What is the force on a moving charge in a magnetic field?

The force on a moving charge in a magnetic field is given by the equation F = q(v × B), where F is the force, q is the charge, v is the velocity of the charge, and B is the magnetic field. This force is perpendicular to both the velocity of the charge and the magnetic field, which is why it is often referred to as the Lorentz force.

How does the direction of the magnetic force on a moving charge get determined?

The direction of the magnetic force on a moving charge is determined using the right-hand rule. Point your thumb in the direction of the velocity of the positive charge, and your fingers in the direction of the magnetic field. The force is directed out of your palm. For a negative charge, the force direction is opposite to that given by the right-hand rule.

Why does a moving charge experience no force in a magnetic field if it moves parallel to the field lines?

A moving charge experiences no force in a magnetic field if it moves parallel to the field lines because the magnetic force depends on the sine of the angle between the velocity vector and the magnetic field vector. When the charge moves parallel to the field lines, this angle is 0 degrees, and sin(0) is zero, resulting in no force.

How does the magnitude of the force on a moving charge change with velocity and magnetic field strength?

The magnitude of the force on a moving charge increases with both the velocity of the charge and the strength of the magnetic field. According to the formula F = qvBsin(θ), the force is directly proportional to the velocity (v) and the magnetic field strength (B), as well as the sine of the angle (θ) between the velocity and the magnetic field.

What role does the charge of a particle play in the force experienced in a magnetic field?

The charge of a particle plays a crucial role in determining the magnitude and direction of the force experienced in a magnetic field. The force is directly proportional to the charge (q), meaning that a larger charge will experience a greater force. Additionally, the sign of the charge affects the direction of the force, as determined by the right-hand rule for positive charges and its opposite for negative charges.

Why is "Saying charge at rest experiences magnetic force" a common mistake in Force on a moving charge?

Why it happens: Treating like electric. How to avoid it: Magnetic force on charge requires MOTION. At v = 0, F = 0. Source: 9702 Examiner Reports 2022-2024

Why is "Using the wrong dimension for $t$ in the Hall equation" a common mistake in Force on a moving charge?

Why it happens: Confusing strip thickness with width. How to avoid it: t is the thickness MEASURED IN THE DIRECTION OF THE MAGNETIC FIELD — the dimension across which V_H develops. Source: 9702 Examiner Reports 2022-2024

Why is "Thinking velocity selector depends on charge sign or magnitude" a common mistake in Force on a moving charge?

Why it happens: Forget both forces scale with q. How to avoid it: Both qE and Bqv scale identically with q, so v = E/B is independent of the charge.

Magnetic FieldsCambridge International A Levels Physics (9702)

Force on a moving charge

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Force on Moving Charge Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

$F = Bqv\sin\theta$ . Circular motion in B field. Hall effect, Hall probes, and velocity selection.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

20.3.1 — Determine the direction of the force on a moving charge.
20.3.2 — Recall and use $F = Bqv\sin\theta$ .
20.3.3 — Derive and use the Hall voltage $V_H = BI/(ntq)$ .
20.3.4 — Use a Hall probe to measure magnetic flux density.
20.3.5 — Describe circular motion of a charge in a perpendicular field.
20.3.6 — Use crossed E and B fields for velocity selection.

Force on a moving charge

$Bqv$ .

Formula. $F = Bqv\sin\theta$ .

Direction (positive charge). Fleming's left-hand rule (current direction = velocity direction).

For NEGATIVE charge (electron). Opposite to FLHR prediction.

Circular motion. Perpendicular $v$ to $B$ : force is centripetal. $Bqv = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{Bq}$

Period. $T = 2\pi m/(Bq)$ , independent of $v$ .

Cambridge tip. Magnetic force does NO WORK (perpendicular to motion). Speed unchanged in circular motion.

$F = Bqv\sin\theta$ .
Charge at rest: no force.
Circular path with $r = mv/(Bq)$ .

See the full worked example for force on a moving charge →

Hall effect and Hall probes

$V_H = BI/(ntq)$ .

Setup. Current $I$ flows along a thin slice of conductor of thickness $t$ (in the direction of $B$ ) and width $w$ . Magnetic field $B$ is applied perpendicular to the current.

Origin. Moving charge carriers experience $F = Bqv$ which pushes them sideways onto one face. Charge accumulates until the resulting transverse electric field $E_H$ pushes them back with equal force: $qE_H = Bqv \Rightarrow E_H = Bv$

Drift speed link. Recall $I = nAve$ , where $A = wt$ . So $v = I/(ntqw)$ .

