Understand Motion in EM Fields — Understand D.3 Motion in electromagnetic fields - IB Physics

SECTION 1

Historical Context & Motivation

For centuries, electricity and magnetism were treated as separate curiosities—static sparks and compass needles had little in common. The story of how scientists unified these forces and learned to predict the paths of charged particles is one of the most exciting narratives in physics. Understanding how charged particles move through electromagnetic fields opened the door to technologies ranging from television screens to particle accelerators, and it remains a cornerstone of modern physics.

1785 • Coulomb's Law

Charles-Augustin de Coulomb quantified the force between electric charges, establishing that the electric force obeys an inverse-square law analogous to gravity.

1831 • Faraday's Induction

Michael Faraday demonstrated electromagnetic induction, showing that changing magnetic fields produce electric fields. He also introduced the concept of 'field lines' to visualize forces.

1865 • Maxwell's Equations

James Clerk Maxwell unified electricity and magnetism into a single theoretical framework and predicted electromagnetic waves, confirming that light is an electromagnetic phenomenon.

1897 • Thomson's Cathode Ray Experiment

J.J. Thomson used crossed electric and magnetic fields to measure the charge-to-mass ratio of the electron, proving that cathode rays were streams of charged particles deflected by EM fields.

1932 • First Cyclotron

Ernest Lawrence built the first cyclotron, using magnetic fields to curve protons in circles and electric fields to accelerate them, launching the era of particle physics.

These breakthroughs posed a fundamental question that IB Physics topic D.3 tackles directly: given a charged particle entering a region of electric or magnetic field, what path does it follow, and why? Answering this question requires combining Newton's second law with the force laws for electric and magnetic fields.

SECTION 2

Core Principles & Definitions

Before analysing trajectories, you need to be comfortable with the forces that electromagnetic fields exert on charged particles. Two distinct force rules govern the behaviour: one for electric fields and one for magnetic fields. When both fields are present simultaneously, the total force is called the Lorentz force. Let's break down the key ideas.

Electric Force on a Charge

A charge q in an electric field E experiences a force F = qE. The force acts parallel (or anti-parallel) to the field direction. Positive charges accelerate with the field; negative charges accelerate opposite to it.

Magnetic Force on a Moving Charge

A charge q moving with velocity v through a magnetic field B feels a force F = qvB sin θ, where θ is the angle between v and B. This force is always perpendicular to both v and B, so it changes the particle's direction but never its speed.

The Lorentz Force

When both fields are present, the total electromagnetic force is F = qE + qv × B. This combined force determines the full trajectory of the particle.

Work and Energy

Electric fields do work on charges because the force has a component along the displacement. Magnetic fields do no work because the force is always perpendicular to the velocity. Therefore, only electric fields change a particle's kinetic energy.

Right-Hand Rule

To find the direction of the magnetic force on a positive charge, point your fingers along v, curl them toward B, and your thumb points in the direction of F = qv × B. For negative charges, reverse the thumb direction.

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KEY TAKEAWAY

Think of an electric field like a hill: a charged particle on a hill gains speed as it rolls downhill (electric field does work and changes kinetic energy). A magnetic field is more like a banked curve on a racetrack—it steers you left or right without speeding you up or slowing you down. That is why magnetic forces change direction, while electric forces change speed.

SECTION 3

Visualising Charged-Particle Trajectories

The diagram below shows three fundamental trajectory types that arise when a charged particle enters a uniform field region. In an electric field oriented perpendicular to the initial velocity, the path curves into a parabola—just like projectile motion under gravity. In a magnetic field perpendicular to the velocity, the particle follows a circle because the force is always centripetal. When the velocity has a component along the magnetic field, the result is a helical (spiral) path.

Three fundamental trajectories: a charge entering perpendicular to E follows a parabola (left); a charge entering perpendicular to B follows a circle (centre); a charge with a velocity component along B traces a helix (right).

