Mechanics foundations for charged-particle motion in magnetic fields — from centripetal kinematics to cyclotron frequency.
When a charged particle moves through a magnetic field, the field pushes it sideways — perpendicular to its velocity at all times. That's exactly the kind of force needed for circular motion. So the bridge equation is:
Everything else in this reference is a consequence of that one statement. Mechanics gives us what circular motion requires; magnetism supplies that requirement.
Magnetic fields are trajectory benders, not speed-changers. They curve the path without doing any work. That's why you get arcs, circles, and helices — not acceleration along a line.
Term interrogation — before deriving anything, know what every symbol means physically.
| Symbol | Physical meaning |
|---|---|
| $m$ | Mass of the particle — resists changes in motion (inertia) |
| $v$ | Speed — magnitude of velocity; stays constant in pure magnetic circular motion |
| $r$ | Radius of curvature — how tight the circular path is |
| $T$ | Period — time to complete one full revolution |
| $f$ | Frequency — revolutions per second; $f = 1/T$ |
| $\omega$ | Angular speed — radians swept per second; how fast the angle changes |
| $a_c$ | Centripetal acceleration — inward acceleration needed to maintain circular path |
| $F_c$ | Centripetal force — the net inward force that causes circular motion (not a new force type!) |
| $q$ | Electric charge of the particle (sign matters for direction) |
| $B$ | Magnetic field strength (magnitude) |
| $\theta$ | Angle between $\textbf{v}$ and $\textbf{B}$; controls the force magnitude via $\sin\theta$ |
Key assumptions for clean circular motion: uniform magnetic field, $\textbf{v} \perp \textbf{B}$ (so $\theta = 90°$), no electric field, no drag, constant speed.
When a particle moves along a circle of radius $r$ through an angle $\theta$ (in radians), the actual distance traveled along the arc is:
This just converts angle into a physical distance by scaling with the radius. Bigger circle → same angle covers more ground.
Angular speed is the rate at which the angle is sweeping. By definition:
For one full revolution, the angle swept is $2\pi$ radians, taking time $T$. So:
All three forms are equivalent — pick whichever you have data for.
The particle's regular speed $v$ and its angular speed $\omega$ are connected by the radius:
Think of two horses on a merry-go-round — the outer horse covers more distance per revolution than the inner horse, even though both complete a revolution in the same time. $v = r\omega$ captures that: same $\omega$, bigger $r$, bigger $v$.
Even at constant speed, circular motion involves changing direction — so the velocity vector is always changing, meaning there IS acceleration. That acceleration points inward toward the center.
We can rewrite this in terms of $\omega$ by substituting $v = r\omega$:
Newton's second law: $F = ma$. The inward force required to maintain circular motion is:
Centripetal force is not a new type of force. It's just a label for whatever net inward force is doing the curving. In a car turning a corner, it's friction. For a satellite, it's gravity. For a charged particle in a magnetic field, it's the magnetic force. The name just tells you the role, not the source.
The magnitude of the magnetic force on a moving charge is:
| Term | What it represents |
|---|---|
| $q$ | Charge magnitude — more charge, stronger the push |
| $v$ | Speed — faster particle, stronger force |
| $B$ | Field strength — stronger field, stronger push |
| $\sin\theta$ | Projection factor — only the velocity component perpendicular to $\textbf{B}$ gets deflected. If the particle moves parallel to the field ($\theta = 0°$), the force is zero — no deflection at all. |
For our circular-motion case, we set $\textbf{v} \perp \textbf{B}$, so $\theta = 90°$ and $\sin 90° = 1$:
Now set that equal to the centripetal force requirement:
This is the bridge equation — it connects magnetic physics to mechanical circular motion. Everything below flows from this single statement.
Starting from $qvB = mv^2/r$, solve for $r$:
Read this directly: bigger $m$ or $v$ → harder to bend → larger radius. Bigger $q$ or $B$ → stronger inward push → tighter circle. This is the load-bearing formula for charged-particle motion in a field.
Use $\omega = v/r$ and substitute $r = mv/(qB)$:
Wild result: the angular speed has NO $v$ in it. Doesn't matter if the particle is slow or fast — it always completes a revolution at the same angular rate. Faster particles just orbit in bigger circles. This is the principle behind the cyclotron.
From $\omega = 2\pi/T$ → $T = 2\pi/\omega$. Substitute $\omega = qB/m$:
Again — no $v$. The period is independent of how fast the particle moves.
Rearranging $r = mv/(qB)$ to solve for $v$:
If you measure the radius of the circular track (say, in a cloud chamber), you can calculate the particle's speed.
This is a critical conceptual point. Work is force dotted with displacement:
Equivalently, instantaneous power is:
The magnetic force is always perpendicular to the velocity of the particle — that's just how the cross product works in $\textbf{F}_B = q(\textbf{v} \times \textbf{B})$. So:
A magnetic field acts like a perfectly frictionless steering wheel. It can turn a particle left or right but can never hit the gas or the brakes. That's why circular (and helical) motion preserves speed perfectly.
What if $\textbf{v}$ is not fully perpendicular to $\textbf{B}$? Decompose velocity into two components:
| Component | What the field does with it |
|---|---|
| $v_\perp$ | Magnetic force acts on this — produces circular motion in the plane perpendicular to $\textbf{B}$ |
| $v_\parallel$ | Magnetic force is zero on this — particle continues to drift along the field direction unchanged |
Combined: circle + drift = helix. The formulas update as:
Helical motion is how charged particles spiral along magnetic field lines in the Earth's magnetosphere — it's what creates the aurora borealis. Particles from the solar wind get trapped along field lines and spiral toward the poles.
In circular motion, centripetal acceleration and force always point toward the center of the circle. For magnetic circular motion:
The force formula in full vector form is $\textbf{F}_B = q(\textbf{v} \times \textbf{B})$. The sign of $q$ is built right in — negative charge automatically reverses the force direction.
Thinking $a = 0$ because speed is constant. Speed is constant, but direction isn't. Changing direction = changing velocity vector = non-zero acceleration. Always inward.
Treating centripetal force as a new physical force. It's just a label for the net inward force — whatever provides it (gravity, tension, magnetic force, etc.).
Dropping $\sin\theta$ from $F_B = qvB\sin\theta$. If $\theta = 0°$, the force is zero. A particle moving parallel to the field experiences no magnetic deflection whatsoever.
Ignoring the sign of charge. Magnitude formulas use $|q|$, but the actual force direction flips for negative charges. Electrons curve opposite to protons in the same field.
Thinking the magnetic force changes kinetic energy. It can't. $\textbf{F}_B \perp \textbf{v}$ always, so $P = \textbf{F}_B \cdot \textbf{v} = 0$. No work, no energy change.
Forgetting $T$ and $\omega$ are independent of speed. $\omega = qB/m$ and $T = 2\pi m/(qB)$ contain no $v$. This is the cornerstone of cyclotron operation.
Using MLT notation: $M$ = mass, $L$ = length, $T$ = time.
From $F_B = qvB$, we get $[qB] = [F]/[v] = MLT^{-2} / LT^{-1} = MT^{-1}$. Then:
Dimensional analysis is your sanity check. If the dimensions don't come out right, something in the algebra broke. Build this habit — it catches algebra errors before they become wrong answers.