accelerated across the gap by an alternating voltage each half turn. Because the period does not depend on speed, a fixed-frequency supply keeps in step.

Figure 13. A cyclotron: particles spiral outwards as they gain energy.

WORKED EXAMPLE 15 Frequency (exam level)

Find the frequency of the alternating supply for protons in a 1.5 T field.

Solution

f = qB/2πmM1

= 2.3 × 10⁷ HzA1

WORKED EXAMPLE 16 Maximum energy (exam level)

The dees have radius 0.50 m. Find the maximum kinetic energy of the protons in MeV.

Solution

v = qBr/m = 7.2 × 10⁷ m s⁻¹M1

Ek = ½mv² = 4.3 × 10⁻¹² J = 27 MeVA1

3.4 Particle detectors

In bubble chambers and modern detectors, a magnetic field curves particle tracks. The radius gives momentum (p = qBr); the direction of curvature reveals the sign of charge; spiralling tracks show particles losing energy.

Figure 14. Tracks of particles entering a magnetic field: curvature depends on charge sign and momentum.

WORKED EXAMPLE 17 Track curvature (standard)

Two particles with the same speed and charge curve with radii 2.0 cm and 6.0 cm in the same field. Compare their masses.

Solution

r ∝ mM1

Second is 3 times as massiveA1

3.5 Measuring e/m with a fine-beam tube

In a fine-beam tube, electrons accelerated through V enter a uniform field B from Helmholtz coils and travel in a visible circle. Combining eV = ½mv² with r = mv/eB gives e/m = 2V/(B²r²). Thomson's similar measurement (1897) showed electrons are particles far lighter than atoms.

Figure 15. Radius of the electron beam against 1/B at 200 V.

WORKED EXAMPLE 18 e/m (exam level)

Electrons accelerated through 200 V move in a circle of radius 3.4 cm in a field of 1.4 mT. Find e/m.

Solution

e/m = 2 × 200/((1.4 × 10⁻³)² × 0.034²)M1

= 1.8 × 10¹¹ C kg⁻¹A1

3.6 Particle accelerators today

Modern synchrotrons, such as the 27 km Large Hadron Collider, accelerate particles with electric fields in radio-frequency cavities and steer them round a ring with thousands of superconducting magnets. As the particles gain energy, the magnetic field is increased so the radius stays fixed.

WORKED EXAMPLE 19 Field for fixed radius (exam level)

Explain why the bending field must increase as the protons' momentum increases.

Solution

r = p/qBM1

To keep r constant, B must increase in proportion to pR1

In the real world

Medical cyclotrons in hospitals produce short-lived radioactive isotopes for PET scans, and proton beams for cancer therapy.

3.7 The Hall effect

When current flows through a thin conductor in a magnetic field, the magnetic force pushes charge carriers to one side, producing a small transverse pd (the Hall voltage). Hall probes use this effect to measure magnetic flux density; Hall sensors detect wheel speed and phone orientation.

Quick check

1. What does a Hall probe measure?

Answers: 1. magnetic flux density

4 Forces on currents

The motor effect.

Think about it

What makes an electric motor spin?

4.1 Force on a current-carrying conductor

A wire carrying current I in a magnetic field B experiences a force F = BIL sin θ, where L is the length in the field and θ the angle between wire and field. The force is the sum of the forces on the moving charges in the wire, and its direction follows Fleming's left-hand rule.

Figure 16. A wire carrying current into the page between the poles of a magnet is pushed upwards.

WORKED EXAMPLE 20 Wire (standard)

A 0.25 m wire carries 3.0 A perpendicular to a 0.40 T field. Find the force.

Solution

F = 0.40 × 3.0 × 0.25M1

= 0.30 NA1

WORKED EXAMPLE 21 Angled wire (standard)

The same wire is at 30° to the field. Find the force.

Solution

F = 0.30 × sin 30°M1

= 0.15 NA1

Quick check

1. What force acts if the wire is parallel to the field?

Answers: 1. zero

4.2 How the force depends on angle

The force on a wire varies as sin θ: it is maximum when the wire is perpendicular to the field and zero when parallel.

Figure 17. Force on a 0.25 m wire carrying 3.0 A in a 0.40 T field against angle.

