Fields

How to Use This Guide

• D.1 Gravitational Fields — field strength, gravitational potential, orbital mechanics, escape velocity, geostationary orbit
• D.2 Electric and Magnetic Fields — Coulomb’s law, electric potential, uniform field between plates, magnetic force, Hall effect
• D.3 Motion in Electromagnetic Fields — parabolic trajectory in $E$ field, circular motion in $B$ field, velocity selector, cyclotron

Aligned to IB Physics 2025 syllabus — Theme D: Fields (first assessment 2025)

---

Jump to section: D.1 Gravitational Fields · D.2 Electric and Magnetic Fields · D.3 Motion in EM Fields · Exam Predictions · Practice Questions

Videos on this page: Watch: Gravitational Fields and Orbits · Watch: Electric Fields and Coulomb’s Law · Watch: Motion in Magnetic Fields

---

Section 1: D.1 Gravitational Fields

Quick Recall — Gravitational Fields

Q: What is gravitational field strength? A: $g = F/m$ — the gravitational force per unit mass at a point in the field. Units: $\text{N kg}^{-1}$ (equivalent to $\text{m s}^{-2}$).

Q: What does the negative sign in $V = -GM/r$ mean? A: Gravitational potential is zero at infinity; it becomes more negative as you move closer to a mass. The field is attractive — work must be done against the field to move away from the mass.

Q: What is Kepler’s Third Law? A: $T^2 \propto r^3$ — the square of the orbital period is proportional to the cube of the orbital radius.

---

Gravitational Field Strength

A gravitational field is a region of space in which a mass experiences a gravitational force. Field strength $g$ at a point is defined as the force per unit mass experienced by a small test mass placed at that point:

$g = \frac{F}{m} \qquad \text{units: N kg}^{-1}$

For a point mass $M$ (or outside a uniform sphere of mass $M$), the field strength at distance $r$ from the centre is:

$g = \frac{GM}{r^2}$

where $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

Newton’s Law of Gravitation

$F = \frac{GMm}{r^2}$

Gravitational Potential

The gravitational potential $V$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point:

$V = -\frac{GM}{r} \qquad \text{units: J kg}^{-1}$

The negative sign reflects the attractive nature of the field — the potential decreases (becomes more negative) as you approach the mass. At $r \to \infty$, $V = 0$.

The gravitational potential energy of mass $m$ at distance $r$ from mass $M$:

$E_p = mV = -\frac{GMm}{r}$

Orbital Mechanics

For a satellite in a circular orbit of radius $r$ around a planet of mass $M$, the gravitational force provides the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Solving for orbital speed:

$v = \sqrt{\frac{GM}{r}}$

Using $v = 2\pi r / T$, the orbital period follows:

$T^2 = \frac{4\pi^2 r^3}{GM}$

This is Kepler’s Third Law for circular orbits: $T^2 \propto r^3$.

Escape Velocity

The minimum launch speed for an object to escape a planet’s gravitational field (reaching $r \to \infty$ with zero kinetic energy remaining):

Setting total energy = 0:

$\tfrac{1}{2}mv_{\text{esc}}^2 + \left(-\frac{GMm}{R}\right) = 0$

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

where $R$ is the planet’s radius. Note that escape velocity depends on the planet’s mass and radius — not on the mass of the escaping object.

Watch: Gravitational Fields and Orbits

Khan Academy · 12 min · Covers gravitational field strength, Newton’s law of gravitation, orbital mechanics, and Kepler’s Third Law with step-by-step calculations matched to IB Paper 2 style.

---

Section 2: D.2 Electric and Magnetic Fields

Quick Recall — Electric and Magnetic Fields

Q: What is the direction of an electric field line? A: From positive to negative — the direction a positive test charge would move. Field lines never cross.

Q: What direction is the magnetic force on a current-carrying wire? A: Use the left-hand rule (for conventional current): thumb = force, index finger = field $B$, middle finger = current $I$.

