What IS a Magnetic Force?
Before we write a single equation, let's build the right mental model. Forget bar magnets for a second. The deep truth is this: magnetism is fundamentally about moving charges interacting with each other. That's it.
You already know the electric force: put a charge $q$ in an electric field $\textbf{E}$ and it feels $\textbf{F} = q\textbf{E}$. The force points along the field. Simple. Now here's the weird twist nature throws at us — if that charge is moving, it also feels a second force, perpendicular to both its velocity and the local magnetic field $\textbf{B}$. That's the magnetic force.
Why perpendicular? Because of how the cross product works. The magnetic force can't do work on the charge (force ⊥ displacement), but it absolutely can change its direction. Think of it as a steering force, not an accelerating one.
Electric vs Magnetic
Electric force acts on any charge. Magnetic force only acts on moving charges. A stationary charge in a pure $\textbf{B}$ field feels nothing.
Always Perpendicular
$\textbf{F}_B \perp \textbf{v}$ always. This means the magnetic force does zero work. It can't speed up or slow down a particle — only curve its path.
No Monopoles
Unlike electric charges (isolated + or −), isolated magnetic poles don't exist. Cut a magnet in half and you get two smaller magnets, each with N and S poles.
Field Lines
Magnetic field lines are always closed loops — they leave the N pole outside the magnet and return through the S pole, then continue inside the magnet.
Here's a Feynman-level insight: if you're sitting in a frame where a charge is stationary, you only see an electric field. If you then move relative to that charge, special relativity transforms part of that electric field into a magnetic field. Magnetism is literally a relativistic correction to electricity. That's why $c$ shows up in Maxwell's equations.
The Lorentz Force Law
When a charge $q$ moves with velocity $\textbf{v}$ through a magnetic field $\textbf{B}$, it experiences a force. The full Lorentz force (including both electric and magnetic) is:
For pure magnetic situations (no electric field), we keep just the magnetic piece:
⬦ Term Interrogation
| Symbol | Name | Units | Physical Meaning |
|---|---|---|---|
| $\textbf{F}_B$ | Magnetic force | N | The push/pull the magnetic field exerts on the moving charge. Always perpendicular to both $\textbf{v}$ and $\textbf{B}$. |
| $q$ | Charge | C (Coulombs) | The "handle" the field grabs onto. Positive or negative — the sign flips the direction of force. $|q|$ appears in the magnitude because direction is handled by the cross product. |
| $\textbf{v}$ | Velocity | m/s | How fast and in what direction the charge is moving. No velocity = no magnetic force. A stationary charge feels nothing. |
| $\textbf{B}$ | Magnetic field (B-field) | T (Tesla) | The field that permeates space and exerts force on moving charges. Direction is defined as where a compass needle's north pole points. Also: $1\,\text{T} = 10^4\,\text{G}$ (Gauss). |
| $\theta$ | Angle between $\textbf{v}$ and $\textbf{B}$ | radians or ° | The angle the velocity vector makes with the $\textbf{B}$ field vector. This is the $\sin\theta$ from the cross product magnitude. When $\theta=0$ or $180°$, the force vanishes. When $\theta = 90°$, force is maximum. |
When $\theta = 0°$ or $180°$
$\sin(0°) = \sin(180°) = 0$ so $F_B = 0$. Moving parallel or anti-parallel to $\textbf{B}$ produces zero force. The charge blasts right through, undeflected.
When $\theta = 90°$
$\sin(90°) = 1$ so $F_B = |q|vB$. Maximum force. Moving perpendicular to $\textbf{B}$ gives the biggest kick. This is the cyclotron case.
Work Done = Zero
Since $\textbf{F}_B \perp \textbf{v}$, $W = \int \textbf{F} \cdot d\textbf{s} = 0$. The kinetic energy $\frac{1}{2}mv^2$ is conserved. Speed is constant. Only direction changes.
The $\sin\theta$ factor is just the cross product telling you: only the component of velocity that is perpendicular to $\textbf{B}$ contributes to the force. Parallel components are invisible to the magnetic field.
⬦ Units of B — The Tesla
From $F_B = |q|vB\sin\theta$, solving for $B$:
Earth's field ≈ $5 \times 10^{-5}\ \text{T} = 0.5\ \text{G}$. An MRI machine: 1–7 T. A neodymium magnet: ~1 T.
⬦ Interactive: Force vs. Angle
The Right-Hand Rule
The cross product $\textbf{v} \times \textbf{B}$ gives us a vector that's perpendicular to both $\textbf{v}$ and $\textbf{B}$. But which perpendicular direction? There are always two choices (up or down, into or out of page...). The Right-Hand Rule tells us which one.
