IB Physics HL Motion in EM Fields — Complete Cheatsheet
Every formula, derivation, trick and trap for IB Physics HL Topic D.3 — parabolic motion in electric fields, circular motion in magnetic fields, the velocity selector, Thomson's experiment, B-fields from currents, and the force between parallel wires.
Topic: D.3 Motion in EM FieldsSyllabus: AHL Topic D.3Read time: ~13 minutesLast updated: Apr 2026
Motion in electromagnetic fields is the topic where Theme D's geometry catches up with everyone. IB Physics HL Topic D.3 takes the field equations from D.1 and D.2 and asks: what does a charged particle do when it enters those fields? The answer is two distinct, exam-friendly behaviours — parabolic paths in $\vec{E}$ fields (just like projectile motion) and circular paths in $\vec{B}$ fields (with a period that depends only on $m$, $q$ and $B$). Add the velocity selector, Thomson's $e/m$ experiment and the parallel-wires force, and you have one of the most algorithmic topics in HL physics.
This cheatsheet condenses every formula, geometric trick and exam-mark trap from D.3 (AHL) into one page you can revise from. Scroll to the bottom for the printable PDF, the full Notes, the Tutorial booklet, and the marked-up solutions in the gated full library.
§1 — Charged Particles in Uniform Electric Fields D.3
Force, acceleration & energy
Force on charge:$\displaystyle F = qE = \frac{qV}{d}$
Acceleration:$\displaystyle a = \frac{qE}{m} = \frac{qV}{md}$
Energy gained:$\displaystyle qV = \tfrac{1}{2}mv^2 \;\Longrightarrow\; v = \sqrt{\frac{2qV}{m}}$ (from rest)
Parabolic motion (perpendicular entry between parallel plates)
TrickThe parabolic path in an $\vec{E}$ field is mathematically identical to projectile motion under gravity. Replace $g$ with $qE/m$ and every projectile-motion formula transfers directly.
TrapHorizontal velocity does not change inside the field — the electric force acts only vertically (perpendicular to motion). The path is parabolic only while the particle is inside the plates; after exit it travels in a straight line. Don't extend the parabola past the plate region.
Deflection in an electric field. Entering horizontally, the charge follows a parabola while between the plates — identical to projectile motion with $g\to qE/m$ — then travels straight after it leaves.
From the Photon question bank
What examiners actually test on this concept:
F = qE = qV/dv = √(2qV/m)Parabola then straightLike projectile motion
A charge accelerated from rest through $V$ gains $qV=\tfrac12 mv^2$, so $v=\sqrt{2qV/m}$.frequent
Perpendicular entry between plates → parabolic path (identical to projectile motion with $g\to qE/m$); after leaving the plates it goes STRAIGHT.
The horizontal velocity is unchanged inside the field — the force acts only across the plates.
§2 — Magnetic Force on Wires & Charges D.3
Key formulas
Force on wire:$\displaystyle F = BIL\sin\theta$
Force on charge:$\displaystyle F = qvB\sin\theta$
$\theta$ is the angle between the current/velocity and $\vec{B}$.
Maximum force when $\theta = 90°$; zero force when $\theta = 0°$ (parallel to $\vec{B}$).
Direction: Fleming's left-hand rule (thuMb = Motion, First finger = Field, seCond finger = Current).
For negative charges: reverse the force direction.
TrapIf a charge moves parallel to $\vec{B}$, the force is zero ($\sin 0° = 0$). This is a very common Paper 1 multiple-choice trap.
NoteThe magnetic force on a moving charge is always $\perp$ to $\vec{v}$, so it does no work. It cannot change the particle's speed or kinetic energy — only its direction.
Force on a current-carrying wire. $F=BIL\sin\theta$, perpendicular to both the current and the field; its direction comes from Fleming's left-hand rule.
From the Photon question bank
What examiners actually test on this concept:
F = BIL sinθF = qvB sinθZero if parallel to BFleming's LHR · no work
Force is maximum at $\theta=90^\circ$ and ZERO when $v$ (or $I$) is parallel to $B$ — a classic Paper 1 trap. frequent
Direction by Fleming's left-hand rule (thuMb=Motion, First=Field, seCond=Current); reverse for negative charges.
The magnetic force is ⊥ to $v$, so it does NO work — it turns the particle, never changes its speed.
§3 — Circular Motion in a Magnetic Field D.3
Circular motion in a magnetic field. With $\vec v \perp \vec B$ (into the page), the magnetic force $qvB$ acts toward the centre and provides the centripetal force, giving radius $r=mv/qB$.
For $\vec{v} \perp \vec{B}$, the magnetic force provides the centripetal force:
$$qvB = \frac{mv^2}{r}$$
Radius:$\displaystyle r = \frac{mv}{qB} = \frac{p}{qB}$
Period:$\displaystyle T = \frac{2\pi m}{qB}$ (independent of speed!)
Charge-to-mass ratio:$\displaystyle q/m = v/(rB)$
Trick$r \propto mv$ (momentum). Heavier or faster particles trace larger circles. Larger $q$ or $B$ tightens the circle. The period depends only on $m$, $q$ and $B$ — not on speed. This is the principle behind the cyclotron.
TrickIn the velocity selector, $q$ cancels. A proton and an electron with the same speed are both undeflected. Different particles, same speed $\Rightarrow$ same result.
Velocity selector. Crossed $E$ and $B$ fields let a charge pass straight through only when $qE=qvB$, selecting the speed $v=E/B$ regardless of charge or mass.