Q: What is the Hall effect used for? A: To determine the sign of charge carriers in a conductor and to measure magnetic field strength.

---

Electric Field Strength

The electric field strength $E$ at a point is the force per unit positive charge experienced by a small test charge:

$E = \frac{F}{q} \qquad \text{units: N C}^{-1} \text{ (equivalent to V m}^{-1}\text{)}$

Coulomb’s Law

The electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$:

$F = \frac{kq_1 q_2}{r^2} = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}$

where $k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$ and $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$.

The electric field of a point charge $q$ at distance $r$:

$E = \frac{kq}{r^2}$

Electric Potential

The electric potential $V$ at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point:

$V = \frac{kq}{r} \qquad \text{units: J C}^{-1} = \text{V (volt)}$

Note: unlike gravitational potential, electric potential can be positive (near positive charge) or negative (near negative charge).

The relationship between electric field and potential:

$E = -\frac{\Delta V}{\Delta r}$

The field points in the direction of decreasing potential.

The electric potential energy of charge $q$ at a point with potential $V$:

$E_p = qV = \frac{kq_1 q_2}{r}$

Uniform Electric Field Between Parallel Plates

For two parallel plates separated by distance $d$ with potential difference $V$ between them:

$E = \frac{V}{d} \qquad \text{units: V m}^{-1}$

The field is uniform (constant magnitude and direction) between the plates, directed from the positive plate to the negative plate.

Magnetic Field and Flux Density

The magnetic flux density $B$ (also called the magnetic field strength) is measured in tesla (T). It characterises the strength of a magnetic field.

Field patterns (qualitative knowledge required for IB SL):

• Straight current-carrying wire: concentric circular field lines around the wire. Direction given by the right-hand rule: curl the fingers in the direction of field, thumb points in the direction of current.
• Solenoid: uniform field inside, similar to a bar magnet outside. Inside, field lines are parallel to the axis. Direction: use the right-hand rule with the coil current.

Force on a Current-Carrying Conductor

A straight conductor of length $L$ carrying current $I$ in a magnetic field $B$:

$F = BIL\sin\theta$

where $\theta$ is the angle between the current direction and the field direction. Maximum force when $\theta = 90°$ (perpendicular); zero force when the current is parallel to the field.

Force on a Moving Charge

A charge $q$ moving with velocity $v$ in a magnetic field $B$:

$F = qvB\sin\theta$

where $\theta$ is the angle between the velocity and the field. This force is always perpendicular to both $v$ and $B$.

The Hall Effect

When a current-carrying conductor is placed in a magnetic field perpendicular to the current, charge carriers are deflected sideways, building up a potential difference across the conductor known as the Hall voltage:

$V_H = \frac{BI}{nqt}$

where:

• $B$ = magnetic flux density (T)
• $I$ = current through the conductor (A)
• $n$ = number density of charge carriers ($\text{m}^{-3}$)
• $q$ = charge on each carrier (C)
• $t$ = thickness of the conductor in the direction of $B$ (m)

Watch: Electric Fields and Coulomb’s Law

Physics Online (David Fairhurst) · 14 min · Covers Coulomb’s law, electric field strength, electric potential, and the uniform field between parallel plates — all directly relevant to IB Physics SL Theme D. Clear worked examples throughout.

---

Section 3: D.3 Motion in Electromagnetic Fields

Quick Recall — Motion in EM Fields

Q: What shape is the path of a charged particle in a uniform electric field (perpendicular entry)? A: Parabolic — the same shape as projectile motion under gravity, because the electric force is constant and perpendicular to the initial velocity.

Q: What determines the radius of circular motion in a magnetic field? A: $r = mv/(qB)$. Greater momentum or smaller charge/field gives a larger radius.

Q: How does a velocity selector work? A: Electric force and magnetic force are balanced for one particular speed: $qE = qvB$, so $v = E/B$. Only particles with that speed travel in a straight line.

