Why is "Using the right hand (or muddling the fingers) for the force on a current" a common mistake in Motion in electromagnetic fields?

Why it happens: There are several hand rules and students mix them up, or forget which finger is field and which is current. How to avoid it: For the force on a current or moving charge use the **LEFT** hand: First finger = Field, seCond finger = Current, thuMb = Motion/force. The right-hand rule is for the induced current in electromagnetic induction, not this. Source: IB Physics Theme D subject reports

Why is "Forgetting the sinθ, or thinking the force is always BIL / qvB" a common mistake in Motion in electromagnetic fields?

Why it happens: Students memorise F = BIL and F = qvB without the angle factor and apply them regardless of orientation. How to avoid it: Always write F = BIL sinθ and F = qvB sinθ. The force is **maximum when perpendicular** (sinθ = 1) and **zero when the current/velocity is parallel to the field** (sinθ = 0). Source: IB Physics Theme D subject reports

Why is "Thinking a magnetic force does work and changes the particle's speed" a common mistake in Motion in electromagnetic fields?

Why it happens: Students see the charge accelerating (curving) and assume it is speeding up, forgetting that a change of direction is also an acceleration. How to avoid it: A magnetic force is **always perpendicular to the velocity**, so W = Fd cos90° = 0. It changes **direction only** — the speed and kinetic energy stay constant. Source: IB Physics Theme D subject reports

Why is "Confusing the electric-field case (parabola, speed changes) with the magnetic-field case (circle, constant speed)" a common mistake in Motion in electromagnetic fields?

Why it happens: Both fields deflect a moving charge, so students blur the two situations together. How to avoid it: Uniform **magnetic** field ⟂ to v → **circular** path, **constant speed**, no work. Uniform **electric** field → **parabolic** path, **speed changes**, work done. Ask: did the speed change? If yes it was electric; if no it was magnetic. Source: IB Physics Theme D subject reports

Why is "Getting the force direction wrong for a negative charge" a common mistake in Motion in electromagnetic fields?

Why it happens: Students apply Fleming's left-hand rule with the second finger along the electron's velocity, forgetting the electron is negative. How to avoid it: Conventional current is opposite to an electron's motion. Work out the direction as if for a **positive** charge, then **reverse it** for a negative charge such as an electron. Source: IB Physics Theme D subject reports

Why is "Unit and power-of-ten slips (separation left in mm/cm, mishandling μ₀ or 10⁻¹⁹)" a common mistake in Motion in electromagnetic fields?

Why it happens: Distances are quoted in centimetres or millimetres, and the many powers of ten (10⁻¹⁹, 10⁻²⁷, 4π × 10⁻⁷) invite arithmetic errors. How to avoid it: Convert every length to **metres** before substituting, use μ₀/(2π) = 2 × 10⁻⁷ for parallel-wire problems, and track powers of ten line by line. Quote a final answer with units and sensible significant figures. Source: IB Physics Theme D subject reports

IB DP Physics (100452)

Motion in electromagnetic fields

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Detailed Study Notes

Detailed notes on Fields for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

D.3 Motion in Electromagnetic Fields — IB Physics SL Study Notes (Theme D: Fields)

How magnetic and electric fields push on currents and moving charges: the force F = BIL sinθ on a current-carrying wire, Fleming's left-hand rule, the force F = qvB sinθ on a moving charge, why a magnetic force makes a charge move in a circle (r = mv/qB) but does no work, the parabolic path of a charge in a uniform electric field, and the force between two parallel current-carrying wires.

At a glance

A current in a magnetic field feels a force: $F = BIL\sin\theta$ . $B$ is the magnetic flux density in tesla (T); $\theta$ is the angle between the current and the field.
Direction comes from Fleming's left-hand rule (use the LEFT hand): thumb = force/motion, First finger = Field, seCond finger = Current.
The force is maximum when the current is perpendicular to the field ( $\theta = 90^\circ$ ) and zero when it is parallel ( $\theta = 0^\circ$ ).
A moving charge in a magnetic field feels $F = qvB\sin\theta$ — the same physics, since a current is just moving charge.
A charge moving perpendicular to a uniform magnetic field moves in a circle: the magnetic force is the centripetal force, so $qvB = \dfrac{mv^2}{r}$ , giving radius $r = \dfrac{mv}{qB}$ .
A magnetic force does NO work — it is always perpendicular to the velocity — so it changes a charge's direction but never its speed or kinetic energy.
In a uniform electric field the force $F = qE$ is constant, giving constant acceleration and a parabolic path (just like projectile motion under gravity) — and this force does change the speed.
Two parallel wires exert a force per unit length $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ : parallel currents attract, antiparallel currents repel.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Calculate the force on a current-carrying conductor in a magnetic field using F = BIL sinθ and determine its direction with Fleming's left-hand rule.
Calculate the force on a charge moving through a magnetic field using F = qvB sinθ and find its direction for both positive and negative charges.
Explain why a charged particle moving perpendicular to a uniform magnetic field follows a circular path and derive the radius r = mv/(qB) by equating the magnetic force to the centripetal force.
Explain, from first principles, why a magnetic force does no work and therefore leaves a charge's speed and kinetic energy unchanged.
Describe the parabolic motion of a charged particle in a uniform electric field and calculate the force per unit length between two parallel current-carrying wires.

The force on a current-carrying conductor: F = BIL sinθ

A wire carrying a current in a magnetic field feels a force; its size is BIL sinθ and its direction follows Fleming's left-hand rule.

When a wire carrying an electric current sits in a magnetic field, the field pushes on the moving charges inside the wire, and the whole wire feels a force. This is the effect behind every electric motor and loudspeaker.

The size of the force is given by:

$F = BIL\sin\theta$

Symbol	Quantity	SI unit
$F$	force on the conductor	newton (N)
$B$	magnetic flux density	tesla (T)
$I$	current in the wire	ampere (A)
$L$	length of wire in the field	metre (m)
$\theta$	angle between the current and the field	degree/radian

The all-important $\sin\theta$ factor:

When the current is perpendicular to the field, $\theta = 90^\circ$ , $\sin\theta = 1$ , and the force is maximum: $F = BIL$ .
When the current is parallel to the field, $\theta = 0^\circ$ , $\sin\theta = 0$ , and the force is zero — a wire lying along the field lines feels no force at all.

Defining magnetic flux density. Rearranging the maximum-force case gives $B = \dfrac{F}{IL}$ . This is how the tesla is defined: a field has a flux density of 1 tesla if a wire carrying 1 A perpendicular to the field feels a force of 1 N on every 1 m of its length. So $1\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1}$ .