From the Photon question bank
What examiners actually test on this concept:
Crossed E and BqE = qvBv = E/B (q, m cancel)Thomson e/m
Velocity selector: crossed $E$ and $B$; undeflected when $qE=qvB$, so $v=E/B$ — independent of $q$ and $m$. frequent (AHL)
Any particle with exactly $v=E/B$ passes straight; a proton and an electron at the same speed are both undeflected.
Thomson: select $v$, then bend in $B$ alone → $e/m = E/(rB^2)$.
§5 — Magnetic Fields from Currents D.3 AHL
Field formulae
Long straight wire:$\displaystyle B = \frac{\mu_0 I}{2\pi r}$
Solenoid (inside):$\displaystyle B = \mu_0 n I = \mu_0 \frac{N}{L} I$
Trap"Same direction currents attract" feels counterintuitive. To verify: use the right-hand grip on wire 1 to find $\vec{B}$ at wire 2; then apply Fleming's left-hand rule to wire 2 sitting in that field. The force on wire 2 always points back toward wire 1.
Parallel wires. Currents in the same direction attract, opposite currents repel, with $F/L=\mu_0 I_1 I_2/2\pi r$ — the basis of the ampere's definition.
From the Photon question bank
What examiners actually test on this concept:
F/L = μ₀I₁I₂/2πrSame direction attractOpposite repelDefinition of the ampere
Particle in uniform $\vec{B}$, $\vec{v} \perp \vec{B}$
Circle, $r = mv/(qB)$, $T = 2\pi m/(qB)$
"Find the deflection angle"
$\tan\theta = v_y/v_x$ at exit
"Particle undeflected by crossed $\vec{E}, \vec{B}$"
Velocity selector: $v = E/B$
"Find the direction of force on a wire in $\vec{B}$"
Fleming's left-hand rule (M / F / C)
"Find the direction of $\vec{B}$ from a wire"
Right-hand grip: thumb along $I$, fingers curl to $\vec{B}$
"Force between two long parallel wires"
$F/L = \mu_0 I_1 I_2 / (2\pi r)$, then attract / repel by direction
"Charge moves parallel to $\vec{B}$"
$F = 0$ — no deflection (Paper 1 trap)
"Period in cyclotron / mass spectrometer"
$T = 2\pi m/(qB)$ — independent of speed and radius
Worked Example — IB-Style Mass Spectrometer
Question (HL Paper 2 style — 7 marks)
A singly-ionised carbon-12 ion ($q = +1.60 \times 10^{-19}\;\mathrm{C}$, $m = 1.99 \times 10^{-26}\;\mathrm{kg}$) enters a velocity selector with crossed fields $E = 1.20 \times 10^{4}\;\mathrm{V\,m^{-1}}$ and $B_1 = 0.040\;\mathrm{T}$. After leaving the selector it enters a second region of uniform magnetic field $B_2 = 0.080\;\mathrm{T}$ perpendicular to its velocity.
(a) Calculate the speed at which the ion passes through the selector. (b) Calculate the radius of the circular path inside $B_2$. (c) State the period of the circular motion.
Solution
Selector condition: $qE = qvB_1 \Rightarrow v = E/B_1$ (R1)
Examiner's note: The most common error here is using $B_1$ instead of $B_2$ in the radius formula — the selector field has no role beyond setting $v$. A second common slip is forgetting that $T$ does NOT depend on $v$ or $r$ — students often plug in the radius from (b) and overcomplicate the algebra. The selector and the circle are decoupled: $v$ is fixed first, then $r$ and $T$ are evaluated separately.
Common Student Questions
Why is the period of circular motion in a $\vec{B}$ field independent of speed?
From $qvB = mv^2/r$, the radius is $r = mv/(qB)$. The period $T = 2\pi r/v = 2\pi m/(qB)$ — the $v$ cancels. Faster particles travel in larger circles, but they cover the larger circumference at proportionally higher speed, so the period is fixed by $m$, $q$ and $B$ alone. This is why cyclotrons work — the AC frequency is the same regardless of how fast the particle is going.
Does the magnetic force change a particle's speed?
No. The magnetic force is always perpendicular to $\vec{v}$, so it does no work. Kinetic energy and speed remain constant; only the direction changes. So in any "particle in a magnetic field" question, the speed at the start equals the speed at the end. Use this to spot wrong multiple-choice options instantly.
Why do parallel currents in the same direction attract?
Use the right-hand grip rule on wire 1 to find the direction of $\vec{B}$ at wire 2. Then apply Fleming's left-hand rule to the current in wire 2 sitting in that $\vec{B}$ field. The force on wire 2 points back toward wire 1. Same-direction currents $\to$ wires pulled together (attraction). Opposite-direction currents $\to$ wires pushed apart (repulsion). Counterintuitive but reliable.
What's the velocity selector formula and why doesn't charge appear?
For zero deflection, the electric and magnetic forces must balance: $qE = qvB$, so $v = E/B$. The charge $q$ cancels — meaning the selector picks particles by speed only, regardless of the sign or magnitude of charge. A proton and an electron with the same speed are both undeflected. This is the key device used inside Thomson's experiment to measure $e/m$.
How is a charged particle's parabolic path in an $\vec{E}$ field similar to projectile motion?
It's identical in mathematical form — just replace gravity $g$ with the electric acceleration $qE/m$. The horizontal velocity is unchanged (no horizontal force), and the vertical motion is uniformly accelerated. Path equation: $y = \dfrac{qE}{2mv_x^2}x^2$, a perfect parabola while inside the field. Once the particle exits the plates, it travels in a straight line — do NOT extend the parabola past the field region.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Motion in EM Fields library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — every concept worked through in full, with derivations and intuition.
Tutorial booklet — IB-style questions sequenced from foundation to AHL difficulty.
Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/R1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