---

Charged Particle in a Uniform Electric Field

A charged particle entering a uniform electric field perpendicular to the field lines (e.g., horizontally between vertical plates) behaves exactly like a projectile:

• Along the field direction: constant acceleration $a = qE/m$, parabolic displacement
• Perpendicular to the field: constant velocity (no force component)

Charged Particle in a Uniform Magnetic Field

A charged particle moving perpendicular to a uniform magnetic field experiences a force $F = qvB$ that is always perpendicular to its velocity. Since the force never does work (no component along $v$), the speed remains constant and the path is a circle.

Setting magnetic force equal to centripetal force:

$qvB = \frac{mv^2}{r}$

$r = \frac{mv}{qB}$

Velocity Selector

A velocity selector uses crossed electric and magnetic fields to select particles of one specific speed, regardless of their mass or charge.

Setup: $\vec{E}$ and $\vec{B}$ are perpendicular to each other and both perpendicular to the particle’s velocity. The electric force $qE$ and magnetic force $qvB$ act in opposite directions.

For straight-line (undeflected) travel, the forces must be equal:

$qE = qvB \implies v = \frac{E}{B}$

Only particles with speed $v = E/B$ pass through undeflected. Faster particles are deflected in the magnetic force direction; slower particles in the electric force direction.

The Cyclotron

A cyclotron is a particle accelerator that uses a combination of:

1. A uniform magnetic field (perpendicular to the plane of motion) to curve the particles in semicircles
2. An alternating electric field across the gap between two D-shaped hollow electrodes (called dees) to accelerate particles each time they cross the gap

Combined Fields: Mass Spectrometry Principles

In a mass spectrometer, ions pass through a velocity selector (ensuring known speed $v = E/B_1$), then enter a region of different magnetic field $B_2$ where they travel in semicircles. The radius of curvature is:

$r = \frac{mv}{qB_2}$

Since $v$, $q$, and $B_2$ are known, measuring $r$ gives the mass $m$ of the ion. Different isotopes (same $q$, different $m$) have different radii and land at different positions on a detector.

Watch: Motion in Magnetic Fields — Circular Motion, Velocity Selector, and Cyclotron

Physics Online (David Fairhurst) · 16 min · Explains circular motion of charged particles in a uniform magnetic field, derives $r = mv/(qB)$, and covers the velocity selector and cyclotron with clear IB-style exam questions and worked solutions.

---

May 2026 Exam Predictions — Theme D

Paper 1 MCQ — Likely Topics

D.1 Gravitational Fields

• Given two planets with different masses and radii, calculate the ratio of surface gravitational field strengths ($g \propto M/R^2$)
• Use $T^2 \propto r^3$ to find the period or radius of a second orbit when the first is known
• State which of four statements about geostationary satellites is correct
• Identify the sign and shape of a gravitational potential graph ($V = -GM/r$, negative and asymptotically approaching zero from below)

D.2 Electric and Magnetic Fields

• Coulomb’s law: if one charge is doubled and separation is halved, find the factor change in force
• Identify the direction of force on a charge moving in a magnetic field (left-hand rule / $\vec{v} \times \vec{B}$)
• Which statement about the Hall effect is correct — expect one option about carrier sign determination
• Compare field lines of a uniform electric field vs a point charge field

D.3 Motion in EM Fields

• Given a diagram of two particle tracks in a magnetic field, identify which has greater momentum (larger radius = greater momentum)
• A proton and an electron enter a magnetic field with the same speed — compare their radii (proton has larger $r$ because greater mass)
• Velocity selector: show that the selected speed is independent of charge — expect a derivation-style MCQ

Paper 2 Multi-Part — Likely Questions

Question type 1 (D.1): Satellite orbit calculation

Typical structure: given a planet’s mass and radius, calculate orbital speed and period at a specified altitude. Then state what happens to orbital speed if the satellite moves to a higher orbit (speed decreases — counter-intuitive but follows from $v = \sqrt{GM/r}$). May ask to explain why a geostationary orbit must be equatorial.