Direction — Fleming's left-hand rule. Hold the thumb and first two fingers of your left hand at right angles to each other:

thuMb → direction of the Motion (force $F$ )
First finger → direction of the magnetic Field $B$ (N to S)
seCond finger → direction of the conventional Current $I$ (+ to −)

A current-carrying wire in a magnetic field feels a force F = BIL sinθ. Here the field is into the page and the current is to the right, so the force is vertically upward — read off with the LEFT hand.

Remember: use the left hand for the force on a current (the motor effect). The right hand is for a different rule (the induced current in electromagnetic induction).

F = BIL sinθ; B is magnetic flux density in tesla (T).
Force is maximum when current ⟂ field (θ = 90°), zero when parallel (θ = 0°).
Direction: Fleming's LEFT-hand rule — thumb = force, First finger = Field, seCond finger = Current.

See the full worked example for motion in electromagnetic fields →

The force on a moving charge: F = qvB sinθ

A single charge moving through a magnetic field feels F = qvB sinθ; a negative charge is pushed the opposite way to a positive one.

A current is simply moving charge, so it is no surprise that a single charged particle moving through a magnetic field also feels a force:

$F = qvB\sin\theta$

Symbol	Quantity	SI unit
$F$	force on the charge	newton (N)
$q$	charge (magnitude)	coulomb (C)
$v$	speed of the charge	m s⁻¹
$B$	magnetic flux density	tesla (T)
$\theta$	angle between the velocity and the field	degree/radian

Just as with a wire, the force is maximum when the velocity is perpendicular to the field ( $\theta = 90^\circ$ , so $F = qvB$ ) and zero when the charge moves along the field lines ( $\theta = 0^\circ$ ).

Direction for a positive charge. Use Fleming's left-hand rule exactly as before, with the second finger pointing along the direction the positive charge moves (that is the conventional current direction).

Direction for a negative charge (e.g. an electron). A negative charge moving to the right is equivalent to a conventional current to the left. So point the second finger opposite to the electron's velocity — or, more simply, work out the force for a positive charge and then reverse it. This sign trap is one of the most common exam slips.

Same field, same velocity: the positive charge is pushed up and the negative charge is pushed down. Always reverse the Fleming result for a negative charge.

F = qvB sinθ for a charge of magnitude q moving at speed v.
Maximum force when v ⟂ B; zero force when v is along B.
For a negative charge, reverse the direction the left-hand rule gives.

See the full worked example for motion in electromagnetic fields →

Deriving circular motion and the radius r = mv/(qB) (first principles)

The magnetic force is always perpendicular to the velocity, so it does no work and acts as a centripetal force — equating qvB to mv²/r gives r = mv/(qB).

This is the heart of D.3, and IB rewards students who can reason it out rather than just quote the result.

Step 1 — a magnetic force does no work. Work done is $W = F d\cos\phi$ , where $\phi$ is the angle between the force and the displacement. The magnetic force $F = qvB$ is always perpendicular to the velocity (it acts at $90^\circ$ to the motion), so $\phi = 90^\circ$ and $\cos 90^\circ = 0$ . Therefore:

$W = Fd\cos 90^\circ = 0$

A magnetic force does zero work. By the work–energy theorem the kinetic energy — and hence the speed — of the charge cannot change. A magnetic field can only change the charge's direction, never its speed.

Step 2 — constant speed + perpendicular force = a circle. If a charge enters a uniform magnetic field moving perpendicular to it, the magnetic force always has the same magnitude ( $qvB$ , since $v$ is constant) and always points at right angles to the velocity, turning towards the same side. A force of constant size, always perpendicular to the motion, is exactly the condition for uniform circular motion. So the charge travels in a circle at constant speed, and the magnetic force provides the centripetal force.

The velocity is always tangent to the circle and the magnetic force always points to the centre. Because the force is perpendicular to v it does no work — the speed stays constant while the direction turns.

Step 3 — derive the radius. For circular motion the centripetal force is $\dfrac{mv^2}{r}$ . Set the magnetic force equal to it:

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

Cancel one factor of $v$ from both sides and rearrange for $r$ :

$qB = \frac{mv}{r} \quad\Rightarrow\quad \boxed{r = \frac{mv}{qB}}$

Reading the equation like a physicist:

Faster particles ( $v$ larger) or heavier particles ( $m$ larger) → bigger circles (more inertia, harder to turn).
Larger charge $q$ or stronger field $B$ → smaller circles (bigger turning force).
Because the speed is constant, the particle keeps going round the same circle forever (in a vacuum) — this is the principle of the cyclotron and of charged particles trapped in planetary magnetic fields.

Magnetic force ⟂ velocity ⇒ W = Fd cos90° = 0 ⇒ speed (and KE) constant.
Constant-magnitude perpendicular force ⇒ uniform circular motion; magnetic force = centripetal force.
qvB = mv²/r ⇒ r = mv/(qB): faster/heavier → bigger circle; larger q or B → smaller circle.

See the full worked example for motion in electromagnetic fields →

A charge in a uniform electric field: parabolic motion

In a uniform electric field the force F = qE is constant, giving constant acceleration and a parabolic path — and, unlike a magnetic force, it does work and changes the speed.

A charge placed in a uniform electric field $E$ (for example between two parallel charged plates) feels a constant force:

$F = qE$

directed along the field for a positive charge, and opposite to the field for a negative charge. Because the mass is constant, this gives a constant acceleration $a = \dfrac{qE}{m}$ .

The projectile analogy. A constant force at right angles to a particle's initial velocity is exactly the situation of a projectile in gravity (topic A.1). So a charge fired across a uniform field follows a parabola, and you solve it the same way — two independent motions sharing only the time:

	Along the initial velocity	Along the field (deflection)
Force / acceleration	none ( $a = 0$ )	constant $a = qE/m$
Velocity	constant $v$	changes: grows from 0
Displacement	$x = vt$	$y = \tfrac{1}{2}\left(\dfrac{qE}{m}\right)t^2$

The time spent in the field is $t = \dfrac{L}{v}$ (length of the plates divided by the horizontal speed); feed that into the deflection equation to find how far the charge is pushed sideways.

A positive charge fired between charged plates is pushed toward the negative plate by the constant force qE, tracing a parabola — the electric-field version of projectile motion.

The crucial contrast with a magnetic field. The electric force $qE$ has a component along the direction of motion, so it does work on the charge — its speed and kinetic energy change. This is the opposite of the magnetic case, where the force is always perpendicular to $v$ , does no work, and leaves the speed unchanged. In short: electric field → parabola, speed changes; magnetic field → circle, speed constant.