Hall voltage. $V_H = E_H \times w = Bv \times w = \dfrac{BI}{ntq}$ .

$\boxed{V_H = \frac{BI}{ntq}}$

Why semiconductor? Carrier density $n$ is much lower than in a metal, so $V_H$ is much larger and easier to measure.

Hall probe. Calibrated semiconductor slice carrying constant $I$ . Place in unknown field → $V_H$ read off is proportional to $B$ . Used to map magnetic field patterns.

Cambridge tip. $t$ in the formula is the thickness measured IN THE DIRECTION OF $B$ , not the strip width. Get this wrong and the answer is out by a factor of $w/t$ .

$V_H = BI/(ntq)$ derives from $E_H = Bv$ balance.
Semiconductors give bigger $V_H$ than metals.
Hall probe measures $B$ .

See the full worked example for force on a moving charge →

Velocity selection with crossed fields

$v = E/B$ .

Idea. A beam of charged particles of mixed speeds enters a region of mutually perpendicular electric and magnetic fields, with $\vec{v}$ perpendicular to both.

Force balance. Electric force $qE$ (in field direction) opposes magnetic force $Bqv$ (the cross product reverses it for the right geometry). Net deflection is zero only when: $qE = Bqv \Rightarrow v = \frac{E}{B}$

Outcome. Particles with the right speed pass straight through; faster particles are deflected one way, slower particles the other.

Independent of $q$ . Both forces scale with $q$ , so the selected speed is the same for any charged particle.

Use case. First stage of a mass spectrometer — produces a mono-energetic beam before the bending region.

Cambridge tip. $v = E/B$ is the only condition examiners want for full marks.

$qE = Bqv$ at the selected speed.
$v = E/B$ , independent of charge.
Used in mass spectrometers.

See the full worked example for force on a moving charge →

How it’s examined

Force on charge on every Paper 4. Most-tested: force (5 marks), radius (7 marks), Hall voltage derivation + numerical (6-8 marks), velocity selector setup (4-5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/42 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Force on a moving charge, ready to print or save as PDF.

Step-by-step worked examples — Force on a moving charge

Step-by-step solutions to past-paper-style questions on force on a moving charge, written exactly the way a tutor would explain them at the board.

1Force on electron (5 marks)
Extended• Bqv
Question
Electron moves at $5.0 \times 10^6$ m/s perpendicular to 0.10 T field. Find force. $e = 1.6 \times 10^{-19}$ C. (5 marks)
Step-by-step solution
1. Step 1
  $F = Bqv$ .
  $F = 0.10 \times 1.6 \times 10^{-19} \times 5.0 \times 10^6 = 8.0 \times 10^{-14} \text{ N}$
Answer
$F = 8.0 \times 10^{-14}$ N.
2Radius of circular path (7 marks)
Extended• radius
Question
Find radius of electron's circular path in same field. $m_e = 9.1 \times 10^{-31}$ kg. (7 marks)
Step-by-step solution
1. Step 1
  Magnetic force = centripetal. $Bqv = mv^2/r$ .
2. Step 2
  Solve for $r$ .
  $r = \dfrac{mv}{Bq} = \dfrac{9.1 \times 10^{-31} \times 5 \times 10^6}{0.10 \times 1.6 \times 10^{-19}} \approx 2.84 \times 10^{-4} \text{ m}$
Answer
$r \approx 0.28$ mm.
3Hall voltage of a thin strip (6 marks)
Extended• Adapted from 9702/42 May/Jun 2024• Hall
Question
A semiconductor strip of thickness $t = 0.10$ mm carries $I = 50$ mA in a field $B = 0.20$ T perpendicular to the strip. Charge density $n = 1.0 \times 10^{22}$ m $^{-3}$ , $q = 1.6 \times 10^{-19}$ C. Find Hall voltage $V_H$ . (6 marks)
Step-by-step solution
1. Step 1
  Hall equation. $V_H = \dfrac{BI}{ntq}$ where $t$ is the thickness in the direction of the field.
  $V_H = \dfrac{0.20 \times 0.050}{1.0 \times 10^{22} \times 1.0 \times 10^{-4} \times 1.6 \times 10^{-19}}$
2. Step 2
  Evaluate.
  $V_H \approx 6.25 \times 10^{-3} \text{ V} = 6.3 \text{ mV}$
Answer
$V_H \approx 6.3$ mV.
4Velocity selector condition (4 marks)
Extended• velocity-selector
Question
Crossed fields $E = 1.0 \times 10^4$ V/m and $B = 0.020$ T select charged particles travelling straight through. Find selected speed. (4 marks)
Step-by-step solution
1. Step 1
  Balance. Electric force = magnetic force: $qE = Bqv \Rightarrow v = E/B$ .
  $v = 1.0 \times 10^{4} / 0.020 = 5.0 \times 10^{5} \text{ m/s}$
Answer
$v = 5.0 \times 10^{5}$ m/s.