Quick check

1. At what angle is the force half its maximum value?

Answers: 1. 30° (or 150°)

4.3 Measuring magnetic flux density

A current balance measures the force on a wire using a sensitive balance: F = BIL, so B = F/IL. A graph of F against I is a straight line with gradient BL.

Figure 18. Force on 5.0 cm of wire against current.

WORKED EXAMPLE 22 Balance (exam level)

Use the graph (L = 5.0 cm) to find B.

Solution

Gradient = 5.88 × 10⁻³ N A⁻¹ = BLM1

B = 5.88 × 10⁻³/0.050 = 0.12 TA1

Try this

Place a horseshoe magnet on a top-pan balance and pass a current through a rigid wire between its poles. Record the change in reading for several currents.

4.4 Force between parallel wires

Two parallel wires carrying currents in the same direction attract; in opposite directions they repel. Each wire sits in the other's magnetic field. The force per unit length is F/L = μ₀I₁I₂/2πr, with μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Figure 19. Parallel currents in the same direction attract.

WORKED EXAMPLE 23 Power lines (exam level)

Two parallel wires 0.10 m apart each carry 50 A in opposite directions. Find the force per metre.

Solution

F/L = 4π × 10⁻⁷ × 50 × 50/(2π × 0.10)M1

= 5.0 × 10⁻³ N m⁻¹ (repulsive)A1

4.5 The electric motor

A rectangular coil carrying current in a magnetic field experiences equal and opposite forces on its two sides, forming a couple that turns it. A split-ring commutator reverses the current every half turn, so the couple keeps acting in the same rotational direction.

Figure 20. A simple DC motor.

WORKED EXAMPLE 24 Torque on a coil (exam level)

A 50-turn coil, 4.0 cm × 6.0 cm, carries 0.50 A in a 0.30 T field with its plane parallel to the field. Find the force on each 6.0 cm side and the torque.

Solution

F = NBIL = 50 × 0.30 × 0.50 × 0.060 = 0.45 NM1

Torque = F × 0.040 = 0.018 N mA1

Quick check

1. Give two ways to make a motor turn faster.

Answers: 1. more current; stronger magnetic field; more turns

4.6 Loudspeakers and other devices

In a loudspeaker, alternating current in a coil attached to a cone sits in a permanent magnet's field. The alternating force makes the cone vibrate, producing sound. Moving-coil meters, maglev trains and railguns also use the motor effect.

Figure 21. A moving-coil loudspeaker.

Quick check

1. Why must a loudspeaker be driven by alternating current?

Answers: 1. the cone must move back and forth

4.7 Magnetic levitation and launchers

Maglev trains use magnetic forces between coils in the guideway and magnets on the train to lift it clear of the track, removing rolling friction, and to propel it forwards. Electromagnetic launchers (on some aircraft carriers) accelerate aircraft using the motor effect.

Figure 22. Magnetic levitation lifts a train above its guideway.

Quick check

1. Why can maglev trains reach higher speeds than conventional trains?

Answers: 1. no wheel–rail contact: no rolling friction and less wear

5 Energy and fields

Work done by electric and magnetic forces.

5.1 Work done by fields

Electric fields can do work on charges, changing their kinetic energy by qΔV. Magnetic forces on free charges are always perpendicular to velocity, so they do no work: they change direction but not speed.

Feature	Electric field	Magnetic field
force on stationary charge	yes (qE)	no
force direction	along field lines	perpendicular to v and B
effect on speed	can change	never changes
path (perpendicular entry)	parabola	circle

Quick check

1. Can a magnetic field increase a particle's speed?

Answers: 1. no

5.2 Earth's magnetic field and charged particles

Earth's magnetic field deflects charged particles in the solar wind, trapping some in the Van Allen belts. Near the poles, particles follow field lines into the atmosphere and excite atoms, which emit light as aurorae.

Quick check

1. What would happen to Earth's atmosphere without its magnetic field?

Answers: 1. the solar wind could gradually strip it away

6 Exam technique and further practice

Applying the ideas.

6.1 Method checklist

•Use energy (qV = ½mv²) for acceleration through a pd.
•In electric deflection, treat motion along and across the plates separately.
•Use r = mv/qB and remember T is independent of v.
•Reverse direction rules for negative charges.
•Include sin θ for angled velocities or wires.

