Apply Motion in EM Fields — Apply D.3 Motion in electromagnetic fields in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

The study of charged particles moving through electromagnetic fields has shaped nearly every branch of modern physics and technology. From the first cathode-ray experiments that revealed the electron to the massive particle accelerators that probe the structure of matter, scientists have relied on the interplay between electric and magnetic forces to steer, accelerate, and identify charged particles. Understanding how charges behave in these fields is not just an abstract exercise — it's the foundation of everything from television screens to cancer-treatment proton beams.

1897 • Thomson Discovers the Electron

J. J. Thomson used crossed electric and magnetic fields to deflect cathode rays and measure the charge-to-mass ratio (e/m) of the electron, proving that atoms contain subatomic particles.

1913 • Thomson's Mass Spectrograph

J. J. Thomson built the first mass spectrometer, separating neon isotopes by bending ion paths in magnetic fields — a direct application of motion in EM fields.

1932 • Lawrence's Cyclotron

Ernest Lawrence invented the cyclotron, using a magnetic field to bend protons in circles while an electric field accelerated them across a gap, reaching energies high enough to split atoms.

1954 • CERN Founded

The European Organization for Nuclear Research began building ever-larger accelerators that use sophisticated EM field configurations to push protons to nearly the speed of light.

2012 • Higgs Boson Discovery

The Large Hadron Collider at CERN, relying on superconducting magnets to curve 7 TeV proton beams around a 27 km ring, confirmed the existence of the Higgs boson — a triumph built on the physics of motion in EM fields.

The central question of IB Physics topic D.3 is deceptively simple: What path does a charged particle follow when it enters an electric field, a magnetic field, or both? Answering this requires combining Coulomb's law, Newton's second law, and the magnetic force rule into a unified toolkit. The sections ahead will equip you with that toolkit and show you how to apply it to real IB-style problems.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a firm grasp of four core ideas that govern how charged particles move through electromagnetic fields. Each principle connects force, acceleration, and trajectory in a specific way. Together, they form the conceptual backbone of every problem you will encounter in D.3.

Electric Force on a Charge

A charge q in an electric field E experiences a force F = qE. This force acts parallel (or antiparallel) to the field, so it can speed the particle up, slow it down, or deflect it like a projectile.

Magnetic Force on a Moving Charge

A charge q moving with velocity v in a magnetic field B feels F = qvB sin θ. Crucially, this force is always perpendicular to the velocity, so it changes direction but never changes speed.

Circular Motion in B Fields

Because the magnetic force is always perpendicular to velocity, a charge entering a uniform magnetic field at right angles traces a circular path. The radius of this circle depends on mass, speed, charge, and field strength: r = mv/(qB).

Parabolic Paths in E Fields

A charge entering a uniform electric field perpendicular to its velocity behaves like a projectile under gravity — constant acceleration in one direction, constant velocity in the other — producing a parabolic trajectory.

Velocity Selector (Crossed Fields)

When electric and magnetic fields are perpendicular to each other and to the particle's velocity, only particles with v = E/B pass through undeflected. This is the basis of the velocity selector, used in mass spectrometers.

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KEY TAKEAWAY

Think of the electric field as a hill: it pushes the particle downhill, changing its speed. Think of the magnetic field as a guardrail on a curved road: it forces the particle to turn but never speeds it up or slows it down. When both fields act together, the particle can be steered with extraordinary precision — which is exactly how particle accelerators and old-style CRT televisions work.

SECTION 3

Visual Explanation — Paths in E and B Fields

The diagram below compares the two fundamental trajectories side by side. On the left, a positive charge enters a uniform electric field between parallel plates and curves in a parabola — exactly like a ball thrown horizontally under gravity. On the right, the same charge enters a region of uniform magnetic field directed into the page; the perpendicular force bends it into a perfect circle. Study the force arrows carefully: in the electric-field case, force is always downward; in the magnetic-field case, force always points toward the center of the circular arc.

Left: A positive charge enters horizontally between parallel plates. The uniform electric field accelerates it downward, producing a parabolic curve. Right: A positive charge enters a region of magnetic field directed into the page (shown by × symbols). The magnetic force is always perpendicular to velocity, producing uniform circular motion with radius r = mv/(qB).

Notice the key difference: in the electric-field diagram, the force arrow (red) always points straight down regardless of where the particle is, just like gravity on a projectile. In the magnetic-field diagram, the force arrow always points radially inward toward the center of the circle. This is why the magnetic force does no work — it is always perpendicular to displacement. The particle's kinetic energy stays constant even though its direction keeps changing.