In the left panel, equally spaced horizontal field lines (pointing downward) create a constant downward force on the positive charge. Because the charge enters horizontally, it behaves exactly like a projectile: constant horizontal velocity, uniformly accelerating vertical velocity, producing a parabolic arc. In the centre panel, the magnetic field points into the page. The magnetic force is always perpendicular to the velocity, acting as a centripetal force that bends the path into a perfect circle. In the right panel, the velocity has a component parallel to B (which is unaffected) and a component perpendicular to B (which creates circular motion), combining into a helix.

SECTION 4

Mathematical Framework

The equations below form the mathematical toolkit for solving D.3 problems. They connect the forces from electric and magnetic fields to the resulting motion via Newton's second law.

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 is parallel to E for positive charges and antiparallel for negative charges.

MAGNETIC FORCE

F = qvB sin θ

v = speed of the charge (m s⁻¹), B = magnetic flux density (T), θ = angle between v and B. When v ⊥ B, sin θ = 1 and the force is maximum. When v ∥ B, sin θ = 0 and there is no force.

RADIUS OF CIRCULAR ORBIT IN B

r = mv / (qB)

Setting the magnetic force equal to the centripetal force: qvB = mv²/r. Solving for r gives r = mv/(qB). Here m = mass (kg). A faster or heavier particle orbits in a larger circle; a stronger field or larger charge produces a tighter circle.

KINETIC ENERGY FROM ELECTRIC POTENTIAL DIFFERENCE

qV = ½mv²

A charge q accelerated from rest through a potential difference V gains kinetic energy equal to qV. This is used frequently to find the speed of a particle before it enters a magnetic field region.

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IB Exam Tip

The IB data booklet provides F = qvB sin θ and the Lorentz force expression. You are expected to derive r = mv/(qB) by equating the magnetic force to the centripetal force. Practise this derivation until it feels automatic—it appears in nearly every D.3 exam question.

SECTION 5

The Velocity Selector & Crossed Fields

One of the most elegant applications of combined electric and magnetic fields is the velocity selector (also called a Wien filter). By arranging E and B perpendicular to each other and perpendicular to the particle's velocity, the electric force and the magnetic force act in opposite directions. Only particles travelling at a specific speed pass through undeflected, because at that speed the two forces balance exactly.

In a velocity selector, the downward electric force F_E = qE and the upward magnetic force F_B = qvB cancel when v = E/B. Faster particles curve one way; slower particles curve the other.

The velocity selector is the foundation of the mass spectrometer. After selecting particles of a known speed, they enter a region of magnetic field only. There, each particle follows a semicircular path whose radius depends on its mass (r = mv/(qB)). By measuring the radius, scientists determine the mass of the ion, which reveals its identity. This is how chemists and physicists identify isotopes and unknown compounds.

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Remember

The velocity selector equation v = E/B is independent of charge and mass. This means it works the same way for protons, electrons, or any ion—only speed matters.

SECTION 6

Worked Example — Proton in a Magnetic Field

A proton is accelerated from rest through a potential difference of 500 V and then enters a region of uniform magnetic field B = 0.20 T directed into the page. The proton's velocity is perpendicular to the field. Find (a) the proton's speed upon entering the field and (b) the radius of its circular path. Use m_p = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C.

Proton Orbit Radius

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Step 1 — Use energy conservation to find speed

The proton starts from rest, so all the electrical potential energy converts to kinetic energy: qV = ½mv². Rearranging for v gives v = √(2qV / m).

v = √(2qV / m)
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Step 2 — Substitute numerical values

v = √(2 × 1.60 × 10⁻¹⁹ × 500 / 1.67 × 10⁻²⁷) = √(1.60 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) = √(9.58 × 10¹⁰)

v ≈ 3.10 × 10⁵ m s⁻¹
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Step 3 — Apply the orbit radius formula

With v ⊥ B, the magnetic force provides centripetal force: qvB = mv²/r, so r = mv / (qB).

r = mv / (qB)
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Step 4 — Substitute and calculate

r = (1.67 × 10⁻²⁷ × 3.10 × 10⁵) / (1.60 × 10⁻¹⁹ × 0.20) = 5.18 × 10⁻²² / 3.20 × 10⁻²⁰

r ≈ 0.016 m = 1.6 cm
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Step 5 — Interpret the result

The proton traces a circle with a radius of about 1.6 cm. This is a small circle, which makes sense because the magnetic field is fairly strong (0.20 T) and the proton's mass is small. Note that the speed does not change—the magnetic field only redirects the proton.