Here's the physical protocol: point your right hand's fingers in the direction of $\textbf{v}$, then curl them toward $\textbf{B}$ (through the smaller angle). Your thumb points in the direction of $\textbf{v} \times \textbf{B}$, which is the direction of force on a positive charge.
SIGN FLIP: For a negative charge (like an electron), the force is in the opposite direction from what the RHR gives. Always apply RHR for a positive charge first, then flip if the charge is negative. The factor of $q$ in $\textbf{F}_B = q\,\textbf{v} \times \textbf{B}$ handles this automatically when you include the sign of $q$.
Cyclic order (x→y→z→x) gives positive. Reverse order gives negative.
⬦ Interactive: Direction Trainer
Here's the algebraic shortcut: write $\textbf{v}$ and $\textbf{B}$ in component form, take the cross product with the determinant method. The resulting unit vector tells you the direction; the magnitude $vB\sin\theta$ tells you how big. Together, that's $\textbf{F}_B = q(\textbf{v} \times \textbf{B})$, fully computed.
Circular Motion & Cyclotron Frequency
Here's where things get beautiful. Suppose a positive charge moves with speed $v$ in a plane perpendicular to a uniform magnetic field $\textbf{B}$. Then $\theta = 90°$ always, and the force magnitude is $F_B = |q|vB$. This force is always perpendicular to the velocity — that's the definition of centripetal force!
So the charge orbits in a perfect circle. The magnetic force IS the centripetal force:
Left side: magnetic force magnitude. Right side: centripetal force needed for circular motion.
Solve for $r$: divide both sides by $v$, then divide by $|q|B$:
| Symbol | Name | Physical Meaning |
|---|---|---|
| $r$ | Orbital radius | Bigger mass or faster speed → bigger circle. Stronger $B$ field or more charge → tighter circle. This is exactly what a mass spectrometer exploits! |
| $m$ | Mass of particle | Heavier particle needs more centripetal force to curve → larger radius at same speed and field. |
| $v$ | Speed | Faster particle → larger orbit. But notice the period doesn't depend on speed! (It cancels out.) |
Now find the period $T$ (time for one full orbit). Circumference divided by speed:
The $v$ cancels! Period is independent of speed. Faster particles travel bigger circles in the same time.
$\omega$ depends only on charge-to-mass ratio $q/m$ and field strength. This is the heart of how cyclotrons and mass spectrometers work.
The cyclotron frequency being independent of velocity is what makes the cyclotron machine possible. You can tune an oscillating electric field to exactly $\omega_c = |q|B/m$ and it will resonantly kick particles up in energy no matter how fast they're going. The circular path just gets bigger and bigger — same period, larger radius.
⬦ Interactive: Cyclotron Orbit Simulator
Force on a Current-Carrying Wire
A current-carrying wire is just a bunch of moving charges. So of course a magnetic field exerts a force on it. Let's derive the force formula from scratch, starting from what we know about individual charges.
Consider a wire of length $L$ carrying current $I$ in a uniform field $\textbf{B}$. Current is charge per unit time: $I = nqv_d A$, where $n$ is carrier density, $q$ is charge per carrier, $v_d$ is drift velocity, and $A$ is cross-sectional area.
Total number of charge carriers in length $L$: $N = nAL$. Force on one carrier: $f = |q|v_dB\sin\theta$. Total force on the wire:
But $I = nA|q|v_d$, so...
| Symbol | Name | Physical Meaning |
|---|---|---|
| $I$ | Current (Amps) | Amount of charge flowing per second. Replaces $|q|v$ from the single-particle case — current IS the collective drift of charges. |
| $\textbf{L}$ | Length vector | Vector of magnitude $L$ pointing in the direction of conventional current flow. This is the analog of $\textbf{v}$ for individual charges. |
| $\theta$ | Angle between wire and $\textbf{B}$ | Same deal as before. Wire parallel to $\textbf{B}$: zero force. Wire perpendicular to $\textbf{B}$: maximum force $F = ILB$. |
The wire-force equation $\textbf{F} = I\textbf{L} \times \textbf{B}$ is structurally identical to $\textbf{F} = q\textbf{v} \times \textbf{B}$. Just swap $q \to I$ and $\textbf{v} \to \textbf{L}$. The RHR applies the same way — point along current direction, curl toward $\textbf{B}$, thumb gives force direction.
Applications of Magnetic Forces
Mass Spectrometer
Ions enter at speed $v$, then curve in B field. Since $r = mv/|q|B$, heavier ions curve less. Measure landing position → measure $m/q$ ratio. How we discovered isotopes and weigh molecules.
Velocity Selector
Cross $\textbf{E}$ and $\textbf{B}$ fields. Particles go straight only when $qE = qvB$, so $v = E/B$. All other speeds get deflected. Produces a monoenergetic beam. Used upstream of mass spectrometers.