6.2 Multi-step problem: mass spectrometer

WORKED EXAMPLE 25 Ion path (exam level)

Singly charged ions of mass 3.3 × 10⁻²⁶ kg are accelerated from rest through 1.2 kV and enter a 0.25 T field at right angles. Find (a) their speed; (b) the radius of their path; (c) the electric field needed in a selector with B = 0.25 T.

Solution

(a) v = √(2qV/m) = 1.08 × 10⁵ m s⁻¹M1 A1

(b) r = mv/qB = 8.9 cmA1

6.3 Multi-step problem: deflection and energy

WORKED EXAMPLE 26 Beam between plates (exam level)

Electrons accelerated through 1.5 kV enter plates 4.0 cm long and 1.0 cm apart with 100 V across them. Find (a) their entry speed; (b) the time between the plates; (c) the deflection on leaving.

Solution

(a) v = √(2eV/m) = 2.3 × 10⁷ m s⁻¹A1

(b) t = 0.040/2.3 × 10⁷ = 1.7 × 10⁻⁹ sA1

6.4 Graded practice

WORKED EXAMPLE 27 Alpha particle (standard)

An alpha particle (q = 3.2 × 10⁻¹⁹ C, m = 6.6 × 10⁻²⁷ kg) moves at 1.5 × 10⁷ m s⁻¹ perpendicular to a 1.2 T field. Find the radius.

Solution

r = mv/qBM1

= 0.26 mA1

WORKED EXAMPLE 28 Wires (standard)

Find the force per metre between two wires 5.0 cm apart carrying 20 A and 30 A.

Solution

F/L = 2 × 10⁻⁷ × 600/0.050M1

= 2.4 × 10⁻³ N m⁻¹A1

WORKED EXAMPLE 29 Doubling speed (standard)

A charged particle in a magnetic field doubles its speed. What happens to the radius and the period?

Solution

r ∝ v: radius doublesA1

T independent of v: unchangedA1

6.5 Graded practice: electric fields

WORKED EXAMPLE 30 Stopping distance (exam level)

An electron moving at 2.0 × 10⁶ m s⁻¹ enters a uniform field of 500 V m⁻¹ directed along its motion. How far does it travel before stopping?

Solution

½mv² = eEdM1

d = 9.11 × 10⁻³¹ × 4.0 × 10¹²/(2 × 1.6 × 10⁻¹⁹ × 500) = 2.3 cmA1

WORKED EXAMPLE 31 Deflection ratio (standard)

By what factor does the deflection change if the electron's entry speed doubles (same plates and pd)?

Solution

y = ½a(L/v)²: y ∝ 1/v²M1

Falls to one quarterA1

6.6 Graded practice: magnetic forces

WORKED EXAMPLE 32 Electron period (standard)

Find the period of an electron in a 0.010 T field.

Solution

T = 2πm/eBM1

= 3.6 × 10⁻⁹ sA1

WORKED EXAMPLE 33 Earth's field (exam level)

A proton moves at 5.0 × 10⁵ m s⁻¹ perpendicular to Earth's field (5.0 × 10⁻⁵ T). Find the radius of its path.

Solution

r = mv/eB = 1.67 × 10⁻²⁷ × 5.0 × 10⁵/(1.6 × 10⁻¹⁹ × 5.0 × 10⁻⁵)M1

= 104 mA1

WORKED EXAMPLE 34 Wire in a field (exam level)

A 0.20 m horizontal wire of mass 5.0 g is supported by the magnetic force in a horizontal 0.50 T field perpendicular to it. Find the current.

Solution

BIL = mgM1

I = 0.049/(0.50 × 0.20) = 0.49 AA1

6.7 Graded practice: devices

WORKED EXAMPLE 35 Cyclotron design (exam level)

A cyclotron must accelerate deuterons (m = 3.34 × 10⁻²⁷ kg, q = e) in a 1.2 T field. Find the supply frequency.

Solution

f = qB/2πmM1

= 9.1 MHzA1

WORKED EXAMPLE 36 Selector and spectrometer (exam level)

Ions leave a selector (E = 4.0 × 10⁴ V m⁻¹, B = 0.20 T) and enter a 0.20 T field, landing 30 cm from the entry slit. Find their mass (charge e).