SECTION 7

Electric vs. Magnetic Fields — Key Differences

Students often confuse the effects of electric and magnetic fields on charged particles. The table below highlights the critical differences that IB examiners love to test.

Property	Electric Field (E)	Magnetic Field (B)
Force direction	Parallel or antiparallel to E	Perpendicular to both v and B
Force on stationary charge	Yes — F = qE always applies	No — charge must be moving
Work done	Yes — changes kinetic energy	No — speed is constant
Typical trajectory	Straight line (parallel to E) or parabola (perpendicular entry)	Circle (v ⊥ B) or helix (v at angle to B)
Depends on charge sign?	Yes — positive and negative deflect opposite ways	Yes — opposite deflection via right-hand rule reversal
Depends on mass?	Acceleration a = qE/m depends on mass	Orbit radius r = mv/(qB) depends on mass

Comparison of electric and magnetic field effects on charged particles

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KEY TAKEAWAY

Here's a quick memory trick: Electric fields are like pushes — they speed you up or slow you down along a straight line. Magnetic fields are like turnstiles — they redirect you without changing how fast you're going. If an IB question asks 'does the kinetic energy change?', the answer is always no for a magnetic field and yes for an electric field (unless the charge moves along an equipotential).

SECTION 8

Connections to Advanced Topics

The principles of D.3 are the building blocks for many technologies and advanced physics topics. Understanding circular motion in a magnetic field, for example, leads directly to how cyclotrons and synchrotrons accelerate particles to near the speed of light. At relativistic speeds, the mass in r = mv/(qB) must be replaced by the relativistic momentum γmv, causing the orbit radius to grow as the particle speeds up.

D.3 Concept	Advanced Extension
r = mv/(qB) for circular orbits	Relativistic momentum: r = γmv/(qB) in synchrotrons; explains why the magnetic field must increase as particles accelerate
Velocity selector v = E/B	Mass spectrometry for isotope identification; used in nuclear physics and chemistry analysis
Helical motion in B	Charged particles spiral along Earth's magnetic field lines, creating the aurora borealis near the poles
Lorentz force F = qE + qv × B	Foundation of the Hall effect, electromagnetic induction (Faraday's law), and plasma confinement in fusion reactors (tokamaks)

From D.3 basics to cutting-edge applications

If you continue to study physics at university level, you'll encounter the full vector form of the Lorentz force and solve differential equations of motion in non-uniform fields. For now, the key insight is that every electromagnetic technology—from MRI machines to particle colliders—relies on the same force laws you've studied in this lesson, just applied in more complex geometries.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A proton moves horizontally to the right and enters a region where the magnetic field points vertically upward. Describe the initial direction of the magnetic force on the proton, and explain whether the proton's speed changes as it moves through the field.

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Using the right-hand rule (fingers point right along v, curl upward toward B), the force initially points out of the page—toward the observer. The speed does not change because the magnetic force is always perpendicular to the velocity, meaning it does no work. Only the direction of the proton's motion changes, causing it to curve into a circular or helical path depending on whether v has a component along B.

Question 2: BASIC CALCULATION

An electron (m = 9.11 × 10⁻³¹ kg, q = 1.60 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m s⁻¹ perpendicular to a uniform magnetic field of 0.050 T. Calculate the radius of its circular path.

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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. The electron traces a tiny circle because its mass is very small. A proton at the same speed and field would have a radius about 1836 times larger, since the proton is that many times more massive.

Question 3: INTERMEDIATE

A velocity selector uses an electric field E = 3.0 × 10⁴ V m⁻¹ and a magnetic field B = 0.15 T. (a) What speed must a singly charged ion have to pass through undeflected? (b) If the ion then enters a region of magnetic field only (B₂ = 0.30 T) and follows a semicircular path of radius 0.12 m, calculate the mass of the ion.