Cyclotron
Uses $\omega_c = |q|B/m$ (speed-independent!). Oscillating electric field at this frequency kicks particles in resonance. Used in hospitals to produce radioisotopes for PET scans.
Aurora Borealis
Solar wind electrons spiral along Earth's dipole field lines (helical motion when $v$ has both parallel and perpendicular components to $\textbf{B}$). They funnel toward the poles, exciting atmospheric gases.
DC Motor
Current-carrying coil in a magnetic field experiences torques: $\tau = NIAB\sin\phi$. The commutator keeps the torque in one direction. Converts electrical to mechanical energy.
Hall Effect
Current in a magnetic field pushes charge carriers sideways, building up a transverse voltage (Hall voltage). Used to measure $B$ fields and determine sign of charge carriers in semiconductors.
⬦ Velocity Selector — Deep Dive
This is elegant. Put $\textbf{E}$ pointing upward and $\textbf{B}$ pointing into the page. A positive charge moving rightward feels:
Electric force: $F_E = qE$ upward. Magnetic force: by RHR ($\textbf{v}$ right, $\textbf{B}$ into page), $\textbf{v} \times \textbf{B}$ points downward, so $F_B = qvB$ downward.
For zero deflection, set them equal:
Only particles with exactly this speed pass through undeflected. Note: the drift velocity is independent of charge! Works for positive and negative charges equally.
Worked Examples
An electron ($q = -1.6\times10^{-19}$ C, $m = 9.11\times10^{-31}$ kg) moves with velocity $v = 3.0\times10^6$ m/s in the $+x$ direction through a uniform magnetic field $\textbf{B} = 0.20\,\hat{z}$ T. Find the magnetic force on the electron.
Identify what we know.
$q = -1.6\times10^{-19}$ C · $\textbf{v} = 3.0\times10^6\,\hat{x}$ m/s · $\textbf{B} = 0.20\,\hat{z}$ T
Compute the cross product $\textbf{v} \times \textbf{B}$ first.
$$\textbf{v} \times \textbf{B} = (3.0\times10^6\,\hat{x}) \times (0.20\,\hat{z}) = (3.0\times10^6)(0.20)\,(\hat{x}\times\hat{z})$$
From the cross product table: $\hat{x}\times\hat{z} = -\hat{y}$ (reverse of cyclic order)
$$\textbf{v}\times\textbf{B} = 6.0\times10^5\,(-\hat{y}) = -6.0\times10^5\,\hat{y}\ \text{m}^2/(\text{T·s})$$
Multiply by charge (including sign!).
$$\textbf{F}_B = q(\textbf{v}\times\textbf{B}) = (-1.6\times10^{-19})\,(-6.0\times10^5\,\hat{y})$$
$$\textbf{F}_B = +9.6\times10^{-14}\,\hat{y}\ \text{N}$$
Interpret physically.
The force is in the $+y$ direction. For a positive charge moving in $+x$ with $\textbf{B}$ in $+z$, the RHR gives force in $-y$. But the electron is negative, so it flips to $+y$. ✓ The double-negative gave us positive.
A proton ($m = 1.67\times10^{-27}$ kg, $q = +1.6\times10^{-19}$ C) moves at $v = 2.5\times10^6$ m/s perpendicular to a magnetic field $B = 0.75$ T. Find: (a) the radius of its circular orbit, (b) the period, (c) the cyclotron angular frequency.
Part (a): Radius. Use $F_B = F_C$:
$$|q|vB = \frac{mv^2}{r} \implies r = \frac{mv}{|q|B}$$
$$r = \frac{(1.67\times10^{-27})(2.5\times10^6)}{(1.6\times10^{-19})(0.75)}$$
Numerator: $(1.67)(2.5) = 4.175$, so $4.175\times10^{-21}$
Denominator: $(1.6)(0.75) = 1.2$, so $1.2\times10^{-19}$
$$r = \frac{4.175\times10^{-21}}{1.2\times10^{-19}} = 0.0348\ \text{m} \approx \mathbf{3.5\ cm}$$
Part (b): Period. Notice $v$ cancels:
$$T = \frac{2\pi m}{|q|B} = \frac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.75)}$$
Numerator: $2\pi(1.67\times10^{-27}) = 1.049\times10^{-26}$
Denominator: $1.2\times10^{-19}$
$$T = \frac{1.049\times10^{-26}}{1.2\times10^{-19}} = 8.74\times10^{-8}\ \text{s} \approx \mathbf{87.4\ \text{ns}}$$
Part (c): Angular frequency.
$$\omega = \frac{|q|B}{m} = \frac{(1.6\times10^{-19})(0.75)}{1.67\times10^{-27}} = \frac{1.2\times10^{-19}}{1.67\times10^{-27}}$$
$$\omega = 7.19\times10^7\ \text{rad/s} \approx \mathbf{71.9\ \text{Mrad/s}}$$
Check: $\omega = 2\pi/T = 2\pi/(8.74\times10^{-8}) = 7.19\times10^7$ ✓
Two uranium isotopes, $^{235}$U ($m_1 = 3.90\times10^{-25}$ kg) and $^{238}$U ($m_2 = 3.95\times10^{-25}$ kg), are singly ionized and sent through a mass spectrometer at $v = 1.05\times10^5$ m/s. The magnetic field is $B = 0.750$ T. Find the separation $\Delta d$ between the two isotopes after a half-orbit.
