The Lorentz Force and Motion in Fields
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with Newton's second law, vector cross products, and basic kinematics in two dimensions. Familiarity with the concept of a magnetic field from previous lectures is assumed.
Why this matters
The Lorentz force is the fundamental interaction that governs how charged particles behave in electromagnetic fields.
Understanding this force is essential because it bridges the gap between abstract field concepts and observable particle motion. It explains phenomena ranging from the deflection of electron beams in cathode-ray tubes to the confinement of plasma in fusion reactors. By mastering the resulting trajectories, you gain the tools to analyze complex systems in accelerator physics, astrophysics, and electronic devices.
The Lorentz Force in Plain Language
Imagine a charged particle moving through space. If there is an electric field, the particle feels a push or pull in the direction of the field (if positive) or opposite to it (if negative). This force depends only on the charge and the field strength, not on how fast the particle is moving. However, if there is also a magnetic field, the situation changes. The magnetic force is unique: it is always perpendicular to both the particle's velocity and the magnetic field direction. Because this force is always sideways, it does no work on the particle. It cannot speed the particle up or slow it down; it can only change the direction of motion. This is why a charged particle entering a uniform magnetic field travels in a circle or a helix, rather than accelerating linearly.
Equation of Motion in Uniform Fields
We begin with the total electromagnetic force acting on a point charge \( q \) of mass \( m \). According to the Lorentz force law, the force is the sum of the electric and magnetic components. In vector notation, this is expressed as:
Applying Newton's second law, \( m\dot{\vec{v}} = \vec{F} \), we obtain the equation of motion. For a uniform static magnetic field \( \vec{B} \) and zero electric field, the magnetic force provides the centripetal acceleration required for circular motion. The magnitude of the force is \( F = qv_\perp B \), where \( v_\perp \) is the component of velocity perpendicular to the field. Equating this to the centripetal force \( mv_\perp^2 / r_c \), we can solve for the radius of curvature.
The angular frequency of this circular motion, known as the cyclotron frequency, is independent of the particle's speed (for speeds well below \(c\); relativistically \(m\) is replaced by \(\gamma m\)). It is given by:
This result is crucial because it allows for the synchronization of electric fields with the particle's natural orbital period, a principle used in cyclotron accelerators.
Helical Trajectories
In general, the velocity vector \( \vec{v} \) can be decomposed into two components: one parallel to the magnetic field \( \vec{B} \) and one perpendicular to it. The component parallel to the field, \( v_\parallel \), experiences no magnetic force because the cross product \( \vec{v}_\parallel \times \vec{B} \) is zero. Consequently, the particle moves with constant velocity along the field lines. The perpendicular component, \( v_\perp \), undergoes uniform circular motion as described above. The superposition of these two motions results in a helical path.
The pitch of the helix, which is the distance traveled along the field direction during one complete revolution, is determined by the parallel velocity and the cyclotron period \( T = 2\pi/\omega_c \). This behavior explains why charged particles in the Earth's magnetosphere spiral along magnetic field lines, creating the auroras when they collide with atmospheric gases near the poles.
The \( \vec{E}\times\vec{B} \) Drift
When both electric and magnetic fields are present and uniform, the motion becomes more complex. The electric field exerts a constant force \( q\vec{E} \), which tends to accelerate the particle. However, the magnetic field deflects this acceleration. If the fields are perpendicular to each other, the particle undergoes a specific drift motion superposed on its gyration. This drift velocity is independent of the charge and mass of the particle, a remarkable result known as the \( \vec{E}\times\vec{B} \) drift.
This drift occurs in a direction perpendicular to both the electric and magnetic fields. It is a single-particle effect: during each gyration the electric field speeds the particle up on one side of its orbit and slows it on the other, so the radius changes and the orbit centre steps sideways. In plasma physics, this drift is significant because it can lead to the transport of plasma across magnetic field lines, which is a major challenge in magnetic confinement fusion devices.
The derivation
Lorentz Force and Cyclotron Motion
For a point charge \( q \) of mass \( m \) moving with velocity \( \vec{v} \) through an electric field \( \vec{E} \) and a magnetic field \( \vec{B} \), the total electromagnetic force is \( \vec{F} = q\left(\vec{E} + \vec{v}\times\vec{B}\right) \). Taking this as the definition of the fields and inserting it into \( m\dot{\vec{v}}=\vec{F} \), we solve the equation of motion for uniform static fields and obtain a helical cyclotron trajectory: circular gyration at angular frequency \( \omega_c = qB/m \) and radius \( r_c = mv_\perp/(qB) \) in the plane perpendicular to \( \vec B \), superposed with uniform glide along \( \vec B \) and a charge-independent \( \vec{E}\times\vec{B}/B^2 \) drift.
