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D.3Motion in electromagnetic fields

Old TV screens, particle accelerators, mass spectrometers and the northern lights all depend on how charged particles move in electric and magnetic fields. An electric field speeds particles up or bends them into parabolas; a magnetic field bends them into circles without changing their speed. This topic works out both, and the forces on wires carrying currents. There is no additional HL content in D.3.

Knowledge and science

Nature of science

ExperimentsModelsMeasurementGlobal impact of science

Experiments. In 1897 J.J. Thomson bent beams of "cathode rays" with electric and magnetic fields and measured their charge-to-mass ratio. It was about 2000 times larger than for hydrogen ions, so the rays had to be tiny particles: he had discovered the electron.

Models. Magnetic field lines give a picture of an invisible field. Combined with the rule $F = qvB\sin\theta$, they let us predict exactly how a particle will curve.

Measurement. Until 2019 the ampere was defined by the force between two parallel wires. Today it is defined by fixing the value of the elementary charge $e$.

Global impact. Magnetic fields steer particles in cancer-treatment accelerators and in the Large Hadron Collider, and mass spectrometers identify substances from drug tests to space missions.

ToK: questions to think about

  • Is a field pattern like a map? A map simplifies the land to make it useful. In what ways are field lines a simplified "map" of reality, and what do they leave out?
  • How do we know about particles we can't see? Thomson never saw an electron; he inferred it from the bending of a beam. How much can be known from indirect evidence?
  • Why so many hand rules? Different countries and textbooks teach Fleming's left-hand rule, a right-hand slap rule or a vector cross product, all for the same physics. Does it matter which convention we learn?
  • Who owns discoveries made with huge machines? Modern particle physics needs thousands of scientists from many countries. How does knowledge production change when no single person can understand the whole experiment?

How do physics, NoS and ToK fit together? →

1. Charged particles in a uniform electric field

In a uniform field $E$ (for example between parallel plates, D.2), a charge $q$ feels a constant force $F = qE$, in the direction of the field for a positive charge and opposite to it for a negative one. A constant force gives a constant acceleration $a = \dfrac{qE}{m}$, so the kinematics equations of A.1 apply.

Accelerating from rest

A charge accelerated from rest through a potential difference $V$ gains kinetic energy equal to the work done on it:

$$\tfrac{1}{2}mv^2 = qV$$

Worked example: an electron gun

Electrons are accelerated from rest through a p.d. of 2000 V. Find their final speed. ($m_e = 9.11 \times 10^{-31}$ kg.)

$v = \sqrt{\dfrac{2qV}{m}} = \sqrt{\dfrac{2 \times 1.60 \times 10^{-19} \times 2000}{9.11 \times 10^{-31}}} = 2.7 \times 10^7\ \text{m s}^{-1}$, about 9% of the speed of light.

Entering a field at right angles

A particle entering the field at right angles to it keeps its original velocity along the plates, but accelerates steadily across them. Its path is a parabola, exactly like a ball thrown horizontally in a gravitational field (A.1). After leaving the field it travels in a straight line.

An electron enters the space between two horizontal parallel plates, moving to the right. The top plate is positive. Between the plates the electron follows a parabola curving upwards towards the positive plate. After leaving the plates it moves in a straight line. +− u E
An electron entering a uniform field at right angles follows a parabola towards the positive plate. (Grey arrows: field direction, from + to −.)

Worked example: deflecting an electron beam

The electrons from the previous example ($u = 2.65 \times 10^7\ \text{m s}^{-1}$) enter a uniform field of $1.0 \times 10^4\ \text{V m}^{-1}$ at right angles. The plates are 5.0 cm long. How far are the electrons deflected sideways by the time they leave the plates?

Time between the plates: $t = \dfrac{0.050}{2.65 \times 10^7} = 1.89 \times 10^{-9}$ s.

Acceleration across the plates: $a = \dfrac{eE}{m} = \dfrac{1.60 \times 10^{-19} \times 1.0 \times 10^4}{9.11 \times 10^{-31}} = 1.76 \times 10^{15}\ \text{m s}^{-2}$.

Deflection: $y = \tfrac{1}{2}at^2 = 0.5 \times 1.76 \times 10^{15} \times (1.89 \times 10^{-9})^2 = 3.1 \times 10^{-3}$ m, about 3 mm.

2. The magnetic force on a moving charge

A magnetic field exerts a force on a charge only if the charge is moving, and not moving parallel to the field:

$$F = qvB\sin\theta$$

$B$ is the magnetic field strength in tesla (T), $v$ the speed and $\theta$ the angle between the velocity and the field. The force is greatest when $v$ is at right angles to $B$, and zero when they are parallel. A stationary charge feels no magnetic force at all.