Uniform E-field: constant force F = qE, constant acceleration a = qE/m.
Charge fired across the field traces a parabola (like a projectile under gravity).
Unlike a magnetic force, the electric force does work — the speed changes.

See the full worked example for motion in electromagnetic fields →

The force between two parallel current-carrying wires

Each wire sits in the other's magnetic field, so they push on each other: parallel currents attract, antiparallel currents repel, with F/L = μ₀I₁I₂/(2πd).

A current-carrying wire creates a magnetic field around itself. If a second current-carrying wire is placed nearby, it sits in that field and therefore feels a force ( $F = BIL$ ). By Newton's third law the first wire feels an equal and opposite force. The two wires exert a force per unit length on each other:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Symbol	Quantity	SI unit
$F/L$	force per unit length	N m⁻¹
$\mu_0$	permeability of free space $= 4\pi\times10^{-7}$	T m A⁻¹
$I_1, I_2$	the two currents	ampere (A)
$d$	separation of the wires	metre (m)

Direction — attract or repel?

Parallel currents (same direction) ATTRACT.
Antiparallel currents (opposite directions) REPEL.

You can confirm this with Fleming's left-hand rule: find the field one wire makes at the position of the other (field circles around a wire), then apply the rule to the second wire's current. A handy memory hook: "same way, come together; opposite ways, push apart."

Two parallel wires with currents in the same direction attract; with currents in opposite directions they repel. The force per unit length is μ₀I₁I₂/(2πd).

Watch the powers of ten and units. $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$ , and the separation $d$ must be in metres — converting a distance given in centimetres or millimetres is where most marks are lost. A useful shortcut: $\dfrac{\mu_0}{2\pi} = 2\times10^{-7}\ \text{T m A}^{-1}$ , so $\dfrac{F}{L} = 2\times10^{-7}\times\dfrac{I_1 I_2}{d}$ .

Each wire sits in the other's magnetic field, so they push on each other.
F/L = μ₀I₁I₂/(2πd); μ₀ = 4π×10⁻⁷ T m A⁻¹; μ₀/(2π) = 2×10⁻⁷.
Parallel currents attract; antiparallel currents repel.

See the full worked example for motion in electromagnetic fields →

Putting it together: magnetic vs electric field motion

One table to lock in the difference: magnetic fields give circles at constant speed (no work); electric fields give parabolas with changing speed (work done).

Almost every exam trap in D.3 comes from confusing the two field cases. Here is the single comparison you must be able to reproduce:

Feature	Charge in a uniform magnetic field (⟂ to v)	Charge in a uniform electric field
Force	$F = qvB$ , always ⟂ to velocity	$F = qE$ , constant direction (along field)
Does it do work?	No ( $F\perp v$ )	Yes ( $F$ has a component along the motion)
Speed / kinetic energy	unchanged	changes
Path	circle (radius $r = mv/qB$ )	parabola
Analogy	motion in a circle at constant speed	projectile motion under gravity

How to decide in the exam:

Is the field magnetic? Then the force is perpendicular to the velocity → the path curves into a circle, and the speed is constant. Use $qvB = mv^2/r$ .
Is the field electric? Then the force is constant in direction → the path is a parabola, and the speed changes. Use $F = qE$ and treat it like a projectile.

A worked mental check. An electron enters a region moving horizontally. If it comes out moving at the same speed but in a new direction, the region contained a magnetic field. If it comes out faster or slower, the region contained an electric field (which did work on it). That one observation — did the speed change? — tells you which field was present.

Keep the language precise: a magnetic force changes direction only; an electric force can change both speed and direction. Examiners specifically look for the statement that the magnetic force does no work.

Magnetic field ⟂ v: circular path, r = mv/qB, constant speed, no work done.
Electric field: parabolic path, speed changes, work is done.
Quick test: if the speed is unchanged it was a magnetic field; if it changed it was electric.

Quick recap

Force on a current-carrying conductor: F = BIL sinθ; maximum when current ⟂ field, zero when parallel; direction from Fleming's left-hand rule.
Force on a moving charge: F = qvB sinθ; reverse the direction for a negative charge.
A charge moving ⟂ to a uniform magnetic field moves in a circle because the magnetic force is the centripetal force: qvB = mv²/r ⇒ r = mv/(qB).
A magnetic force does no work (it is ⟂ to v), so the charge's speed and kinetic energy are unchanged — only its direction changes.
In a uniform electric field the constant force qE gives constant acceleration and a parabolic path, and it DOES change the speed (work is done).
Two parallel wires exert F/L = μ₀I₁I₂/(2πd) on each other: parallel currents attract, antiparallel currents repel; μ₀ = 4π×10⁻⁷ T m A⁻¹.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Force on a wire: F = BIL sinθ (max when ⟂, zero when parallel).
Fleming's LEFT hand: thuMb = Motion/force, First = Field, seCond = Current.
Force on a moving charge: F = qvB sinθ (reverse direction for a negative charge).
Circular motion: qvB = mv²/r ⇒ r = mv/(qB).
A magnetic force does NO work → speed and KE constant, direction only changes.
Uniform E-field: F = qE, constant a, parabolic path, speed changes (work done).
Parallel wires: F/L = μ₀I₁I₂/(2πd); μ₀ = 4π×10⁻⁷ T m A⁻¹; same direction attract, opposite repel.
1 tesla: 1 N of force per metre on a 1 A current placed ⟂ to the field (B = F/IL).

How it’s examined

D.3 sits in Theme D (Fields) and appears on both papers. Paper 1A (MCQ, no calculator): identifying the direction of a force with Fleming's left-hand rule, recognising that a magnetic force does no work / keeps the speed constant, spotting whether parallel wires attract or repel, and reading the effect of the sinθ factor. Paper 1B (data-based): substituting into F = BIL, F = qvB or r = mv/qB with careful powers of ten. Paper 2: multi-step questions (4–10 marks) such as calculating the radius of a charged particle's circular path, comparing the motion in electric and magnetic fields, or working out the force per unit length between two wires — often with a 'describe/explain' part asking WHY the path is a circle or WHY the speed is unchanged. Command terms: state, determine, calculate, describe, explain, show that. Examiner reports repeatedly flag: using the right hand instead of the left, dropping the sinθ, claiming a magnetic force changes the speed, confusing the electric-field (parabola, speed changes) case with the magnetic-field (circle, constant speed) case, getting the direction wrong for a negative charge, and unit slips (mm → m, μ₀). Always state your sign/direction reasoning and show the substitution — method marks are given even when the final arithmetic slips.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme D: Fields (D.3 Motion in electromagnetic fields); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Motion in electromagnetic fields

Step-by-step solutions to past-paper-style questions on motion in electromagnetic fields, written exactly the way a tutor would explain them at the board.