SECTION 4

Mathematical Framework

Three core equations form the mathematical backbone of D.3. Each one connects force, motion, and field quantities in a different scenario. You should be able to recall these from memory and identify when each applies.

ELECTRIC FORCE

F = qE

F = force on the charge (N), q = charge (C), E = electric field strength (N C⁻¹ or V m⁻¹). The force acts parallel to the field for positive charges and antiparallel for negative charges.

MAGNETIC FORCE

F = qvB sin θ

v = speed of the charge (m s⁻¹), B = magnetic field strength (T), θ = angle between velocity and field. Maximum force occurs when v ⊥ B (sin 90° = 1); zero force when v ∥ B (sin 0° = 0). Use the right-hand rule for positive charges; reverse the result for negative charges.

RADIUS OF CIRCULAR MOTION IN B FIELD

r = mv / (qB)

Derived by setting the magnetic force equal to the centripetal force: qvB = mv²/r. Solving for r gives this result. m = mass (kg). A heavier or faster particle traces a larger circle; a stronger field or larger charge gives a tighter circle.

VELOCITY SELECTOR CONDITION

v = E / B

When perpendicular E and B fields are arranged so that the electric force (qE) exactly balances the magnetic force (qvB), only particles with this specific velocity travel in a straight line. All others are deflected. This is independent of charge and mass.

💡

Right-Hand Rule Reminder

Point your right-hand fingers in the direction of the velocity v, then curl them toward the magnetic field B. Your thumb points in the direction of the force on a positive charge. For a negative charge (like an electron), the force is in the opposite direction.

SECTION 5

Detailed Breakdown — Key Scenarios

IB D.3 problems typically fall into a handful of recognizable scenarios. Knowing which scenario you are dealing with is half the battle. The table below summarizes the most common setups, the shape of the resulting path, and the key equation you need.

Scenario	Field(s) Present	Path Shape	Key Equation
Charge at rest in E	Uniform E only	Straight line (accelerating)	F = qE, then a = F/m
Charge enters E ⊥ to v	Uniform E only	Parabola (projectile analogy)	a = qE/m in field direction
Charge enters B ⊥ to v	Uniform B only	Circle	r = mv/(qB)
Charge enters B at angle θ	Uniform B only	Helix (spiral)	r = mv sin θ/(qB); pitch from v cos θ
Crossed E and B (velocity selector)	E ⊥ B ⊥ v	Straight line (if v = E/B)	v = E/B

Summary of common D.3 motion scenarios

A velocity selector uses crossed E and B fields. The electric force pushes the positive charge downward while the magnetic force pushes it upward. Only when v = E/B do these forces balance, letting the particle pass straight through (green). Faster particles (pink) curve upward (magnetic force dominates) and slower particles (orange) curve downward (electric force dominates).

The velocity selector diagram illustrates a concept that appears frequently in IB exams. Notice how the selector is mass-independent: any particle with the correct speed passes through, regardless of its mass or charge magnitude. Once the selected particles exit, they often enter a pure magnetic field region where they separate by mass — this is the principle behind the mass spectrometer. The radius of curvature in that second region, r = mv/(qB), depends on mass, so different isotopes land at different positions on a detector.

SECTION 6

Worked Example

Let's work through a full IB-style problem that combines several of the ideas we've covered. Pay close attention to the strategy of identifying the scenario, selecting the right equation, and checking units at each step.

Proton in a Magnetic Field

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Step 1 — Read and Identify the Scenario

A proton (mass = 1.67 × 10⁻²⁷ kg, charge = 1.60 × 10⁻¹⁹ C) is accelerated from rest through a potential difference of 500 V and then enters a uniform magnetic field of 0.20 T directed perpendicularly to its velocity. Find (a) the speed of the proton as it enters the field, and (b) the radius of its circular path.
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Step 2 — Find the Speed Using Energy Conservation

The proton gains kinetic energy equal to the work done by the electric field: qV = ½mv². Rearranging for v: v = √(2qV / m) Substitute: v = √(2 × 1.60 × 10⁻¹⁹ × 500 / 1.67 × 10⁻²⁷) v = √(1.60 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) v = √(9.58 × 10¹⁰)

v ≈ 3.10 × 10⁵ m s⁻¹
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Step 3 — Apply the Radius Formula