Key Formulae — Force on a moving charge

The formulae you need to memorise for force on a moving charge on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Force on moving charge
$F = Bqv\sin\theta$
When to use
Charge $q$ moving at $v$ at angle $\theta$ to field $B$ .
Radius in magnetic field
$r = \dfrac{mv}{Bq}$
When to use
Charge moving perpendicular to field undergoes uniform circular motion.
Hall voltage
$V_H = \dfrac{BI}{ntq}$
When to use
Voltage across the thickness $t$ of a current-carrying conductor in a perpendicular field. $n$ = charge-carrier density.
Velocity selector
$v = \dfrac{E}{B}$
When to use
Crossed $E$ and $B$ fields. Particles with $v = E/B$ pass straight through; others deflect.

Key Definitions and Keywords — Force on a moving charge

Definitions to memorise and the exact keywords mark schemes credit for force on a moving charge answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Magnetic force on moving charge
Examiner keyword
$F = Bqv\sin\theta$ . Perpendicular to both $v$ and $B$ .
Hall voltage
Examiner keyword
Steady transverse PD that develops across a current-carrying conductor placed in a magnetic field, perpendicular to both the current and the field. Charge carriers are deflected to one face until the resulting electric field balances the magnetic force.
Hall probe
Examiner keyword
Thin slice of semiconductor used to MEASURE magnetic flux density. A constant current flows through it; the Hall voltage produced is proportional to $B$ .
Velocity selector
Examiner keyword
Region of crossed (perpendicular) electric and magnetic fields. Only charged particles with $v = E/B$ travel undeflected; faster or slower particles are deflected out of the beam.

Common Mistakes and Misconceptions — Force on a moving charge

The traps other students keep falling into on force on a moving charge questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Saying charge at rest experiences magnetic force
9702 Examiner Reports 2022-2024
Why it happens
Treating like electric.
How to avoid it
Magnetic force on charge requires MOTION. At $v = 0$ , $F = 0$ .
✕Using the wrong dimension for $t$ in the Hall equation
9702 Examiner Reports 2022-2024
Why it happens
Confusing strip thickness with width.
How to avoid it
$t$ is the thickness MEASURED IN THE DIRECTION OF THE MAGNETIC FIELD — the dimension across which $V_H$ develops.
✕Thinking velocity selector depends on charge sign or magnitude
Why it happens
Forget both forces scale with $q$ .
How to avoid it
Both $qE$ and $Bqv$ scale identically with $q$ , so $v = E/B$ is independent of the charge.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Force on a moving charge — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Magnetic FieldsCambridge International A Levels Physics (9702)

Force on a moving charge

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Study Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Force on a moving charge, ready to print or save offline.

Force on Moving Charge Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

$F = Bqv\sin\theta$ . Circular motion in B field. Hall effect, Hall probes, and velocity selection.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

20.3.1 — Determine the direction of the force on a moving charge.
20.3.2 — Recall and use $F = Bqv\sin\theta$ .
20.3.3 — Derive and use the Hall voltage $V_H = BI/(ntq)$ .
20.3.4 — Use a Hall probe to measure magnetic flux density.
20.3.5 — Describe circular motion of a charge in a perpendicular field.
20.3.6 — Use crossed E and B fields for velocity selection.

Force on a moving charge

$Bqv$ .

Formula. $F = Bqv\sin\theta$ .

Direction (positive charge). Fleming's left-hand rule (current direction = velocity direction).

For NEGATIVE charge (electron). Opposite to FLHR prediction.

Circular motion. Perpendicular $v$ to $B$ : force is centripetal. $Bqv = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{Bq}$

Period. $T = 2\pi m/(Bq)$ , independent of $v$ .

Cambridge tip. Magnetic force does NO WORK (perpendicular to motion). Speed unchanged in circular motion.

$F = Bqv\sin\theta$ .
Charge at rest: no force.
Circular path with $r = mv/(Bq)$ .