Question type:

1Force on a current-carrying wire
Getting startedDirect calculation• motor effect, AO2
Question
A straight wire of length 0.10 m carries a current of 5.0 A at right angles to a uniform magnetic field of flux density 0.20 T. Calculate the force on the wire. (2 marks)
Step-by-step solution
1. Step 1
  The current is perpendicular to the field, so $\theta = 90^\circ$ and $\sin\theta = 1$ . Use $F = BIL\sin\theta$ .
  $F = BIL = (0.20)(5.0)(0.10)$
2. Step 2
  Evaluate.
  $F = 0.10\,\text{N}$
Answer
F = 0.10 N.
Examiner tip
Mark scheme: (1) correct substitution into F = BIL; (2) F = 0.10 N with unit. Because the current is perpendicular to the field, sinθ = 1 and the full BIL applies.
2Including the sinθ factor
Getting startedDirect calculation• motor effect, AO2
Question
The same wire (length 0.10 m, current 5.0 A, field 0.20 T) is now turned so that the current makes an angle of 30° with the field. Calculate the new force. (2 marks)
Step-by-step solution
1. Step 1
  Now $\theta = 30^\circ$ , so include the sine factor: $F = BIL\sin\theta$ .
  $F = (0.20)(5.0)(0.10)\sin30^\circ$
2. Step 2
  Since $\sin30^\circ = 0.5$ , the force is half the perpendicular value.
  $F = 0.10 \times 0.5 = 0.050\,\text{N}$
Answer
F = 0.050 N.
Examiner tip
Mark scheme: (1) include sin30°; (2) F = 0.050 N. The whole point of the question is the sinθ: at 30° the force is half its perpendicular maximum, and it would be zero if the current lay along the field.
3Force on a moving charge
Getting startedDirect calculation• moving charge, AO2
Question
A proton (charge 1.6 × 10⁻¹⁹ C) moves at 6.0 × 10⁶ m s⁻¹ perpendicular to a uniform magnetic field of flux density 0.50 T. Calculate the magnetic force on the proton. (2 marks)
Step-by-step solution
1. Step 1
  Velocity is perpendicular to the field, so $\theta = 90^\circ$ . Use $F = qvB$ .
  $F = (1.6\times10^{-19})(6.0\times10^{6})(0.50)$
2. Step 2
  Multiply the powers of ten carefully.
  $F = 4.8\times10^{-13}\,\text{N}$
Answer
F = 4.8 × 10⁻¹³ N.
Examiner tip
Mark scheme: (1) correct substitution into F = qvB; (2) F = 4.8 × 10⁻¹³ N with unit. Handling the powers of ten (10⁻¹⁹ × 10⁶ = 10⁻¹³) is where marks are commonly lost.
4Radius of a proton's circular path
Building confidenceDirect calculation• circular motion, AO2
Question
The proton from the previous example (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C, v = 6.0 × 10⁶ m s⁻¹) moves in a circle in the 0.50 T field. Calculate the radius of its path. (3 marks)
Step-by-step solution
1. Step 1
  The magnetic force is the centripetal force: $qvB = \dfrac{mv^2}{r}$ . Rearranging gives $r = \dfrac{mv}{qB}$ .
  $r = \frac{mv}{qB}$
2. Step 2
  Substitute the values.
  $r = \frac{(1.67\times10^{-27})(6.0\times10^{6})}{(1.6\times10^{-19})(0.50)}$
3. Step 3
  Evaluate numerator and denominator.
  $r = \frac{1.00\times10^{-20}}{8.0\times10^{-20}} = 0.13\,\text{m}$
Answer
r ≈ 0.13 m (about 13 cm).
Examiner tip
Mark scheme: (1) equate qvB to mv²/r and rearrange for r; (2) correct substitution; (3) r ≈ 0.13 m with unit. Cancelling one factor of v from qvB = mv²/r is the key algebra step.
5Radius of an electron's path
Building confidenceDirect calculation• circular motion, moving charge, AO2
Question
An electron (m = 9.11 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C) enters a uniform magnetic field of 4.0 × 10⁻³ T at right angles, moving at 5.0 × 10⁶ m s⁻¹. Calculate the radius of its circular path. (3 marks)
Step-by-step solution
1. Step 1
  Use the circular-motion result $r = \dfrac{mv}{qB}$ .
  $r = \frac{mv}{qB}$
2. Step 2
  Substitute the values for the electron.
  $r = \frac{(9.11\times10^{-31})(5.0\times10^{6})}{(1.6\times10^{-19})(4.0\times10^{-3})}$
3. Step 3
  Evaluate.
  $r = \frac{4.56\times10^{-24}}{6.4\times10^{-22}} = 7.1\times10^{-3}\,\text{m}$
Answer
r ≈ 7.1 × 10⁻³ m (about 7.1 mm).
Examiner tip
Mark scheme: (1) r = mv/qB; (2) correct substitution with the electron's mass; (3) r ≈ 7.1 mm. The much smaller mass of the electron gives a far smaller radius than a proton in the same field.
6Force and acceleration in a uniform electric field
Building confidenceDirect calculation• electric field, AO2
Question
A proton (q = 1.6 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) sits in a uniform electric field of strength 1.2 × 10⁴ N C⁻¹. Calculate (a) the electric force on it and (b) its acceleration. (3 marks)
Step-by-step solution
1. Step 1
  The electric force on a charge is $F = qE$ .
  $F = qE = (1.6\times10^{-19})(1.2\times10^{4}) = 1.92\times10^{-15}\,\text{N}$
2. Step 2
  The acceleration follows from Newton's second law $a = F/m$ .
  $a = \frac{F}{m} = \frac{1.92\times10^{-15}}{1.67\times10^{-27}} = 1.15\times10^{12}\,\text{m s}^{-2}$
Answer
(a) F = 1.9 × 10⁻¹⁵ N; (b) a = 1.2 × 10¹² m s⁻².
Examiner tip
Mark scheme: (1) F = qE with substitution; (2) F = 1.9 × 10⁻¹⁵ N; (3) a = F/m = 1.2 × 10¹² m s⁻². Because the force qE is constant, the acceleration is constant — this is what makes the path parabolic when the charge is fired across the field.
7Force per unit length between parallel wires
StretchMulti-step problem• parallel wires, AO2
Question
Two long parallel wires are 5.0 cm apart. One carries a current of 10 A and the other 10 A in the same direction. Calculate the force per unit length between them and state whether they attract or repel. (μ₀ = 4π × 10⁻⁷ T m A⁻¹) (4 marks)
Step-by-step solution
1. Step 1
  Convert the separation to metres: $d = 5.0\,\text{cm} = 0.050\,\text{m}$ . Use $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ .
  $\frac{F}{L} = \frac{(4\pi\times10^{-7})(10)(10)}{2\pi(0.050)}$
2. Step 2
  The factor $\dfrac{\mu_0}{2\pi} = 2\times10^{-7}$ simplifies the arithmetic.
  $\frac{F}{L} = 2\times10^{-7}\times\frac{(10)(10)}{0.050}$
3. Step 3
  Evaluate.
  $\frac{F}{L} = 2\times10^{-7}\times 2000 = 4.0\times10^{-4}\,\text{N m}^{-1}$
4. Step 4
  The currents are in the same direction, so the wires attract.
Answer
F/L = 4.0 × 10⁻⁴ N m⁻¹; the wires attract (currents parallel).
Examiner tip
Mark scheme: (1) convert cm → m and select the correct formula; (2) substitution; (3) F/L = 4.0 × 10⁻⁴ N m⁻¹; (4) attract because the currents are parallel. Forgetting to convert 5.0 cm to 0.050 m is the most common error.
8Parabolic deflection in a uniform electric field
StretchMulti-step problem• electric field, projectile analogy, AO2
Question
An electron (m = 9.11 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C) enters midway between two horizontal plates, travelling horizontally at 2.0 × 10⁷ m s⁻¹. The plates are 0.060 m long and produce a uniform field of 5.0 × 10³ N C⁻¹. Calculate the vertical deflection of the electron as it leaves the plates. (5 marks)
Step-by-step solution
1. Step 1
  Treat this like a projectile: horizontal motion is constant velocity, vertical motion has constant acceleration. First find the vertical force and acceleration.
  $F = qE = (1.6\times10^{-19})(5.0\times10^{3}) = 8.0\times10^{-16}\,\text{N}$
2. Step 2
  Vertical acceleration from Newton's second law.
  $a = \frac{F}{m} = \frac{8.0\times10^{-16}}{9.11\times10^{-31}} = 8.78\times10^{14}\,\text{m s}^{-2}$
3. Step 3
  Time spent between the plates = plate length ÷ horizontal speed.
  $t = \frac{L}{v} = \frac{0.060}{2.0\times10^{7}} = 3.0\times10^{-9}\,\text{s}$
4. Step 4
  Vertical deflection using $y = \tfrac{1}{2}at^2$ (starts with zero vertical velocity).
  $y = \tfrac{1}{2}(8.78\times10^{14})(3.0\times10^{-9})^2$
5. Step 5
  Evaluate.
  $y = \tfrac{1}{2}(8.78\times10^{14})(9.0\times10^{-18}) = 4.0\times10^{-3}\,\text{m}$
Answer
Vertical deflection y ≈ 4.0 × 10⁻³ m (about 4.0 mm).
Examiner tip
Mark scheme: (1) F = qE; (2) a = F/m; (3) time t = L/v; (4) select y = ½at²; (5) y ≈ 4.0 mm. The grade-9 move is recognising that this is projectile motion with a constant electric force instead of gravity — horizontal and vertical motions are independent and share only the time.