The proton enters the B field perpendicularly (θ = 90°), so it traces a circular path. The radius is: r = mv / (qB) Substitute: r = (1.67 × 10⁻²⁷ × 3.10 × 10⁵) / (1.60 × 10⁻¹⁹ × 0.20) r = (5.18 × 10⁻²²) / (3.20 × 10⁻²⁰)

r ≈ 0.016 m = 1.6 cm
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Step 4 — Interpret and Check

The proton travels in a small circle of radius 1.6 cm. This is physically reasonable — protons are relatively light, and 0.20 T is a moderately strong field. Notice that the speed (3.10 × 10⁵ m s⁻¹) is about 0.1% of the speed of light, so we do not need to worry about relativistic effects. Also note that while the proton circulates, its kinetic energy remains constant because the magnetic force does no work.

SECTION 7

Electric vs. Magnetic Fields — A Comparison

Students often confuse the effects of electric and magnetic fields on charged particles. The table below highlights the most important differences. Memorizing these distinctions will help you quickly identify which equations to use and what path to expect.

Property	Electric Field (E)	Magnetic Field (B)
Force direction	Parallel (or anti-parallel) to field	Perpendicular to both v and B
Force depends on speed?	No — F = qE regardless of v	Yes — F = qvB sin θ
Acts on stationary charges?	Yes	No — particle must be moving
Work done on charge	Can do positive or negative work (changes KE)	Zero (F ⊥ v always)
Typical path	Straight line or parabola	Circle or helix
Changes speed?	Yes	No — only direction changes

Comparison of electric and magnetic field effects on charged particles

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KEY TAKEAWAY

If an IB question asks whether the particle's kinetic energy changes, the answer depends on which field is doing the work. Electric fields can accelerate or decelerate a charge — they change its energy. Magnetic fields only steer — they redirect without adding or removing energy. This single distinction will earn you marks in many D.3 exam questions.

SECTION 8

Connection to Advanced Theory

The physics of D.3 lays the groundwork for more advanced topics you may encounter in university physics or the IB HL extension. The table below maps key D.3 ideas to their advanced counterparts, showing how the same principles scale up to more complex and powerful models.

D.3 Concept	Advanced Extension
F = qvB (magnetic force)	Lorentz force: F = q(E + v × B), combining both fields in a single vector equation
r = mv/(qB) at low speed	Relativistic radius: r = γmv/(qB), where γ is the Lorentz factor — needed when v approaches c
Circular motion in uniform B	Cyclotron frequency ω = qB/m, which is independent of radius and speed (non-relativistic limit)
Velocity selector (E/B)	Wien filter in mass spectrometry; Hall effect in solid-state physics
Helical motion (v at angle to B)	Magnetic mirroring and plasma confinement in fusion reactors (tokamaks)

From D.3 to advanced electromagnetic theory

One of the most elegant extensions is the idea of cyclotron resonance. In a cyclotron, the time to complete each semicircle is always the same (T = 2πm/(qB)) regardless of the radius, because as the particle speeds up and its radius grows, it also has farther to travel — the two effects exactly cancel. This means you can use a fixed-frequency alternating voltage to accelerate the particle every half-turn. The breakdown of this elegant feature at relativistic speeds motivated the invention of the synchrotron, which adjusts the frequency as the particle gains energy.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A proton moves in a straight line through a region containing both an electric field and a magnetic field. The fields are perpendicular to each other and to the proton's velocity. Explain why the proton's kinetic energy remains constant even though two forces act on it.

Show answer

The magnetic force is always perpendicular to the proton's velocity, so it does zero work. For the proton to travel in a straight line, the electric and magnetic forces must be equal and opposite (F_E = F_B), which means qE = qvB. In this balanced condition, the net force on the proton is zero, so no net work is done and the kinetic energy stays constant. This is the velocity selector condition, where v = E/B.

Question 2: BASIC CALCULATION

An electron (mass = 9.11 × 10⁻³¹ kg, charge = 1.60 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m s⁻¹ perpendicular to a magnetic field of 0.050 T. Calculate the radius of the electron's circular path.

Show answer

Using r = mv/(qB): r = (9.11 × 10⁻³¹ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.050) = (1.822 × 10⁻²⁴) / (8.0 × 10⁻²¹) = 2.28 × 10⁻⁴ m ≈ 0.23 mm. This is a very small circle, which makes sense because the electron's mass is tiny. The electron spirals tightly in even a modest field.