Solution

v = E/B = 2.0 × 10⁵ m s⁻¹; r = 0.15 mM1

m = qBr/v = 1.6 × 10⁻¹⁹ × 0.20 × 0.15/2.0 × 10⁵ = 2.4 × 10⁻²⁶ kgA1

WORKED EXAMPLE 37 Motor coil (standard)

Explain why the torque on a motor coil is zero when its plane is perpendicular to the field.

Solution

Forces on the two sides act along the same line through the axisR1

No perpendicular distance, so no turning effectR1

6.8 Graded practice: explanations

WORKED EXAMPLE 38 Speed unchanged (exam level)

Explain why a charged particle moving in a uniform magnetic field keeps a constant speed.

Solution

Magnetic force is always perpendicular to velocityR1

So it does no work and kinetic energy is unchangedR1

WORKED EXAMPLE 39 Period independent of speed (exam level)

Explain why the period of a charged particle in a magnetic field does not depend on its speed.

Solution

r ∝ v, so the circumference ∝ vR1

T = 2πr/v = 2πm/qB is independent of vR1

WORKED EXAMPLE 40 Parallel cables (standard)

Why are the two wires of a heavy-current cable held firmly together?

Solution

They carry opposite currents and repelR1

Large currents give large forces, especially during surgesR1

6.9 Self-test

Try these without notes, then check the answers.

Quick check

1. Find the energy in J of a 10 keV electron.

2. What path does a charge take entering an electric field at right angles?

3. What is the force on a stationary charge in a magnetic field?

4. Write the radius formula for circular motion in a B field.

5. Which way do parallel wires with currents in the same direction move?

6. What does a split-ring commutator do?

Answers: 1. 1.6 × 10⁻¹⁵ J 2. a parabola 3. zero 4. r = mv/qB 5. towards each other (attract) 6. reverses the coil current every half turn

6.10 Graded practice: parallel wires and balances

WORKED EXAMPLE 41 Force per metre ratio (standard)

Two wires carry currents I and 2I a distance r apart. Compare the forces on each wire.

Solution

Newton's third lawR1

Forces are equal in size and opposite in directionA1

WORKED EXAMPLE 42 Distance for a given force (exam level)

Two wires each carry 15 A in the same direction. At what separation is the force per metre 1.0 × 10⁻³ N m⁻¹?

Solution

r = μ₀I₁I₂/(2π × F/L) = 2 × 10⁻⁷ × 225/1.0 × 10⁻³M1

= 4.5 cmA1

WORKED EXAMPLE 43 Balance reading (exam level)

A current balance reading changes by 1.8 g when 3.0 A flows in a 4.0 cm wire. Find B.

Solution

F = 0.0018 × 9.81 = 0.0177 NM1

B = 0.0177/(3.0 × 0.040) = 0.15 TA1

6.11 Summary of instruments

Instrument	Fields used	Key equation	Measures or does
electron gun	electric	eV = ½mv²	accelerates electrons
deflecting plates	electric	y = ½at²	steers beams
velocity selector	crossed E and B	v = E/B	selects one speed
mass spectrometer	magnetic	r = mv/qB	separates by mass
cyclotron	electric (gap) + magnetic	f = qB/2πm	accelerates ions
fine-beam tube	electric + magnetic	e/m = 2V/B²r²	charge-to-mass ratio

WORKED EXAMPLE 44 Choosing an instrument (standard)

Which instrument would you use to find the ratio of two isotopes in a sample, and why?

Solution

Mass spectrometerA1

Different masses land at different positions; detector counts give relative abundancesR1

6.12 Key terms

Term	Meaning
electronvolt	energy gained by an electron through 1 V
magnetic force on a charge	F = qvB sin θ
velocity selector	crossed E and B fields passing v = E/B
mass spectrometer	separates ions by mass using r = mv/qB
cyclotron	accelerator using dees and a fixed-frequency supply
motor effect	force F = BIL sin θ on a current in a field

6.13 Common exam errors

•Forgetting to reverse the force direction for electrons.
•Saying the magnetic force speeds particles up.
•Using the full speed instead of the perpendicular component in qvB sin θ.
•Mixing up attraction and repulsion for parallel currents.
•Forgetting that the deflection inside plates is parabolic but straight afterwards.