Set up the geometry. Both ions enter at the same point. After a half-circle, each hits the detector wall at a horizontal distance of $2r$ from the entry point. The separation is:
$$\Delta d = 2r_{238} - 2r_{235} = 2(r_{238} - r_{235})$$
Calculate each radius. Both ions are singly ionized, so $|q| = 1.6\times10^{-19}$ C.
$$r = \frac{mv}{|q|B}$$
The $v$, $|q|$, and $B$ are the same for both, so:
$$r_{235} = \frac{(3.90\times10^{-25})(1.05\times10^5)}{(1.6\times10^{-19})(0.750)} = \frac{4.095\times10^{-20}}{1.2\times10^{-19}} = 0.341\ \text{m}$$
$$r_{238} = \frac{(3.95\times10^{-25})(1.05\times10^5)}{1.2\times10^{-19}} = \frac{4.148\times10^{-20}}{1.2\times10^{-19}} = 0.346\ \text{m}$$
Find the separation.
$$\Delta d = 2(0.346 - 0.341) = 2(0.005\ \text{m}) = \mathbf{1.0\ \text{cm}}$$
Just 1 cm separates uranium-235 from uranium-238. That's why nuclear isotope separation is so difficult — you need extremely precise detectors and long drift paths.
A wire of length $L = 0.50$ m and mass $m = 10$ g is suspended horizontally by flexible leads. It is in a magnetic field $B = 0.50$ T directed into the page. What current $I$ must flow in the wire to remove the tension in the supporting leads (i.e., levitate the wire)?
Draw the force balance. To levitate, the upward magnetic force must equal the downward gravitational force:
$$F_{\text{mag}} = mg$$
Determine current direction using RHR. $\textbf{B}$ is into the page ($-\hat{z}$). We need force upward ($+\hat{y}$). We need $\textbf{L} \times \textbf{B}$ to point upward.
If $\textbf{L} = +\hat{x}$ (current rightward): $\hat{x} \times (-\hat{z}) = -(\hat{x}\times\hat{z}) = -(-\hat{y}) = +\hat{y}$ ✓
Current must flow in the $+x$ direction (rightward).
Solve for $I$. The wire is perpendicular to $\textbf{B}$, so $\sin\theta = 1$:
$$ILB = mg \implies I = \frac{mg}{LB}$$
$$I = \frac{(0.010\ \text{kg})(9.8\ \text{m/s}^2)}{(0.50\ \text{m})(0.50\ \text{T})} = \frac{0.098}{0.25} = \mathbf{0.39\ \text{A}}$$
An electron beam enters a crossed-field velocity selector with $B = 2.0$ mT and $E = 6.0\times10^3$ N/C. (a) What speed passes undeflected? (b) If E is turned off, what is the acceleration? (c) What is the radius of the resulting circular motion?
Part (a): Velocity selector condition.
$$v_d = \frac{E}{B} = \frac{6.0\times10^3\ \text{N/C}}{2.0\times10^{-3}\ \text{T}} = 3.0\times10^6\ \text{m/s}$$
Only electrons at $v = 3.0\times10^6$ m/s pass undeflected.
Part (b): Acceleration with E off. Only $B$ remains. Newton's 2nd law:
$$ma = |q|vB$$
$$a = \frac{|q|vB}{m} = \frac{(1.6\times10^{-19})(3.0\times10^6)(2.0\times10^{-3})}{9.1\times10^{-31}}$$
Numerator: $(1.6)(3.0)(2.0) = 9.6$, giving $9.6\times10^{-16}$
$$a = \frac{9.6\times10^{-16}}{9.1\times10^{-31}} \approx 1.1\times10^{15}\ \text{m/s}^2$$
Part (c): Radius of circular orbit.
$$r = \frac{mv}{|q|B} = \frac{(9.1\times10^{-31})(3.0\times10^6)}{(1.6\times10^{-19})(2.0\times10^{-3})}$$
Numerator: $2.73\times10^{-24}$ Denominator: $3.2\times10^{-22}$
$$r = \frac{2.73\times10^{-24}}{3.2\times10^{-22}} = 8.5\times10^{-3}\ \text{m} \approx \mathbf{8.5\ \text{mm}}$$