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D.3.6 Charged Particles in Electric and Magnetic Fields — IB Physics HL Revision Notes & Topical Questions

5 min read 10 parts Lesson 33 of 39 1 quiz

📚 IB Physics HL Hub / Topical Revision
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Lesson summary

D.3.6 Charged Particles in Electric and Magnetic Fields — IB PHYSICS SL notes: Analyse the motion of charges in perpendicular electric and magnetic fields, including the velocity selector and Thomson's measurement of e/m.Two forces — in a region with both fields the charge feels …

  • Crossed fields
  • Positive particle
  • Negative particle
  • Reversing the sign
  • Selected speed
  • Faster particles

Part 1 of 8Combined fields

Two forces
  • Two forces in a region with both fields the charge feels a force parallel to E and a force perpendicular to B
  • The electric part qE does not depend on motion.
  • The magnetic part has size qvB sin θ.
Crossed fields
  • E, B and the velocity are mutually perpendicular
  • ExampleE up the page, B out of the page and the particle moving to the right.
Net force
  • Net force find it by adding the two forces as vectors

Part 2 of 8Positive and negative particles

Positive and negative particles
Positive particle
  • Positive particle the electric force acts upward along E and the magnetic force acts downward
  • The magnetic direction comes from Fleming's left-hand rule with the second finger along the velocity.
  • The two forces are opposite.
Negative particle
  • Negative particle the electric force acts downward against E and the magnetic force acts upward
  • Use the rule with the current opposite to the velocity.
  • The two forces are again opposite.
Reversing the sign
  • Reversing the sign both forces reverse, so any balance is unchanged

Part 3 of 8The velocity selector

The velocity selector
Balance
  • Balance the particle moves in a straight line at constant speed when the electric and magnetic forces are equal
  • FE = FB, so qE = Bqv.
Selected speed
  • Selected speed v = E/B, whatever the charge or mass
Faster particles
  • Faster particles the magnetic force is larger and they are deflected one way
Slower particles
  • Slower particles the electric force wins and they are deflected the other way
Slit
  • Slit a slit at the exit passes only particles of speed E/B
Use
  • Use it is the first stage of many mass spectrometers, giving a single-speed beam
  • A magnetic field alone then sorts ions by r = mv/(qB), according to m/q.
Between plates
  • Between plates with E = V/d, v = V/(Bd)

Part 4 of 8Charge-to-mass ratio

Charge-to-mass ratio
  • Charge-to-mass ratio the ratio of the total charge of a particle to its mass, q/m
Electron
  • Electron e/me = 1.60 × 10−19/9.11 × 10−31 = 1.76 × 1011 C kg−1
Proton
  • Proton e/mp = 9.58 × 107 C kg−1

Part 5 of 8Thomson's measurement of e/m

1
Send an electron beam through perpendicular electric and magnetic fields, using parallel plates for E and Helmholtz coils for B.
2
Adjust the p.d. across the plates until the beam on the fluorescent screen is straight and horizontal.
3
At balance the speed is v = V/(Bd) = E/B.
4
Switch off the electric field so the beam follows a circle in the magnetic field alone.
5
Measure the radius r of the circular path.
6
Use r = mv/(eB) to get e/m = v/(rB).
Result
  • Result e/m = E/(rB2) = V/(rB2d), so V, d, r and B are the four quantities needed
Alternative
  • Alternative accelerating electrons through a known V and measuring r in B gives e/m = 2V/(B2r2)
Significance
  • Significance about 1.76 × 1011 C kg−1 showed that cathode rays are particles much lighter than atoms

Part 6 of 8Other combined-field motion

Unbalanced fields
  • Unbalanced fields a charge starting from rest in perpendicular E and B follows a cycloid
  • It drifts steadily sideways at speed E/B while circling.
Hall effect
  • Hall effect charge carriers in a conductor in a perpendicular field build up charge on one side until qEH = qvB
  • The Hall voltage VH = Bvd, with d the strip width, is used to measure B.
  • Its sign shows the sign of the charge carriers: electrons in a metal strip and positive holes in some semiconductors.
Hall probe
  • Hall probe for a fixed current the Hall voltage is proportional to B

Part 7 of 8Problem strategy

1
Draw the directions of both forces for the sign of charge given.
2
Decide whether the particle goes straight (forces balanced) or is deflected.
3
If straight, equate the force magnitudes.
4
If it follows a circular arc, find the radius from the centripetal equation.
Problem strategy
Chaining
  • Chaining accelerate with qV = ½mv2, select with v = E/B, then bend with r = mv/(qB)
Restoring a straight beam
  • Restoring a straight beam increase E, decrease B or alter the speed so that qE = qvB
Units check
  • V m−1 divided by T gives m s−1

Part 8 of 8More from the specification

1Deflection in a Uniform Electric Field
  • A charged particle entering perpendicular to uniform electric field lines experiences a constant electric force F = qE,
  • resulting in parabolic deflection;
    • for instance, a positively charged boron nucleus accelerates toward the negative plate with a = qE/m.
2Motion in a Uniform Magnetic Field
  • When a charged particle moves through a uniform magnetic field of flux density B, a perpendicular magnetic Lorentz force F = qvB provides centripetal acceleration, forcing the particle into a circular trajectory of radius r = mv / (qB).
3Magnetic Force on a Current-Carrying Conductor
  • When a current-carrying conductor of length L is positioned within a uniform magnetic field of flux density B, the cumulative force on moving charge carriers is given by F = BIL sin θ, directed perpendicularly as defined by Fleming's left-hand rule.
4Deflection in a Uniform Electric Field
  • A charged particle entering perpendicular to uniform electric field lines experiences a constant electric force F = qE,
  • resulting in parabolic deflection;
    • for instance, a positively charged boron nucleus accelerates toward the negative plate with a = qE/m.
5Motion in a Uniform Magnetic Field
  • When a charged particle moves through a uniform magnetic field of flux density B, a perpendicular magnetic Lorentz force F = qvB provides centripetal acceleration, forcing the particle into a circular trajectory of radius r = mv / (qB).
6Magnetic Force on a Current-Carrying Conductor
  • When a current-carrying conductor of length L is positioned within a uniform magnetic field of flux density B, the cumulative force on moving charge carriers is given by F = BIL sin θ, directed perpendicularly as defined by Fleming's left-hand rule.
7Uniform
  • Same density throughout, so the weight acts at the centre.
8Length
  • Resistance is directly proportional to the length of a wire.
9Density
  • Mass per unit volume.
10Field lines
  • Lines showing the direction and strength of a magnetic field.
11Field strength
  • G = Fg/m.
12F = BIL
What the symbols mean: F = force on the wire (N) · B = magnetic flux density (T) · I = current (A) · L = length of wire in the field (m) · θ = angle between wire and field

Key terms

TermMeaning
Crossed fieldsElectric and magnetic fields perpendicular to each other and to the beam.
Velocity selectorDevice passing only particles with v = E/B.
Thomson experimentMeasured e/m of the electron using crossed fields.
Charge-to-mass ratioq/m of a particle; 1.76 × 10¹¹ C kg⁻¹ for the electron.
Hall voltageTransverse p.d. across a current-carrying strip in a perpendicular field.

Avoid these mistakes

1
Thinking the selector speed depends on the charge or mass — v = E/B for any particle.
2
Getting the two forces in the same direction by misusing the hand rule for the sign of the charge.
3
Using the selector formula with E and B not perpendicular.
4
Forgetting that after the selector only a magnetic force acts in the next region.
5
Mixing up which quantities are measured and which are calculated in Thomson's experiment.

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