Question 3: INTERMEDIATE

A velocity selector uses an electric field of 3.0 × 10⁴ V m⁻¹ and a magnetic field of 0.15 T. Particles that pass through undeflected then enter a second region where only the same magnetic field exists. A singly-charged ion (q = 1.60 × 10⁻¹⁹ C) curves with a radius of 0.12 m in the second region. Calculate the mass of the ion.

Show answer

First, find the selected speed: v = E/B = (3.0 × 10⁴) / (0.15) = 2.0 × 10⁵ m s⁻¹. In the second region, r = mv/(qB), so m = rqB/v. Substituting: m = (0.12 × 1.60 × 10⁻¹⁹ × 0.15) / (2.0 × 10⁵) = (2.88 × 10⁻²¹) / (2.0 × 10⁵) = 1.44 × 10⁻²⁶ kg. Dividing by the atomic mass unit (1.66 × 10⁻²⁷ kg), this gives roughly 8.7 u, which is close to the mass of a beryllium-9 or boron isotope. In practice, a mass spectrometer reading of this type identifies the isotope.

Question 4: APPLIED

In a proton therapy facility, protons are accelerated through a potential difference of 1.5 × 10⁷ V and then steered into a patient's tumor using a bending magnet of 1.2 T. Calculate (a) the speed of the protons (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) and (b) the radius of curvature in the bending magnet. State any assumptions.

Show answer

(a) Using qV = ½mv²: v = √(2 × 1.60 × 10⁻¹⁹ × 1.5 × 10⁷ / 1.67 × 10⁻²⁷) = √(2.87 × 10¹⁵) ≈ 5.36 × 10⁷ m s⁻¹. This is about 18% of the speed of light, so relativistic effects would be significant in a real facility. (b) Assuming non-relativistic: r = mv/(qB) = (1.67 × 10⁻²⁷ × 5.36 × 10⁷) / (1.60 × 10⁻¹⁹ × 1.2) = 0.466 m ≈ 47 cm. The assumption of non-relativistic speeds introduces some error; a relativistic treatment would give a larger effective mass (γm) and thus a larger radius. Nevertheless, the calculation shows why proton therapy bending magnets must be large and powerful.

Question 5: CRITICAL THINKING

Two particles — one an alpha particle (q = 3.20 × 10⁻¹⁹ C, m = 6.64 × 10⁻²⁷ kg) and one a proton (q = 1.60 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) — are accelerated from rest through the same potential difference and then enter the same uniform magnetic field perpendicularly. Derive an expression for the ratio of their radii of curvature r_α /

r_p

and evaluate it numerically.

Show answer

After acceleration: v = √(2qV/m). In the B field: r = mv/(qB) = m√(2qV/m) / (qB) = √(2mV/q) / B. Therefore the ratio r_α/

r_p

= √(m_α

q_p

/ (

m_p

q_α)). Substituting: r_α/

r_p

= √((6.64 × 10⁻²⁷ × 1.60 × 10⁻¹⁹) / (1.67 × 10⁻²⁷ × 3.20 × 10⁻¹⁹)) = √((1.062 × 10⁻⁴⁵) / (5.344 × 10⁻⁴⁶)) = √(1.988) ≈ 1.41. The alpha particle curves with a radius about 1.41 times (or √2 times) that of the proton. This result elegantly shows that even though both have different masses and charges, the ratio simplifies to a clean factor because m_α ≈ 4m_p and q_α = 2q_p, giving √(4/2) = √2.

SUMMARY

Lesson Summary

Summary

Charged particles in electric fields experience a force F = qE that acts parallel to the field, changing both speed and kinetic energy. A charge entering perpendicular to a uniform E field follows a parabolic path, exactly analogous to projectile motion. In contrast, the magnetic force F = qvB sin θ is always perpendicular to velocity, so it does no work and produces uniform circular motion with radius r = mv/(qB) when v is perpendicular to B.

When both fields are present and perpendicular, the velocity selector condition v = E/B allows only particles of one specific speed to pass through undeflected. This principle underpins the mass spectrometer and many particle accelerators. To solve D.3 problems, identify the scenario (E only, B only, or crossed fields), select the correct equation, apply Newton's second law or energy conservation, and always check whether the magnetic force does work (it doesn't). These tools will carry you confidently through any IB exam question on motion in electromagnetic fields.