Free Cheatsheet · Topic D.3 · AHL

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 Fields Syllabus: AHL Topic D.3 Read time: ~13 minutes Last 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)

  • Horizontal: $x = v_x t$ — constant velocity, no horizontal force.
  • Vertical: $y = \tfrac{1}{2}at^2 = \dfrac{qE}{2m}t^2$ — constant acceleration.
Path equation:$\displaystyle y = \frac{qE}{2mv_x^2}\,x^2$   (parabola, valid only inside the field)
Exit $v_y$:$\displaystyle v_y = \frac{qEL}{mv_x}$   (after travelling length $L$)
Deflection angle:$\tan\theta = v_y / v_x$
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.

§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.

§3 — Circular Motion in a Magnetic Field D.3

Charged particle moving in a circle inside a uniform magnetic field directed into the page, with crosses indicating field direction
A charged particle in a uniform $\vec{B}$ field (into page, shown by crosses) follows a circular path; magnetic force $= qvB$ provides centripetal force.

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.

§4 — Velocity Selector & Thomson's Experiment D.3 AHL

Velocity selector

Crossed (perpendicular) $\vec{E}$ and $\vec{B}$ fields. For zero deflection: $qE = qvB$.

Selected speed:$\displaystyle v = \frac{E}{B}$   (independent of $q$ and $m$!)

Only particles with this exact speed pass through undeflected.

Thomson's experiment ($e/m$)

  1. Use the velocity selector to fix the speed $v = E/B$.
  2. Switch off $E$; the electron then follows a circular path of radius $r$ in $\vec{B}$ alone.
  3. Combine: $\dfrac{e}{m_e} = \dfrac{v}{rB} = \dfrac{E}{rB^2}$.

Result: $e/m_e = 1.76 \times 10^{11}\;\mathrm{C\,kg^{-1}}$.

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.

§5 — Magnetic Fields from Currents D.3 AHL

Field formulae

Long straight wire:$\displaystyle B = \frac{\mu_0 I}{2\pi r}$
Solenoid (inside):$\displaystyle B = \mu_0 n I = \mu_0 \frac{N}{L} I$
Constant:$\mu_0 = 4\pi \times 10^{-7}\;\mathrm{T\,m\,A^{-1}}$

Direction — right-hand grip rule

  • Straight wire: thumb along the current; fingers curl in the direction of $\vec{B}$.
  • Solenoid: fingers curl in the direction of the current; thumb points to the N pole (i.e. the direction of $\vec{B}$ along the axis).

§6 — Parallel Wires & the Ampere D.3 AHL

Force between two parallel wires

Force / length:$\displaystyle \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$
  • Same-direction currents $\Rightarrow$ attract.
  • Opposite-direction currents $\Rightarrow$ repel.

Definition of the ampere

One ampere: two infinite parallel wires, $1\;\mathrm{m}$ apart, with a force per unit length of $2 \times 10^{-7}\;\mathrm{N\,m^{-1}}$.

Check: $F/L = \mu_0 \times 1 \times 1 / (2\pi \times 1) = 2 \times 10^{-7}\;\mathrm{N\,m^{-1}}$  ✓

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.

§7 — Exam Attack Plan All sections

When you see this in the question — reach for that:

Question triggerReach for
Charged particle accelerated from rest through $V$$qV = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2qV/m}$
Particle enters parallel plates perpendicular to $\vec{E}$Parabolic motion: $x = v_x t$, $y = \tfrac{1}{2}(qE/m)t^2$
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

  1. Selector condition: $qE = qvB_1 \Rightarrow v = E/B_1$  (R1)
  2. Substitute (a): $v = \dfrac{1.20\times 10^4}{0.040} = 3.00 \times 10^{5}\;\mathrm{m\,s^{-1}}$  (M1)(A1)
  3. Circular-path condition: $qvB_2 = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB_2}$  (M1)
  4. Substitute (b): $r = \dfrac{(1.99\times10^{-26})(3.00\times10^5)}{(1.60\times10^{-19})(0.080)} = 4.66 \times 10^{-4}\;\mathrm{m} \approx 0.47\;\mathrm{mm}$  (A1)
  5. Period (c): $T = \dfrac{2\pi m}{qB_2} = \dfrac{2\pi (1.99\times10^{-26})}{(1.60\times10^{-19})(0.080)} = 9.77 \times 10^{-6}\;\mathrm{s}$  (A1)(A1)

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.

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Related IB Physics HL Topics

D.1 Gravitational Fields D.2 Electric & Magnetic Fields D.4 Induction A.2 Forces & Momentum IB Physics Tuition (Singapore)