Question type 2 (D.2): Parallel plate deflection

Given plate separation, voltage, and particle initial speed, calculate: (a) electric field between plates; (b) acceleration of particle; (c) vertical deflection after travelling the length of the plates. Expect 5–7 marks total with a follow-up asking about the shape of the trajectory.

Question type 3 (D.3): Cyclotron or mass spectrometer

Describe how a cyclotron accelerates protons — IB mark schemes want at minimum: role of the dees, role of the alternating electric field at the gap, role of the magnetic field causing circular motion, and why the radius increases. Alternatively, a mass spectrometer calculation: select speed, find radius, determine mass.

---

Exam-Style Practice Questions

Paper 1 Style (MCQ)

Q1. A satellite orbits a planet in a circular orbit of radius $r$. The gravitational field strength at radius $r$ is $g$. Which expression gives the orbital speed of the satellite?

A. $\sqrt{gr}$

B. $\sqrt{g/r}$

C. $gr$

D. $g/r$

Answer

A. Setting gravitational force equal to centripetal force: $mg = mv^2/r$, so $v^2 = gr$, giving $v = \sqrt{gr}$. Alternatively, use $g = GM/r^2$ and $v = \sqrt{GM/r}$ to get $v = \sqrt{gr}$.

---

Q2. A proton and an electron enter a region of uniform magnetic field with the same speed, both moving perpendicular to the field. How do their radii of curvature compare?

A. The proton has a smaller radius because it has greater charge.

B. The radii are equal because the charges are equal in magnitude.

C. The electron has a larger radius because the magnetic force on it is directed opposite to the field.

D. The proton has a larger radius because it has greater mass.

Answer

D. From $r = mv/(qB)$, with the same $v$, $q$ (same magnitude), and $B$, the radius is proportional to mass $m$. The proton’s mass is approximately 1836 times the electron mass, so the proton curves much less (much larger radius).

---

Q3. An electric field $E$ and magnetic field $B$ are perpendicular to each other in a velocity selector. A particle of charge $q$ and mass $m$ travels undeflected through the selector. The speed of the particle is:

A. $qB/E$

B. $E/B$

C. $qE/mB$

D. $B/E$

Answer

B. For straight-line travel: $qE = qvB$, so $v = E/B$. The charge $q$ and mass $m$ cancel — the speed selected is independent of the particle’s charge or mass.

---

Q4. Two identical point charges, each of charge $+Q$, are separated by distance $d$. The electric force between them is $F$. The separation is increased to $3d$ and one charge is changed to $-2Q$. What is the new force?

A. $\frac{2F}{9}$

B. $\frac{F}{9}$

C. $\frac{2F}{3}$

D. $2F$

Answer

A. Original: $F = kQ^2/d^2$. New: $F' = k(Q)(2Q)/(3d)^2 = 2kQ^2/(9d^2) = 2F/9$. The sign of the force changes from repulsive to attractive, but the magnitude is $2F/9$.

---

Paper 2 Style (Structured Response)

Q5. A satellite of mass $m$ orbits Earth (mass $M$, radius $R$) at an altitude $h$ above the surface.

(a) Show that the orbital speed is given by $v = \sqrt{GM/(R+h)}$. [2]

(b) The satellite is at altitude $h = 200 \text{ km}$. Given $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$, $M = 5.97 \times 10^{24} \text{ kg}$, $R = 6.37 \times 10^6 \text{ m}$, calculate the orbital period in minutes. [3]

(c) State two conditions that a geostationary satellite must satisfy in addition to having the correct orbital period. [2]

(d) Explain why a geostationary orbit must be at one specific altitude and no other. [2]

Mark-scheme answers

(a) The gravitational force provides the centripetal force:

$\frac{GMm}{(R+h)^2} = \frac{mv^2}{(R+h)}$

Cancel $m$ and one factor of $(R+h)$:

$v^2 = \frac{GM}{R+h} \implies v = \sqrt{\frac{GM}{R+h}} \checkmark$

[2 marks: 1 for equating gravitational and centripetal forces; 1 for correct rearrangement]