7 Common misconceptions and links

Students often think…	Actually…
magnetic forces change a particle's speed	they change direction only
stationary charges feel magnetic forces	only moving charges do
the left-hand rule applies directly to electrons	reverse direction for negative charges
a faster particle in a B-field completes its circle faster	the period is independent of speed
parallel currents in the same direction repel	they attract

How D.3 connects to the rest of the course

Topic	Connection
D.2 Electric and magnetic fields	field definitions
A.2 Forces and momentum	circular motion
D.4 Induction (HL)	generators and Lenz's law
E.1 Structure of the atom	mass spectrometry, isotopes

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Singly charged ions of mass 3.3 × 10⁻²⁶ kg are accelerated from rest through 1.2 kV and then enter a uniform magnetic field of 0.25 T at right angles.

(a) Calculate the speed of the ions. [2]

(b) Calculate the radius of their path in the magnetic field. [2]

(d) Before entering, the ions pass through a velocity selector with B = 0.25 T. Calculate the electric field needed. [2]

(e) State the effect on the radius of using doubly charged ions of the same mass and speed. [1]

Markscheme

Part	Answer	Marks
(a)	½mv² = qV	M1
	v = 1.08 × 10⁵ m s⁻¹	A1
(b)	r = mv/qB = 3.3 × 10⁻²⁶ × 1.08 × 10⁵/(1.6 × 10⁻¹⁹ × 0.25)	M1
	8.9 cm	A1
(c)	magnetic force always perpendicular to velocity	R1
	constant magnitude provides centripetal force	R1
(d)	E = vB = 1.08 × 10⁵ × 0.25	M1
	2.7 × 10⁴ V m⁻¹	A1
(e)	radius halves	A1

Examiner's comments. In (c), the best answers stated that the force does no work so speed is constant.

9 Check your understanding

Section A: multiple choice

1. The magnetic force on a charge moving parallel to B is

A. zero B. maximum C. qvB D. qE

2. The radius of a charged particle's circular path is

A. qB/mv B. qvB C. mv/qB D. mB/qv

3. In a velocity selector, undeflected particles have speed

A. B/E B. E/B C. EB D. qE/m

4. The force on a 0.5 m wire carrying 2 A perpendicular to 0.1 T is

A. 1 N B. 10 N C. 0.01 N D. 0.1 N

5. Parallel wires with currents in opposite directions

A. attract B. repel C. feel no force D. rotate

6. A charge entering an electric field at right angles follows

A. a parabola B. a circle C. a straight line D. a helix

7. The work done by a magnetic force on a free charge is

A. qvB B. qBr C. positive D. zero

8. In a mass spectrometer, heavier ions

A. have smaller radii B. are undeflected C. have larger radii D. move faster

Section B: structured

9. An alpha particle moves at 1.5 × 10⁷ m s⁻¹ perpendicular to a 1.2 T field. Find the radius (q = 3.2 × 10⁻¹⁹ C, m = 6.6 × 10⁻²⁷ kg). [2]

10. Describe how a mass spectrometer separates isotopes. [4]

11. Explain how a split-ring commutator keeps a motor turning. [2]

12. Calculate the force per metre between two wires 5.0 cm apart carrying 20 A and 30 A. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. r = 0.26 m.

10. Ionise; accelerate; velocity select; enter B field; r ∝ m; detected at different positions.

11. Reverses current each half turn so the couple acts in the same rotational sense.

12. F/L = 2 × 10⁻⁷ × 600/0.050 = 2.4 × 10⁻³ N m⁻¹.

10 Summary and topic map

Idea	Remember
E fields	qV = ½mv²; parabola for perpendicular entry
B on charge	F = qvB sin θ
Circles	r = mv/qB; T = 2πm/qB
Selector	v = E/B
B on wire	F = BIL sin θ
Parallel wires	F/L = μ₀I₁I₂/2πr

Before the exam, make sure you…

•reverse direction rules for negative charges
•use energy (qV) for acceleration
•remember B fields do no work
•include sin θ for angled fields

Motion in Electromagnetic Fields on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 23. Topic map for Motion in Electromagnetic Fields.

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