Model Answers — Motion in electromagnetic fields

High-scoring sample answers for motion in electromagnetic fields on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define magnetic flux density and state its SI unit.
Model answer
Magnetic flux density $B$ is defined through the force on a current-carrying conductor: the force per unit length per unit current on a wire placed perpendicular to the field, i.e. $B = \dfrac{F}{IL}$ .

Its SI unit is the tesla (T), where $1\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1}$ .
Why this scores
Why this scores 2/2. (1) B = F/(IL) with the wire perpendicular to the field (the definition, not just 'field strength'); (2) unit tesla, equivalent to N A⁻¹ m⁻¹. Saying merely 'the strength of a magnetic field' is too vague to earn the definition mark.
Question 2
3 marks
Q (3 marks). A wire of length 0.25 m carries a current of 4.0 A perpendicular to a uniform magnetic field of flux density 0.15 T. Calculate the force on the wire and state one way the force could be made zero without switching off the current.
Model answer
Using $F = BIL\sin\theta$ with $\theta = 90^\circ$ (so $\sin\theta = 1$ ):

$F = (0.15)(4.0)(0.25) = 0.15\,\text{N}$

The force could be made zero by rotating the wire so that the current is parallel to the field ( $\theta = 0^\circ$ , so $\sin\theta = 0$ ).
Why this scores
Why this scores 3/3. (1) correct substitution into F = BIL; (2) F = 0.15 N with unit; (3) force is zero when the current lies along the field (θ = 0°). The final part tests understanding of the sinθ factor.
Question 3
3 marks
Q (3 marks). A horizontal wire carries a conventional current from west to east. A uniform magnetic field points vertically downward. (a) State the rule used to find the direction of the force on the wire. (b) Determine the direction of the force.
Model answer
(a) Fleming's left-hand rule is used: with the left hand, the First finger points along the Field, the seCond finger along the Current, and the thuMb gives the direction of the force (Motion).

(b) With the field pointing down (first finger) and the current pointing east (second finger), the thumb points to the north (horizontally). So the force on the wire is directed horizontally toward the north.
Why this scores
Why this scores 3/3. (a) names Fleming's LEFT-hand rule (1); (b) correctly aligns fingers (1) and states the force is horizontal, toward the north (1). Using the right hand, or muddling which finger is which, is the usual error here.
Question 4
4 marks
Q (4 marks). An electron travels horizontally to the east through a uniform magnetic field of 0.40 T that points vertically upward, at a speed of 3.0 × 10⁶ m s⁻¹. (a) Calculate the magnitude of the magnetic force on the electron. (b) Determine the direction of this force. (charge magnitude = 1.6 × 10⁻¹⁹ C)
Model answer
(a) The velocity is perpendicular to the field, so $F = qvB$ :

$F = (1.6\times10^{-19})(3.0\times10^{6})(0.40) = 1.9\times10^{-13}\,\text{N}$

(b) First find the direction for a positive charge moving east in an upward field, using Fleming's left-hand rule: first finger up (field), second finger east (current) gives a force pointing south. Because the electron is negative, the force is reversed, so the actual force on the electron is directed toward the north.
Why this scores
Why this scores 4/4. (a) F = qvB with substitution (1) and F ≈ 1.9 × 10⁻¹³ N (1); (b) apply Fleming's rule for a positive charge (1) then reverse for the negative electron (1). The assessed idea is remembering to flip the direction for a negative charge.
Question 5
4 marks
Q (4 marks). A charged particle enters a uniform magnetic field at right angles to the field. Explain why the particle moves in a circle at constant speed, and state what happens to its kinetic energy.
Model answer
The magnetic force on the moving charge is always perpendicular to its velocity (and perpendicular to the field).

