Physics · Electrodynamics

Lorentz force

Q: How do you calculate the magnitude of the Lorentz force?

The magnitude of the magnetic contribution is $|F_B|=|q|vB\sin\alpha$, where the angle lies between velocity and magnetic field. Insert charge in coulombs, speed in metres per second and flux density in tesla. An electron with $v=10^6\,\mathrm{m/s}$ perpendicular to $B=0.1\,\mathrm T$ experiences about $1.60\cdot10^{-14}\,\mathrm N$. Perpendicular motion gives maximum magnetic force; parallel motion gives zero. Charge sign determines direction: find the cross product first, then reverse it for negative charge. Include any electric field separately. The complete Lorentz force is $\vec F=q(\vec E+\vec v\times\vec B)$; its magnitude does not generally follow from the magnetic shortcut.

Q: What is the difference between the electric force and the Lorentz force?

Electric force $\vec F_E=q\vec E$ acts on a charge even when it is stationary in the chosen reference frame. For positive charge it points along the electric field, and for negative charge in the opposite direction. The magnetic contribution $\vec F_B=q\vec v\times\vec B$ requires motion transverse to the magnetic field and is perpendicular to velocity. It does no work itself and, acting alone, does not change speed. Both contributions form the complete Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$. Use the same frame for fields and velocity. A problem without relevant electric field may calculate only the magnetic part; electric and magnetic contributions need not remain the same when changing frames.

The upper equation gives the magnetic contribution to the Lorentz force on a moving charge; an electric field adds electric force.

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Formula

Calculate

F = q \cdot (v \times B)

F in N · q in C · v in m/s · B in T (tesla)

Variables & units

Variables & units
Symbol	Meaning	Unit
F⃗	Magnetic contribution to the Lorentz force (vector)	N
$q$	Electric charge	C
v⃗	Velocity of the charge	m/s
B⃗	Magnetic flux density	T

LaTeX

\vec{F} = q \cdot (\vec{v} \times \vec{B})

QUANTA STUDYquanta-study.de

Two charges +e and −e moving right in a field out of the plane. Force on +e points down, on −e up, with equal magnitudes. — Transverse speed 300 km/s rightward and B=0.200 T out of the plane give force magnitude 9.61·10⁻¹⁵ N for ±e. Negative charge reverses direction on the right. Purely magnetic power is zero; arrow lengths are schematic.

Lorentz force calculator

Find the magnitude of the magnetic Lorentz force on a moving charge, or the speed and flux density from the force; without an electric field.

Find

Charge q

Charge with its sign, for an electron −1.602 · 10⁻¹⁹ C (type -1.602e-19).

= −1.602 · 10⁻¹⁹

Speed magnitude v

Speed, e.g. 10⁷ m/s for an electron after an accelerating voltage of about 300 V.

= 10⁷

Magnetic flux density B

Magnetic flux density, e.g. 2 mT between Helmholtz coils; Earth's field is only about 50 µT.

Angle α

Angle between velocity and magnetic field, 0° to 180°; the force is largest at 90°.

Filled with example values. Type your own and the result updates straight away.

Result: Force magnitude |F|

|F| = 3.204 · 10⁻¹⁵ N

With q = -1.602 · 10⁻¹⁹ C, v = 10 000 000 m/s, B = 2 mT and α = 90 °, |F| = 3.204 · 10⁻¹⁵ N.

Working · rearrange for |F|, substitute, evaluate

1. Rearrange for |F|
$|F|=|q|vB\sin\alpha$
2. Substitute values (in SI units)
- B = 2 mT = 0.002 T
- α = 90 ° = 1.5708 rad
$|F|=|(-1.602\times10^{-19})|\cdot 10000000\cdot 0.002\cdot \sin\left(1.57079632679\right)=3.204\times10^{-15}$
3. Result
|F| = 3.204 · 10⁻¹⁵ N

Quick tips

If the charge moves parallel to the field (α = 0° or 180°), there is no magnetic force. At right angles (α = 90°) simply F = |q| · v · B.
Unit check: 1 T = 1 N·s/(C·m). So coulombs times m/s times teslas gives exactly newtons.

Assumptions and limits

Only the magnetic contribution |F| = |q|vB sin α, with no electric field. v and B are magnitudes; α lies between 0° and 180°. Force direction and signed force components are not calculated.

Derivation & background

The complete electromagnetic force is $\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)$ . The cross product determines magnetic magnitude and direction. Determine direction first for positive charge, then reverse for negative charge. Magnetic force is perpendicular to velocity and itself does no work. A circle additionally needs uniform steady field, purely perpendicular entry and no other forces.

Exam blueprint

Validity range

Local magnetic force on a charge in a specified reference frame. An electric field adds to the complete force. Circle and helix formulas need uniform steady field, no other forces and nonrelativistic motion.

Derivation steps

Evaluate the empirical magnetic force expression through cross product, energy check and orbit model.