See the full worked example for force on a moving charge →

Hall effect and Hall probes

$V_H = BI/(ntq)$ .

Setup. Current $I$ flows along a thin slice of conductor of thickness $t$ (in the direction of $B$ ) and width $w$ . Magnetic field $B$ is applied perpendicular to the current.

Drift speed link. Recall $I = nAve$ , where $A = wt$ . So $v = I/(ntqw)$ .

Hall voltage. $V_H = E_H \times w = Bv \times w = \dfrac{BI}{ntq}$ .

$\boxed{V_H = \frac{BI}{ntq}}$

Why semiconductor? Carrier density $n$ is much lower than in a metal, so $V_H$ is much larger and easier to measure.

Hall probe. Calibrated semiconductor slice carrying constant $I$ . Place in unknown field → $V_H$ read off is proportional to $B$ . Used to map magnetic field patterns.

Cambridge tip. $t$ in the formula is the thickness measured IN THE DIRECTION OF $B$ , not the strip width. Get this wrong and the answer is out by a factor of $w/t$ .

$V_H = BI/(ntq)$ derives from $E_H = Bv$ balance.
Semiconductors give bigger $V_H$ than metals.
Hall probe measures $B$ .

See the full worked example for force on a moving charge →

Velocity selection with crossed fields

$v = E/B$ .

Idea. A beam of charged particles of mixed speeds enters a region of mutually perpendicular electric and magnetic fields, with $\vec{v}$ perpendicular to both.

Outcome. Particles with the right speed pass straight through; faster particles are deflected one way, slower particles the other.

Independent of $q$ . Both forces scale with $q$ , so the selected speed is the same for any charged particle.

Use case. First stage of a mass spectrometer — produces a mono-energetic beam before the bending region.

Cambridge tip. $v = E/B$ is the only condition examiners want for full marks.

$qE = Bqv$ at the selected speed.
$v = E/B$ , independent of charge.
Used in mass spectrometers.

See the full worked example for force on a moving charge →

How it’s examined

Force on charge on every Paper 4. Most-tested: force (5 marks), radius (7 marks), Hall voltage derivation + numerical (6-8 marks), velocity selector setup (4-5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/42 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Force on a moving charge, ready to print or save as PDF.

Step-by-step worked examples — Force on a moving charge

Step-by-step solutions to past-paper-style questions on force on a moving charge, written exactly the way a tutor would explain them at the board.

1Force on electron (5 marks)
Extended• Bqv
Question
Electron moves at $5.0 \times 10^6$ m/s perpendicular to 0.10 T field. Find force. $e = 1.6 \times 10^{-19}$ C. (5 marks)
Step-by-step solution
1. Step 1
  $F = Bqv$ .
  $F = 0.10 \times 1.6 \times 10^{-19} \times 5.0 \times 10^6 = 8.0 \times 10^{-14} \text{ N}$
Answer
$F = 8.0 \times 10^{-14}$ N.
2Radius of circular path (7 marks)
Extended• radius
Question
Find radius of electron's circular path in same field. $m_e = 9.1 \times 10^{-31}$ kg. (7 marks)
Step-by-step solution
1. Step 1
  Magnetic force = centripetal. $Bqv = mv^2/r$ .
2. Step 2
  Solve for $r$ .
  $r = \dfrac{mv}{Bq} = \dfrac{9.1 \times 10^{-31} \times 5 \times 10^6}{0.10 \times 1.6 \times 10^{-19}} \approx 2.84 \times 10^{-4} \text{ m}$
Answer
$r \approx 0.28$ mm.
3Hall voltage of a thin strip (6 marks)
Extended• Adapted from 9702/42 May/Jun 2024• Hall
Question
A semiconductor strip of thickness $t = 0.10$ mm carries $I = 50$ mA in a field $B = 0.20$ T perpendicular to the strip. Charge density $n = 1.0 \times 10^{22}$ m $^{-3}$ , $q = 1.6 \times 10^{-19}$ C. Find Hall voltage $V_H$ . (6 marks)
Step-by-step solution
1. Step 1
  Hall equation. $V_H = \dfrac{BI}{ntq}$ where $t$ is the thickness in the direction of the field.
  $V_H = \dfrac{0.20 \times 0.050}{1.0 \times 10^{22} \times 1.0 \times 10^{-4} \times 1.6 \times 10^{-19}}$
2. Step 2
  Evaluate.
  $V_H \approx 6.25 \times 10^{-3} \text{ V} = 6.3 \text{ mV}$
Answer
$V_H \approx 6.3$ mV.
4Velocity selector condition (4 marks)
Extended• velocity-selector
Question
Crossed fields $E = 1.0 \times 10^4$ V/m and $B = 0.020$ T select charged particles travelling straight through. Find selected speed. (4 marks)
Step-by-step solution
1. Step 1
  Balance. Electric force = magnetic force: $qE = Bqv \Rightarrow v = E/B$ .
  $v = 1.0 \times 10^{4} / 0.020 = 5.0 \times 10^{5} \text{ m/s}$
Answer
$v = 5.0 \times 10^{5}$ m/s.