A common misconception is that the magnetic force does work on the particle. Since the force is always perpendicular to the velocity, the power delivered by the magnetic field, \( P = \vec{F}_B \cdot \vec{v} \), is strictly zero. Therefore, the kinetic energy of the particle remains constant in a purely magnetic field. Any change in speed must be attributed to the electric field component. C
Worked example
Problem. An electron (charge \( q = -1.60 \times 10^{-19} \ \mathrm{C} \), mass \( m = 9.11 \times 10^{-31} \ \mathrm{kg} \)) enters a uniform magnetic field of magnitude \( B = 0.50 \ \mathrm{T} \) with a velocity of \( v = 2.0 \times 10^6 \ \mathrm{m/s} \) perpendicular to the field. Calculate the radius of its circular path and its cyclotron frequency.
- Identify the given values: \( |q| = 1.60 \times 10^{-19} \ \mathrm{C} \), \( m = 9.11 \times 10^{-31} \ \mathrm{kg} \), \( B = 0.50 \ \mathrm{T} \), \( v_\perp = 2.0 \times 10^6 \ \mathrm{m/s} \).
- Calculate the radius using \( r_c = \frac{mv_\perp}{|q|B} \): \[ r_c = \frac{(9.11 \times 10^{-31} \ \mathrm{kg})(2.0 \times 10^6 \ \mathrm{m/s})}{(1.60 \times 10^{-19} \ \mathrm{C})(0.50 \ \mathrm{T})} \] \[ r_c = \frac{1.822 \times 10^{-24}}{8.0 \times 10^{-20}} \ \mathrm{m} = 2.28 \times 10^{-5} \ \mathrm{m} \]
- Calculate the cyclotron frequency using \( \omega_c = \frac{|q|B}{m} \): \[ \omega_c = \frac{(1.60 \times 10^{-19} \ \mathrm{C})(0.50 \ \mathrm{T})}{9.11 \times 10^{-31} \ \mathrm{kg}} \] \[ \omega_c = \frac{8.0 \times 10^{-20}}{9.11 \times 10^{-31}} \ \mathrm{rad/s} \approx 8.78 \times 10^{10} \ \mathrm{rad/s} \]
Units check. For radius: \( \frac{\mathrm{kg} \cdot \mathrm{m/s}}{\mathrm{C} \cdot \mathrm{T}} = \frac{\mathrm{kg} \cdot \mathrm{m/s}}{\mathrm{A \cdot s} \cdot \mathrm{N/(A \cdot m)}} = \frac{\mathrm{kg} \cdot \mathrm{m/s}}{\mathrm{N/s}} = \frac{\mathrm{kg} \cdot \mathrm{m/s}}{\mathrm{kg \cdot m/s^2} \cdot \mathrm{s}^{-1}} = \mathrm{m} \). Correct. For frequency: \( \frac{\mathrm{C} \cdot \mathrm{T}}{\mathrm{kg}} = \frac{\mathrm{A \cdot s} \cdot \mathrm{N/(A \cdot m)}}{\mathrm{kg}} = \frac{\mathrm{s} \cdot \mathrm{N/m}}{\mathrm{kg}} = \frac{\mathrm{s} \cdot \mathrm{kg \cdot m/s^2} / \mathrm{m}}{\mathrm{kg}} = \mathrm{s/s^2} = \mathrm{s^{-1}} \). Correct.
Three quick tests
- Does the magnetic force change the kinetic energy of a charged particle?
Answer
No, because the force is always perpendicular to the velocity, so it does no work. - What is the direction of the \( \vec{E}\times\vec{B} \) drift if \( \vec{E} \) is in the +x direction and \( \vec{B} \) is in the +y direction?
Answer
The drift is in the \(+z\) direction, since \(\hat x \times \hat y = \hat z\) — for positive and negative charges alike. - How does the cyclotron frequency depend on the particle's speed?
Answer
It is independent of the particle's speed, depending only on the charge-to-mass ratio and the magnetic field strength.
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.