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(a) For the forces to balance: qE = qvB, so v = E/B = 3.0 × 10⁴ / 0.15 = 2.0 × 10⁵ m s⁻¹. (b) In the deflection region, r = mv/(qB₂), so m = qB₂r/v = (1.60 × 10⁻¹⁹ × 0.30 × 0.12) / (2.0 × 10⁵) = 5.76 × 10⁻²¹ / 2.0 × 10⁵ = 2.88 × 10⁻²⁶ kg. Dividing by the atomic mass unit (1.66 × 10⁻²⁷ kg) gives roughly 17 u, suggesting the ion could be OH⁻ or NH₃⁺. This two-stage process is the basis of mass spectrometry.

Question 4: APPLIED

In a cathode ray tube (old-style television), electrons are accelerated from rest through a potential difference of 2.0 kV and then enter a region of uniform electric field between two horizontal plates 5.0 cm long and 2.0 cm apart, with a voltage of 200 V across the plates. (a) Find the electron's speed upon entering the plates. (b) Find the vertical deflection of the electron as it exits the plates. Treat the motion like a projectile in a uniform gravitational field.

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(a) qV = ½mv² gives v = √(2qV/m) = √(2 × 1.60 × 10⁻¹⁹ × 2000 / 9.11 × 10⁻³¹) = √(7.02 × 10¹⁴) ≈ 2.65 × 10⁷ m s⁻¹. (b) The electric field between the plates is E = 200/0.020 = 1.0 × 10⁴ V m⁻¹. The vertical acceleration is a = qE/m = (1.60 × 10⁻¹⁹ × 1.0 × 10⁴) / 9.11 × 10⁻³¹ = 1.76 × 10¹⁵ m s⁻². The time spent between the plates is t = L/v = 0.050 / 2.65 × 10⁷ = 1.89 × 10⁻⁹ s. The vertical deflection is y = ½at² = 0.5 × 1.76 × 10¹⁵ × (1.89 × 10⁻⁹)² ≈ 3.1 × 10⁻³ m = 3.1 mm. This parabolic-trajectory analysis mirrors projectile motion, with the electric force replacing gravity.

Question 5: CRITICAL THINKING

Two ions—one with mass m and charge q, and another with mass 2m and charge q—are both accelerated from rest through the same potential difference V and then enter the same uniform magnetic field B perpendicular to their velocity. Show that the ratio of their circular orbit radii is r₁/r₂ = 1/√2, and explain physically why the heavier ion has a larger radius despite both ions having the same kinetic energy.

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Both ions gain kinetic energy qV = ½mv², so v = √(2qV/m). For ion 1: v₁ = √(2qV/m). For ion 2: v₂ = √(2qV/(2m)) = √(qV/m). The orbit radius r = mv/(qB). For ion 1: r₁ = m√(2qV/m)/(qB) = √(2mV/q)/B. For ion 2: r₂ = 2m√(qV/m)/(qB) = 2√(mV/q)/B. The ratio r₁/r₂ = √(2mV/q) / (2√(mV/q)) = √2/2 = 1/√2 ≈ 0.707. Physically, although both ions have the same kinetic energy, the heavier ion moves more slowly (KE = ½mv²). However, its momentum p = mv = m√(2qV/m) = √(2mqV) is larger by a factor of √2. Since the orbit radius depends on momentum (r = p/(qB)), the heavier ion makes a wider circle. This demonstrates that it is momentum, not energy, that determines the orbit radius in a magnetic field.

SUMMARY

Lesson Summary

Summary

Charged particles in electric fields experience a force F = qE that acts along the field direction, doing work and changing kinetic energy. This produces straight-line acceleration (when v is parallel to E) or parabolic trajectories (when v is perpendicular to E). In magnetic fields, the force F = qvB sin θ is always perpendicular to the velocity, so it does no work and the particle's speed stays constant. When v is perpendicular to B, the result is uniform circular motion with radius r = mv/(qB); when v has a component along B, a helical path results.

When both fields are present, the Lorentz force F = qE + qv × B governs the motion. A key application is the velocity selector (v = E/B), which allows only particles of a specific speed to pass undeflected. Combined with a magnetic deflection stage, this forms the basis of the mass spectrometer. Remember: electric fields change speed; magnetic fields change direction. Master the right-hand rule for finding force directions and you'll be well prepared for any D.3 exam question.