Key Formulae — Force on a moving charge

The formulae you need to memorise for force on a moving charge on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Force on moving charge
$F = Bqv\sin\theta$
When to use
Charge $q$ moving at $v$ at angle $\theta$ to field $B$ .
Radius in magnetic field
$r = \dfrac{mv}{Bq}$
When to use
Charge moving perpendicular to field undergoes uniform circular motion.
Hall voltage
$V_H = \dfrac{BI}{ntq}$
When to use
Voltage across the thickness $t$ of a current-carrying conductor in a perpendicular field. $n$ = charge-carrier density.
Velocity selector
$v = \dfrac{E}{B}$
When to use
Crossed $E$ and $B$ fields. Particles with $v = E/B$ pass straight through; others deflect.

Key Definitions and Keywords — Force on a moving charge

Magnetic force on moving charge
Examiner keyword
$F = Bqv\sin\theta$ . Perpendicular to both $v$ and $B$ .
Hall voltage
Examiner keyword
Steady transverse PD that develops across a current-carrying conductor placed in a magnetic field, perpendicular to both the current and the field. Charge carriers are deflected to one face until the resulting electric field balances the magnetic force.
Hall probe
Examiner keyword
Thin slice of semiconductor used to MEASURE magnetic flux density. A constant current flows through it; the Hall voltage produced is proportional to $B$ .
Velocity selector
Examiner keyword
Region of crossed (perpendicular) electric and magnetic fields. Only charged particles with $v = E/B$ travel undeflected; faster or slower particles are deflected out of the beam.

Common Mistakes and Misconceptions — Force on a moving charge

✕Saying charge at rest experiences magnetic force
9702 Examiner Reports 2022-2024
Why it happens
Treating like electric.
How to avoid it
Magnetic force on charge requires MOTION. At $v = 0$ , $F = 0$ .
✕Using the wrong dimension for $t$ in the Hall equation
9702 Examiner Reports 2022-2024
Why it happens
Confusing strip thickness with width.
How to avoid it
$t$ is the thickness MEASURED IN THE DIRECTION OF THE MAGNETIC FIELD — the dimension across which $V_H$ develops.
✕Thinking velocity selector depends on charge sign or magnitude
Why it happens
Forget both forces scale with $q$ .
How to avoid it
Both $qE$ and $Bqv$ scale identically with $q$ , so $v = E/B$ is independent of the charge.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Force on a moving charge — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Force on a moving charge

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Force on electron (5 marks)

2Radius of circular path (7 marks)

3Hall voltage of a thin strip (6 marks)

4Velocity selector condition (4 marks)

Force on moving charge

Radius in magnetic field

Hall voltage

Velocity selector

Magnetic force on moving charge

Hall voltage

Hall probe

Velocity selector

✕Saying charge at rest experiences magnetic force

✕Using the wrong dimension for ttt in the Hall equation

✕Thinking velocity selector depends on charge sign or magnitude

Practice questions

Past paper style quiz

Assess your understanding

Force on a moving charge — frequently asked questions

Force on a moving charge

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Force on electron (5 marks)

2Radius of circular path (7 marks)

3Hall voltage of a thin strip (6 marks)

4Velocity selector condition (4 marks)

Force on moving charge

Radius in magnetic field

Hall voltage

Velocity selector

Magnetic force on moving charge

Hall voltage

Hall probe

Velocity selector

✕Saying charge at rest experiences magnetic force

✕Using the wrong dimension for ttt in the Hall equation

✕Thinking velocity selector depends on charge sign or magnitude

Practice questions

Past paper style quiz

Assess your understanding

Force on a moving charge — frequently asked questions

✕Using the wrong dimension for $t$ in the Hall equation

✕Using the wrong dimension for $t$ in the Hall equation