Direction: the force is perpendicular to both $v$ and $B$. Use Fleming's left-hand rule: hold the thumb, first finger and second finger of your left hand at right angles. The First finger points along the Field, the seCond finger along the Current (the direction a positive charge moves), and the thuMb gives the force (the Motion). For a negative charge such as an electron, the "current" is opposite to its velocity, so the force is reversed.

Because the force is always at right angles to the velocity, it does no work. The magnetic force changes the particle's direction but never its speed, so its kinetic energy stays constant.

3. Circular motion in a magnetic field

When a charged particle moves at right angles to a uniform magnetic field, the force has a constant size and is always perpendicular to the velocity. That is exactly the condition for uniform circular motion (A.2). The magnetic force provides the centripetal force:

$$qvB = \frac{mv^2}{r} \qquad\Rightarrow\qquad r = \frac{mv}{qB}$$
A uniform magnetic field pointing out of the page, shown by a grid of dots. A positive particle at the top of a circle moves to the right. The magnetic force on it points down towards the centre of the circle, so it moves clockwise in a circle. vF r B out of the page
A positive particle in a field out of the page (dots). The magnetic force always points to the centre, so the particle moves round a circle, clockwise.

Worked example: a proton in a magnetic field

A proton moving at $3.0 \times 10^6\ \text{m s}^{-1}$ enters a uniform magnetic field of 0.20 T at right angles. Find the radius of its path and the time for one orbit. ($m_p = 1.67 \times 10^{-27}$ kg.)

$r = \dfrac{mv}{qB} = \dfrac{1.67 \times 10^{-27} \times 3.0 \times 10^6}{1.60 \times 10^{-19} \times 0.20} = 0.16$ m.

$T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.20} = 3.3 \times 10^{-7}$ s.

An electron at the same speed would curve the other way, in a circle about 1800 times smaller (because its mass is 1800 times smaller).

4. Measuring the charge-to-mass ratio

Rearranging $r = \dfrac{mv}{qB}$ gives $\dfrac{q}{m} = \dfrac{v}{Br}$. So measuring the radius of a particle's path in a known field, and its speed, gives its charge-to-mass ratio, which identifies the particle.

In a common school experiment, electrons are accelerated through a p.d. $V$ and then bent into a circle by the field of a pair of coils. Combining $\tfrac{1}{2}mv^2 = eV$ with $v = \dfrac{eBr}{m}$:

$$\frac{e}{m} = \frac{2V}{B^2r^2}$$
A glass bulb in a darkened room. An electron gun near the bottom fires a beam that glows violet-pink as it travels round an almost complete circle inside the bulb. A vertical ladder of measuring rungs stands in the bulb behind the beam.
A fine-beam tube. Electrons from the electron gun (bottom) are bent into a circle by the uniform magnetic field of two large coils outside the bulb (not in the picture). The bulb holds a little gas, which glows where the electrons hit it, so the path is visible. The rungs inside the bulb are used to measure the radius $r$. Photo: Marcin Białek, Wikimedia Commons, CC BY-SA 4.0. Resized.

Worked example: measuring e/m

Electrons accelerated through 250 V move in a circle of radius 4.5 cm in a field of 1.2 mT. Find $\dfrac{e}{m}$ and compare it with the accepted value, $1.76 \times 10^{11}\ \text{C kg}^{-1}$.

$\dfrac{e}{m} = \dfrac{2 \times 250}{(1.2 \times 10^{-3})^2 \times (0.045)^2} = 1.7 \times 10^{11}\ \text{C kg}^{-1}$.

That's within about 3% of the accepted value. The radius is hard to measure precisely, and since it is squared, its percentage uncertainty is doubled.

A mass spectrometer uses the same idea. Ions with the same charge and speed curve by different amounts according to their mass, so they land in different places. This separates isotopes and identifies molecules.

5. Crossed electric and magnetic fields

If a uniform electric field and a uniform magnetic field are at right angles to each other and to the particle's velocity, the two forces can point in opposite directions. When they are equal in size, the particle passes straight through, undeflected:

$$qE = qvB \qquad\Rightarrow\qquad v = \frac{E}{B}$$
A velocity selector. A positive particle moves to the right between horizontal plates; the top plate is positive, so the electric field points down. A magnetic field points into the page, shown by crosses. The electric force on the particle is downwards and the magnetic force upwards. When they are equal, the particle goes straight through. +− qvBqE v
A velocity selector: E down (plates), B into the page (crosses). For a positive particle, the electric force qE is down and the magnetic force qvB is up. Only particles with speed $v = E/B$ go straight through.