Because the force is perpendicular to the velocity, the work done is $W = Fd\cos 90^\circ = 0$ — the magnetic force does no work. Therefore the particle's speed (and kinetic energy) cannot change.

The force has a constant magnitude ( $qvB$ , since $v$ is constant) and always points at right angles to the motion, turning the velocity toward the same side. A constant-magnitude force always perpendicular to the velocity is the condition for uniform circular motion, so the particle travels in a circle at constant speed.

Its kinetic energy is unchanged throughout, because no work is done on it.
Why this scores
Why this scores 4/4. (1) force is perpendicular to velocity; (2) therefore no work is done (W = Fd cos90° = 0); (3) constant-magnitude perpendicular force ⇒ circular motion / magnetic force provides the centripetal force; (4) kinetic energy (and speed) unchanged. Examiners specifically want the statement that the magnetic force does no work.
Question 6
5 marks
Q (5 marks). A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 4.0 × 10⁶ m s⁻¹ perpendicular to a uniform magnetic field of 0.30 T. (a) Show that the radius of its circular path is given by r = mv/(qB). (b) Calculate the radius.
Model answer
(a) The magnetic force provides the centripetal force needed for circular motion. Equate them:

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

Cancel one factor of $v$ from both sides:

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

Rearrange for $r$ :

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

(b) Substitute the values:

$r = \frac{(1.67\times10^{-27})(4.0\times10^{6})}{(1.6\times10^{-19})(0.30)} = \frac{6.68\times10^{-21}}{4.8\times10^{-20}} = 0.14\,\text{m}$
Why this scores
Why this scores 5/5. (a) equate qvB to mv²/r (1), cancel v (1), rearrange to r = mv/qB (1); (b) correct substitution (1) and r ≈ 0.14 m with unit (1). The 'show that' requires the algebra to be laid out, not just the final formula quoted.
Question 7
6 marks
Q (6 marks). Two long straight parallel wires are 4.0 cm apart. Wire X carries a current of 8.0 A and wire Y carries a current of 8.0 A in the same direction. (a) Calculate the force per unit length that each wire exerts on the other. (b) State and explain whether the wires attract or repel. (c) State what happens to this force if the current in Y is reversed. (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
Model answer
(a) Convert the separation: $d = 4.0\,\text{cm} = 0.040\,\text{m}$ . Using $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ with $\dfrac{\mu_0}{2\pi} = 2\times10^{-7}$ :

$\frac{F}{L} = 2\times10^{-7}\times\frac{(8.0)(8.0)}{0.040} = 2\times10^{-7}\times 1600 = 3.2\times10^{-4}\,\text{N m}^{-1}$

(b) The wires attract. Each wire lies in the magnetic field produced by the other; applying Fleming's left-hand rule to wire Y in wire X's field gives a force pointing toward X (and vice versa). Currents in the same direction attract.

(c) If the current in Y is reversed, the currents are now antiparallel, so the force reverses direction and the wires repel. Its magnitude is unchanged (still 3.2 × 10⁻⁴ N m⁻¹), since the formula depends only on the sizes of the currents.
Why this scores
Why this scores 6/6. (a) convert cm → m (1), substitution (1), F/L = 3.2 × 10⁻⁴ N m⁻¹ (1); (b) attract, with a field-plus-Fleming justification (1); (c) reversing one current makes them repel (1) with the same magnitude (1). The unit conversion and the attract/repel reasoning are the discriminators.
Question 8
6 marks
Q (6 marks). An electron (m = 9.11 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C) enters midway between two horizontal parallel plates, moving horizontally at 1.5 × 10⁷ m s⁻¹. The plates are 0.040 m long and set up a uniform electric field of strength 8.0 × 10³ N C⁻¹. (a) Calculate the vertical deflection of the electron as it leaves the plates. (b) State and explain the shape of the electron's path between the plates.
Model answer
(a) Treat this like a projectile: constant horizontal velocity, constant vertical acceleration from the electric force.

Vertical force and acceleration: $F = qE = (1.6\times10^{-19})(8.0\times10^{3}) = 1.28\times10^{-15}\,\text{N}$ $a = \frac{F}{m} = \frac{1.28\times10^{-15}}{9.11\times10^{-31}} = 1.405\times10^{15}\,\text{m s}^{-2}$

Time between the plates: $t = \frac{L}{v} = \frac{0.040}{1.5\times10^{7}} = 2.67\times10^{-9}\,\text{s}$

Vertical deflection (zero initial vertical velocity): $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(1.405\times10^{15})(2.67\times10^{-9})^2 = 5.0\times10^{-3}\,\text{m}$

So the electron is deflected about 5.0 mm.

(b) The path is a parabola. The electron has constant velocity horizontally (no horizontal force) and constant acceleration vertically (constant force qE along the field), which is exactly the condition — identical to projectile motion under gravity — that produces a parabolic path.
Why this scores
Why this scores 6/6. (a) F = qE (1), a = F/m (1), t = L/v (1), y = ½at² ≈ 5.0 mm (1); (b) parabola (1) because horizontal velocity is constant while the vertical force/acceleration is constant (1). Recognising the projectile analogy and keeping the two motions independent are the assessed skills.
Question 9
10 marks
Q (10 marks — extended response). A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) travels at 6.0 × 10⁶ m s⁻¹ and enters a region of uniform magnetic field of flux density 0.50 T, moving perpendicular to the field. (a) Calculate the magnitude of the magnetic force on the proton. (b) Explain why the proton follows a circular path and why its speed does not change. (c) Calculate the radius of the circular path. The proton is then removed and instead sent into a region of uniform electric field of strength 3.0 × 10⁴ N C⁻¹, entering perpendicular to the field. (d) Calculate the acceleration of the proton in the electric field and describe the shape of its path. (e) Compare the effect of the magnetic field and the electric field on the proton's speed, explaining the difference in terms of work done.
Model answer
(a) Magnetic force (velocity perpendicular to field, so $F = qvB$ ): $F = (1.6\times10^{-19})(6.0\times10^{6})(0.50) = 4.8\times10^{-13}\,\text{N}$

(b) Why a circle at constant speed. The magnetic force is always perpendicular to the velocity. A perpendicular force does no work ( $W = Fd\cos 90^\circ = 0$ ), so the proton's speed and kinetic energy cannot change. The force has constant magnitude ( $qvB$ ) and always points at right angles to the motion, which is exactly the condition for uniform circular motion — the magnetic force acts as the centripetal force.