(b) $r = R + h = 6.37 \times 10^6 + 2.00 \times 10^5 = 6.57 \times 10^6 \text{ m}$

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.57 \times 10^6}} = \sqrt{6.06 \times 10^7} \approx 7790 \text{ m s}^{-1}$

$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.57 \times 10^6}{7790} \approx 5300 \text{ s} \approx 88 \text{ min}$

[3 marks: 1 for correct $r$; 1 for $v$; 1 for $T$ converted to minutes]

(c) Any two of: (1) the orbit must be in the equatorial plane; (2) the direction of orbit must match Earth’s rotation direction (west to east); (3) the orbit must be circular.

[2 marks: 1 per valid condition]

(d) The orbital period is determined by the orbital radius through Kepler’s Third Law ($T^2 = 4\pi^2r^3/GM$). For a geostationary orbit, $T$ must equal exactly 24 hours (Earth’s rotation period). Since $T$ uniquely determines $r$, there is only one radius — and therefore one altitude — that gives a 24-hour period.

[2 marks: 1 for linking $T = 24 \text{ h}$ to a unique $r$ via Kepler’s Third Law; 1 for explaining that one $T$ value gives one $r$ value]

---

Q6. A student investigates the motion of electrons in a magnetic field. Electrons are accelerated through a potential difference $V_a = 2.5 \text{ V}$ from rest, then enter a uniform magnetic field of $B = 8.0 \times 10^{-3} \text{ T}$ perpendicular to their velocity.

Data: $m_e = 9.11 \times 10^{-31} \text{ kg}$, $e = 1.60 \times 10^{-19} \text{ C}$

(a) Calculate the speed of the electrons as they enter the magnetic field. [2]

(b) Calculate the radius of the circular path. [2]

(c) Explain why the magnetic field does no work on the electron as it moves in the circle. [2]

(d) The electron beam is adjusted so that the same electrons also pass through a region of uniform electric field directed perpendicular to both the electron velocity and $B$. State the value of $E$ required so that the electrons travel in a straight line, and state which direction $E$ must point. [2]

Mark-scheme answers

(a) Using energy conservation: $eV_a = \tfrac{1}{2}m_e v^2$

$v = \sqrt{\frac{2eV_a}{m_e}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 2.5}{9.11 \times 10^{-31}}} = \sqrt{8.78 \times 10^{11}} \approx 9.4 \times 10^5 \text{ m s}^{-1}$

[2 marks: 1 for $eV_a = \tfrac{1}{2}m_ev^2$; 1 for correct answer]

(b) $r = m_e v / (eB) = (9.11 \times 10^{-31} \times 9.4 \times 10^5) / (1.60 \times 10^{-19} \times 8.0 \times 10^{-3})$

$r = \frac{8.56 \times 10^{-25}}{1.28 \times 10^{-21}} \approx 6.7 \times 10^{-4} \text{ m} = 0.67 \text{ mm}$

[2 marks: 1 for formula; 1 for correct answer]

(c) The magnetic force is always perpendicular to the electron’s velocity. Work done = $F \cdot d = Fd\cos 90° = 0$. Since no work is done, there is no change in kinetic energy, so the electron’s speed (and hence kinetic energy) remains constant throughout the circular motion.

[2 marks: 1 for stating force is perpendicular to velocity; 1 for linking this to zero work done and constant speed]

(d) For straight-line travel: $eE = evB \implies E = vB = 9.4 \times 10^5 \times 8.0 \times 10^{-3} \approx 7.5 \times 10^3 \text{ V m}^{-1}$.

The electric force must oppose the magnetic force on the electron. Since the electron is negative, the electric force is opposite to $\vec{E}$. The direction of $E$ must be determined by the geometry of the specific setup — $E$ should point such that the electric force on the electron is opposite to the magnetic force (toward the centre of the original circular path).

[2 marks: 1 for $E = vB$ with correct numerical answer; 1 for recognising the directions must oppose and for the correct direction statement]

---