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D.3 Motion in electromagnetic fields

Charged particles in uniform electric fields (parabolic paths), in uniform magnetic fields (circles), in crossed fields (velocity selectors), the force on a current-carrying wire, and the force between parallel wires.

Charges in E fieldsF = qvB sin θCircular motion in BCrossed fieldsF = BIL sin θParallel wires
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No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • In a uniform electric field a charge feels a constant force qE, so it moves like a projectile: a straight line along the field, or a parabola across it.
  • A charge moving through a magnetic field feels F = qvB sin θ, always perpendicular to its velocity, so in a uniform field it moves in a circle of radius r = mv/qB.
  • A current-carrying wire in a field feels F = BIL sin θ. Two parallel currents attract (same direction) or repel (opposite), with F/L = μ0I1I2/2πr.
  • D.3 is the same at SL and HL.

By the end you can

  • Describe and calculate the motion of a charge in a uniform electric field.
  • Use F = qvB sin θ and explain circular motion of a charge in a uniform magnetic field.
  • Analyse charges in perpendicular electric and magnetic fields.
  • Use F = BIL sin θ and find its direction.
  • Calculate the force per unit length between parallel wires.

1Charges in electric fields

  • Accelerating a charge: moving through a potential difference V, a charge gains kinetic energy qV. For an electron from rest: ½mv2 = eV.
Worked example

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

  1. ½mv2 = eV ⇒ v = √(2eV/m) M1
  2. = √(2 × 1.6 × 10−19 × 500/9.11 × 10−31) = 1.3 × 107 m s−1 A1
electronconstant force upwards (towards +): parabola, like a projectilestraight lineafter the plates
Constant velocity along the plates, constant acceleration across them: a parabola.
Worked example

An electron enters plates 5.0 cm long at 2.0 × 107 m s−1, at right angles to a field of 2.0 × 104 V m−1. Find its sideways deflection.

  1. time between the plates t = 0.050/2.0 × 107 = 2.5 × 10−9 s M1
  2. a = eE/m = 3.5 × 1015 m s−2 M1
  3. y = ½at2 = 1.1 cm A1

2Charges in magnetic fields

F = qvB sin θ DATA BOOKLET
θ = angle between v and B; F is perpendicular to both
  • Direction: use the left-hand rule (or F = qv × B) for positive charges; reverse it for electrons. A charge moving parallel to B feels no force.
−vFB into page (×)electron: F = qvBalways ⟂ to v,so circular motionqvB = mv²/rr = mv/(qB)
The force is always at 90° to the velocity, so it changes direction but never speed.
r = mvqB
from qvB = mv2/r (derive it, it is not in the booklet)
Worked example

An electron moves at 2.0 × 107 m s−1 perpendicular to a 5.0 mT field. Find the radius of its path.

  1. r = mv/qB = 9.11 × 10−31 × 2.0 × 107/(1.6 × 10−19 × 5.0 × 10−3) M1
  2. = 2.3 cm A1
  • No work is done by the magnetic force (it is perpendicular to the motion), so the speed and kinetic energy stay constant.
  • Uses: mass spectrometers separate ions by r ∝ m/q; particle accelerators bend beams; the Earth's field steers charged particles into the auroras.
Common mistakeForgetting to reverse the rule for electrons. The left-hand rule gives the direction for conventional current (positive charges).

3Crossed electric and magnetic fields

  • Velocity selector: with E and B perpendicular to each other and to v, the electric force qE and the magnetic force qvB act in opposite directions. Only particles with v = E/B go straight through, whatever their charge or mass.
Worked example

Find the speed selected by E = 2.0 × 104 V m−1 and B = 0.10 T.

  1. qE = qvB ⇒ v = E/B M1
  2. = 2.0 × 105 m s−1 A1

4Forces on currents

F = BIL sin θ DATA BOOKLET
force on a length L of wire carrying current I, at angle θ to B
NSI outFF = BIL sin θF ⟂ to both B and I(left-hand rule:First finger Field,seCond finger Current,thuMb Motion)
The force is perpendicular to both the field and the current.
Worked example

A 20 cm length of wire carries 3.0 A at right angles to a 0.15 T field. Find the force.

  1. F = BIL = 0.15 × 3.0 × 0.20 M1
  2. = 0.090 N A1
FL = μ0I1I22πr DATA BOOKLET
force per unit length between parallel wires; μ0 = 4π × 10−7 T m A−1
I₁I₂attractrsame direction: attractopposite: repel
Each wire sits in the field of the other, so each feels a force.
Worked example

Two long parallel wires 5.0 cm apart each carry 10 A in the same direction. Find the force per metre.

  1. F/L = 4π × 10−7 × 10 × 10/(2π × 0.050) M1
  2. = 4.0 × 10−4 N m−1, attractive A1
Exam tipThe forces on the two wires are a Newton's third law pair: equal and opposite, even if the currents are different.
Linking questionThe magnetic force on moving charges is the start of induction in D.4: moving a wire through a field pushes its free electrons along it, creating an emf.
Idea for your IA / explorationMeasure the force on a current-carrying wire with a top-pan balance and a magnet (current balance), and test F ∝ I and F ∝ L.

Test yourself — cover the answers first

Quick check

  1. What path does a charge follow when it enters a uniform E field at right angles?
  2. What force acts on a charge moving parallel to a magnetic field?
  3. Why doesn't a magnetic field change a particle's speed?
  4. If v doubles, what happens to the radius in a magnetic field?
  5. What speed passes undeflected through a velocity selector?
  6. Two wires carry currents in opposite directions. Do they attract or repel?

Exam-style

  1. A proton (m = 1.67 × 10−27 kg) moves in a circle of radius 0.20 m in a 0.50 T field. Find its speed and the period.
  2. A horizontal 0.50 m wire of mass 20 g is held at right angles to a horizontal 0.40 T field. What current makes it float?
Show answers
  1. a parabola
  2. none
  3. the force is perpendicular to the velocity, so it does no work
  4. it doubles
  5. v = E/B
  6. repel
  7. v = qBr/m = 1.6 × 10−19 × 0.50 × 0.20/1.67 × 10−27 = 9.6 × 106 m s−1
    T = 2πr/v = 2πm/qB
    = 1.3 × 10−7 s (independent of speed)
  8. BIL = mg
    I = 0.020 × 9.8/(0.40 × 0.50)
    = 0.98 A
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.