(c) Radius (equate the magnetic force to the centripetal force, $qvB = mv^2/r$ ): $r = \frac{mv}{qB} = \frac{(1.67\times10^{-27})(6.0\times10^{6})}{(1.6\times10^{-19})(0.50)} = \frac{1.00\times10^{-20}}{8.0\times10^{-20}} = 0.13\,\text{m}$

(d) In the electric field. The force is $F = qE = (1.6\times10^{-19})(3.0\times10^{4}) = 4.8\times10^{-15}\,\text{N}$ , giving acceleration $a = \frac{F}{m} = \frac{4.8\times10^{-15}}{1.67\times10^{-27}} = 2.9\times10^{12}\,\text{m s}^{-2}$ This force is constant in direction (along the field), so — entering perpendicular to it — the proton follows a parabolic path, just like a projectile under gravity.

(e) Comparison.

In the magnetic field the force is always perpendicular to the velocity, so it does no work: the proton's speed is unchanged — only its direction changes (a circle).

In the electric field the constant force has a component along the direction of motion, so it does work on the proton: its speed (and kinetic energy) changes as it follows the parabola.

The single deciding factor is the work done: a magnetic force does none (perpendicular to v), an electric force does (not perpendicular), which is why one keeps the speed constant and the other does not.
Why this scores
Why this scores 10/10. (a) F = qvB = 4.8 × 10⁻¹³ N (2); (b) perpendicular force → no work → constant speed, and constant-magnitude perpendicular force → circular motion (2); (c) r = mv/qB = 0.13 m (2); (d) a = qE/m = 2.9 × 10¹² m s⁻² and a parabolic path (2); (e) magnetic force does no work so speed constant, electric force does work so speed changes (2). This is a model IB Paper 2 extended response: the grade-9 discriminators are the explicit 'no work' argument for the magnetic case and the parabola-versus-circle contrast, tied back to whether work is done.

Key Formulae — Motion in electromagnetic fields

The formulae you need to memorise for motion in electromagnetic fields on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Force on a current-carrying conductor
$F = BIL\sin\theta$
$F$
force on the conductor (N)
$B$
magnetic flux density (T)
$I$
current (A)
$L$
length of wire in the field (m)
$\theta$
angle between the current and the field
When to use
For the force on a wire in a magnetic field. Maximum (F = BIL) when the current is perpendicular to the field; zero when parallel.
Example
0.20 T, 5.0 A, 0.10 m, perpendicular: F = (0.20)(5.0)(0.10) = 0.10 N.
Force on a moving charge
$F = qvB\sin\theta$
$F$
force on the charge (N)
$q$
charge magnitude (C)
$v$
speed of the charge (m s⁻¹)
$B$
magnetic flux density (T)
$\theta$
angle between the velocity and the field
When to use
For a single charge moving through a magnetic field. Reverse the direction (from Fleming's rule) for a negative charge.
Example
Proton at 6.0 × 10⁶ m s⁻¹ ⟂ to 0.50 T: F = (1.6 × 10⁻¹⁹)(6.0 × 10⁶)(0.50) = 4.8 × 10⁻¹³ N.
Magnetic force as centripetal force
$qvB = \frac{mv^2}{r}$
$q$
charge magnitude (C)
$v$
speed of the charge (m s⁻¹)
$B$
magnetic flux density (T)
$m$
mass of the charge (kg)
$r$
radius of the circular path (m)
When to use
When a charge moves perpendicular to a uniform magnetic field. Equating the magnetic force to the centripetal force is the starting point for finding the radius.
Example
Set qvB = mv²/r, cancel v, and rearrange to get r = mv/(qB).
Radius of a charged particle's circular path
$r = \frac{mv}{qB}$
$r$
radius of the circular path (m)
$m$
mass of the charge (kg)
$v$
speed (m s⁻¹)
$q$
charge magnitude (C)
$B$
magnetic flux density (T)
When to use
To find the radius of the circle a charge follows in a uniform magnetic field. Faster/heavier → bigger circle; larger q or B → smaller circle.
Example
Proton, v = 6.0 × 10⁶ m s⁻¹, B = 0.50 T: r = (1.67 × 10⁻²⁷ × 6.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.50) ≈ 0.13 m.
Force on a charge in an electric field
$F = qE$
$F$
force on the charge (N)
$q$
charge magnitude (C)
$E$
electric field strength (N C⁻¹ or V m⁻¹)
When to use
For a charge in a uniform electric field: the force is constant, giving constant acceleration a = qE/m and a parabolic path when fired across the field.
Example
Proton in E = 1.2 × 10⁴ N C⁻¹: F = (1.6 × 10⁻¹⁹)(1.2 × 10⁴) = 1.9 × 10⁻¹⁵ N.
Force per unit length between parallel wires
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$
$F/L$
force per unit length (N m⁻¹)
$\mu_0$
permeability of free space = 4π × 10⁻⁷ T m A⁻¹
$I_1, I_2$
the two currents (A)
$d$
separation of the wires (m)
When to use
For the force between two long parallel current-carrying wires. Parallel currents attract; antiparallel currents repel. Keep d in metres.
Example
10 A and 10 A, 0.050 m apart: F/L = 2 × 10⁻⁷ × (100/0.050) = 4.0 × 10⁻⁴ N m⁻¹ (attract).