Slower particles feel a weaker magnetic force and are pushed towards the electric force's direction; faster ones are bent the other way. This arrangement, a velocity selector, picks out particles of a single speed, for example before a mass spectrometer. Notice that the result doesn't depend on the charge or mass.

Worked example: choosing the speed

A velocity selector has $E = 3.0 \times 10^4\ \text{V m}^{-1}$ and $B = 0.15$ T. What speed of particle passes straight through? What happens to an alpha particle moving twice as fast?

$v = \dfrac{E}{B} = \dfrac{3.0 \times 10^4}{0.15} = 2.0 \times 10^5\ \text{m s}^{-1}$.

At twice this speed the magnetic force is twice the electric force, so the (positive) alpha particle is deflected in the direction of the magnetic force.

6. The force on a current-carrying wire

A current is a flow of moving charges, so a wire carrying a current in a magnetic field feels a force (the motor effect):

$$F = BIL\sin\theta$$

$I$ is the current, $L$ the length of wire in the field and $\theta$ the angle between the wire and the field. The direction comes from Fleming's left-hand rule, with the second finger along the current. There's no force if the wire is parallel to the field.

This follows from $F = qvB\sin\theta$: in a time $t$, charge $q = It$ moves a distance $L = vt$, so $qv = IL$.

A wire, seen end-on and carrying current into the page, sits between the north pole of a magnet on the left and the south pole on the right. The magnetic field points from left to right. The force on the wire is downwards. NS FB
Current into the page (⊗) in a field from N to S: by Fleming's left-hand rule the force is downwards. Reverse the current or the field and the force reverses.

Worked example: a wire at an angle

A 0.25 m length of wire carries 3.0 A in a uniform field of 0.40 T. The wire makes an angle of 30° with the field. Find the force on it.

$F = BIL\sin\theta = 0.40 \times 3.0 \times 0.25 \times \sin 30^\circ = 0.15$ N.

7. The force between parallel wires

Each current-carrying wire sits in the magnetic field of the other, so two parallel wires exert forces on each other:

$$\frac{F}{L} = \mu_0\frac{I_1I_2}{2\pi r}$$

$\frac{F}{L}$ is the force per metre of wire, $r$ the separation and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ the permeability of free space. Currents in the same direction attract; currents in opposite directions repel. By Newton's third law, the forces on the two wires are equal and opposite, even if the currents are different.

Two parallel wires seen end-on, both carrying current out of the page. Each is surrounded by circular field lines. At each wire, the field of the other wire points vertically, and the resulting forces point towards each other: the wires attract. FF I₁I₂
Two currents out of the page, each in the other's magnetic field. The forces pull them together.

Worked example: power cables

Two long parallel wires 5.0 cm apart carry currents of 20 A and 15 A in the same direction. Find the force on each 1.0 m length of wire, and its direction.

$\dfrac{F}{L} = \dfrac{4\pi \times 10^{-7} \times 20 \times 15}{2\pi \times 0.050} = 1.2 \times 10^{-3}\ \text{N m}^{-1}$, so 1.2 mN on each metre, attracting the wires towards each other.

The forces are small for ordinary currents, but the huge currents in a short circuit can make cables jump violently.

The old definition of the ampere was based on this: two long wires 1 m apart, each carrying 1 A, attract with a force of $2 \times 10^{-7}$ N per metre.

8. Common mistakes

9. Check your understanding

An electron and a proton enter the same magnetic field at the same speed, at right angles to it. Compare their paths.

They curve in opposite directions (opposite charges). The electron's radius is about 1800 times smaller, because $r = \frac{mv}{qB}$ and its mass is 1800 times smaller.

The speed of a charged particle moving in a circle in a magnetic field is doubled. What happens to the radius and to the time for one orbit?

The radius doubles ($r \propto v$). The period stays the same ($T = \frac{2\pi m}{qB}$).

Why does a magnetic field do no work on a moving charge?

The magnetic force is always perpendicular to the velocity, so there is no component of force in the direction of motion. Work = force × displacement in the direction of the force = 0.

A wire carrying current due north lies in a vertical magnetic field pointing downwards. In which direction is the force on the wire?

West. Fleming's left-hand rule: first finger down (field), second finger north (current), thumb points west.

The current in both of two parallel wires is doubled, and their separation is halved. By what factor does the force per metre change?

$\frac{F}{L} \propto \frac{I_1I_2}{r}$: $\frac{2 \times 2}{0.5} = 8$ times larger.

Practise D.3 questions