Key Definitions and Keywords — Motion in electromagnetic fields

Definitions to memorise and the exact keywords mark schemes credit for motion in electromagnetic fields answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Magnetic flux density (B)
Examiner keyword
A measure of the strength of a magnetic field, defined by the force on a current-carrying conductor: B = F/(IL) for a wire perpendicular to the field. A vector. SI unit: tesla (T).
Tesla (T)
Examiner keyword
The SI unit of magnetic flux density. A field of 1 T exerts 1 N of force on each 1 m of a wire carrying 1 A perpendicular to the field: 1 T = 1 N A⁻¹ m⁻¹.
Motor effect
Examiner keyword
The force experienced by a current-carrying conductor placed in a magnetic field, given by F = BIL sinθ. It is the effect behind electric motors.
Fleming's left-hand rule
Examiner keyword
A rule for the direction of the force on a current (or moving positive charge) in a magnetic field: using the left hand, First finger = Field, seCond finger = Current, thuMb = force/Motion.
Force on a moving charge
Examiner keyword
The magnetic force on a charge q moving at speed v through a field B, F = qvB sinθ. It is maximum when v is perpendicular to B and zero when v is along B.
Centripetal force
Examiner keyword
The net inward force required for circular motion, of size mv²/r directed toward the centre. For a charge in a magnetic field the magnetic force provides this centripetal force.
Magnetic force does no work
Examiner keyword
Because the magnetic force is always perpendicular to the velocity, W = Fd cos90° = 0. It therefore changes a charge's direction but never its speed or kinetic energy.
Radius of circular motion (r = mv/qB)
Examiner keyword
The radius of the circle a charge follows in a uniform magnetic field, obtained by equating qvB to mv²/r. Larger mass or speed gives a bigger circle; larger charge or field gives a smaller one.
Electric field strength (E)
Examiner keyword
The force per unit positive charge at a point, E = F/q. In a uniform field the force on a charge, F = qE, is constant. SI unit: N C⁻¹ (equivalently V m⁻¹).
Uniform field
A field with the same magnitude and direction everywhere. A uniform magnetic field gives circular charge motion; a uniform electric field gives constant force and parabolic motion.
Parabolic motion (in an electric field)
Examiner keyword
The curved path of a charge fired across a uniform electric field: constant velocity along the entry direction and constant acceleration along the field, exactly like a projectile under gravity.
Permeability of free space (μ₀)
Examiner keyword
A constant that appears in the magnetic force between wires, μ₀ = 4π × 10⁻⁷ T m A⁻¹. The shortcut μ₀/(2π) = 2 × 10⁻⁷ T m A⁻¹ speeds up calculations.
Force between parallel wires
Examiner keyword
Two long parallel current-carrying wires exert a force per unit length F/L = μ₀I₁I₂/(2πd) on each other: parallel currents attract, antiparallel currents repel.
Conventional current
The direction of current taken as the flow of positive charge (from + to −). Fleming's left-hand rule uses conventional current, so for electrons the current is opposite to their motion.

Common Mistakes and Misconceptions — Motion in electromagnetic fields

The traps other students keep falling into on motion in electromagnetic fields questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the right hand (or muddling the fingers) for the force on a current
IB Physics Theme D subject reports
Why it happens
There are several hand rules and students mix them up, or forget which finger is field and which is current.
How to avoid it
For the force on a current or moving charge use the LEFT hand: First finger = Field, seCond finger = Current, thuMb = Motion/force. The right-hand rule is for the induced current in electromagnetic induction, not this.
✕Forgetting the sinθ, or thinking the force is always BIL / qvB
IB Physics Theme D subject reports
Why it happens
Students memorise F = BIL and F = qvB without the angle factor and apply them regardless of orientation.
How to avoid it
Always write F = BIL sinθ and F = qvB sinθ. The force is maximum when perpendicular (sinθ = 1) and zero when the current/velocity is parallel to the field (sinθ = 0).
✕Thinking a magnetic force does work and changes the particle's speed
IB Physics Theme D subject reports
Why it happens
Students see the charge accelerating (curving) and assume it is speeding up, forgetting that a change of direction is also an acceleration.
How to avoid it
A magnetic force is always perpendicular to the velocity, so W = Fd cos90° = 0. It changes direction only — the speed and kinetic energy stay constant.
✕Confusing the electric-field case (parabola, speed changes) with the magnetic-field case (circle, constant speed)
IB Physics Theme D subject reports
Why it happens
Both fields deflect a moving charge, so students blur the two situations together.
How to avoid it
Uniform magnetic field ⟂ to v → circular path, constant speed, no work. Uniform electric field → parabolic path, speed changes, work done. Ask: did the speed change? If yes it was electric; if no it was magnetic.
✕Getting the force direction wrong for a negative charge
IB Physics Theme D subject reports
Why it happens
Students apply Fleming's left-hand rule with the second finger along the electron's velocity, forgetting the electron is negative.
How to avoid it
Conventional current is opposite to an electron's motion. Work out the direction as if for a positive charge, then reverse it for a negative charge such as an electron.
✕Unit and power-of-ten slips (separation left in mm/cm, mishandling μ₀ or 10⁻¹⁹)
IB Physics Theme D subject reports
Why it happens
Distances are quoted in centimetres or millimetres, and the many powers of ten (10⁻¹⁹, 10⁻²⁷, 4π × 10⁻⁷) invite arithmetic errors.
How to avoid it
Convert every length to metres before substituting, use μ₀/(2π) = 2 × 10⁻⁷ for parallel-wire problems, and track powers of ten line by line. Quote a final answer with units and sensible significant figures.

Motion in electromagnetic fields

Detailed Study Notes

At a glance

What you’ll learn

The force on a current-carrying conductor: F = BIL sinθ

The force on a moving charge: F = qvB sinθ

Deriving circular motion and the radius r = mv/(qB) (first principles)

A charge in a uniform electric field: parabolic motion

The force between two parallel current-carrying wires

Putting it together: magnetic vs electric field motion

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Force on a current-carrying wire

2Including the sinθ factor

3Force on a moving charge

4Radius of a proton's circular path

5Radius of an electron's path

6Force and acceleration in a uniform electric field

7Force per unit length between parallel wires

8Parabolic deflection in a uniform electric field

Force on a current-carrying conductor

Force on a moving charge

Magnetic force as centripetal force

Radius of a charged particle's circular path

Force on a charge in an electric field

Force per unit length between parallel wires

Magnetic flux density (B)

Tesla (T)

Motor effect

Fleming's left-hand rule

Force on a moving charge

Centripetal force

Magnetic force does no work

Radius of circular motion (r = mv/qB)

Electric field strength (E)

Uniform field

Parabolic motion (in an electric field)

Permeability of free space (μ₀)

Force between parallel wires

Conventional current

✕Using the right hand (or muddling the fingers) for the force on a current

✕Forgetting the sinθ, or thinking the force is always BIL / qvB

✕Thinking a magnetic force does work and changes the particle's speed

✕Confusing the electric-field case (parabola, speed changes) with the magnetic-field case (circle, constant speed)

✕Getting the force direction wrong for a negative charge

✕Unit and power-of-ten slips (separation left in mm/cm, mishandling μ₀ or 10⁻¹⁹)