1Separate magnetic contribution from electric force; the upper equation contains only the magnetic contribution.
2Specify velocity and field in the same frame; find their mutual angle.
3The cross product gives magnitude $|F_B|=|q|vB\sin\alpha$ .
4Determine direction first for positive charge, then reverse for negative charge.
5The scalar product $\mathbf F_B\cdot\mathbf v=0$ gives zero magnetic work.
6Uniform steady field with no other forces gives a circle for transverse motion, or a helix with additional parallel motion.

Rearrangements

Magnetic field from force

B=\frac{|F_B|}{|q|v\sin\alpha}

Use magnetic force magnitude only. Absolute charge and transverse velocity must be known and nonzero; determine direction separately.

Task variant

An electron moves parallel to the magnetic field. What is the Lorentz force?

$\alpha=0^\circ,\quad F_{\mathrm{mag}}=|q|vB\sin0^\circ=0$ Parallel motion produces no magnetic force; no additional electric field is specified here.

Common mistakes

Ignoring the charge sign for direction.

The right-hand rule directly applies to positive charges; reverse it for electrons.

Exam context

Often appears in circular-path, mass-spectrometer and motor-principle tasks.

Formula cluster

Electrodynamics

Links moving charge, magnetic field and induction.

F = ke·|q₁q₂|/r²

Coulomb's Law

ℰ = −N dΦB/dt

Faraday’s law of induction

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

Worked example

An electron with rounded $q=-1.6\times10^{-19}\,\mathrm C$ moves at $v=10^6\,\mathrm{m/s}$ perpendicular to $B=0.1\,\mathrm T$ . Magnetic force magnitude is $|F_B|=|q|vB=1.6\times10^{-14}\,\mathrm N$ . Deflection direction additionally requires velocity and field directions.

Applications

Particle deflection and beam guidance, mass spectrometers and cyclotrons; motor conductor forces need the appropriate wire-force model. A compass responds to magnetic-dipole torque.

How do you calculate with Lorentz Force?

Here is how to work through a typical Lorentz Force (F = q(v × B)) task step by step:

1
Task
An electron moves parallel to the magnetic field. What is the Lorentz force?
Solution path
$\alpha=0^\circ,\quad F_{\mathrm{mag}}=|q|vB\sin0^\circ=0$ Parallel motion produces no magnetic force; no additional electric field is specified here.

Understand and apply the equation

A magnetic field deflects a moving charge sideways. Charge magnitude sets the strength and its sign the orientation. We separate magnetic force from the complete Lorentz force, follow a charge entering obliquely and check why purely magnetic deflection supplies no kinetic energy.

Magnetic contribution and complete force

A moving charge can change direction in a magnetic field without speeding up. Magnetic force acts sideways to the instantaneous motion. Its direction also depends on whether the charge is positive or negative. We first separate this magnetic contribution from any electric force. Force directions, an energy check and an orbit calculation then reveal different aspects of the same interaction.[1]

\begin{aligned}\mathbf F_B&=q\,\mathbf v\times\mathbf B\\{}\mathbf F_{\mathrm{em}}&=q(\mathbf E+\mathbf v\times\mathbf B)\end{aligned}

The upper formula describes only the magnetic contribution. All fields and velocity belong to the same reference frame.[1]

A stationary charge experiences no magnetic force in this frame but can experience electric force. An electric field must also be included separately for moving charges. We consider local force on a small test charge without including its own influence on the field sources. Gravity or collisions are additional forces, not terms within the electromagnetic force expression.[1]

How do you find the Lorentz force direction for an electron?

\begin{aligned}|\mathbf F_B|&=|q|vB\sin\alpha=|q|Bv_\perp\\{}\mathbf v\times\mathbf B&=(v_yB_z-v_zB_y,\ v_zB_x-v_xB_z,\ v_xB_y-v_yB_x)\end{aligned}

The angle is between velocity and field; the magnitude equation uses absolute charge.[1]

First find the cross-product direction for a positive charge, then reverse it for negative charge. In right-handed axes, for example, motion along +x and field along +z produce a cross product along −y. An electron is therefore deflected along +y. A finger mnemonic is useful only with explicit assignments; components provide the same checkable direction. Parallel or antiparallel motion makes the magnetic contribution vanish.[1]

Why does the magnetic Lorentz force do no work?

\begin{aligned}P_B&=\mathbf F_B\cdot\mathbf v=0\\{}\frac{dK}{dt}&=q\mathbf E\cdot\mathbf v\end{aligned}

The cross product has zero scalar product with velocity. This follows from the force expression; it does not derive the empirical force law.[2]

With no other forces, kinetic energy stays constant while the velocity vector rotates. Acceleration is therefore possible at constant speed. This does not forbid energy transfer from a changing magnetic field through an induced electric field: the electric contribution then does work. Radiation losses of accelerated charges are neglected in the simple orbit model and require reconsideration at relevant energies or observation times.[2]

Circle or helix?

\begin{aligned}|q|v_\perp B&=\frac{mv_\perp^2}{r}\\{}r&=\frac{mv_\perp}{|q|B}\\{}T&=\frac{2\pi m}{|q|B}\\{}h&=v_\parallel T\end{aligned}

These nonrelativistic relations apply to nonzero transverse motion in a uniform, time-independent field with no electric or other forces.[2]

In a uniform, time-independent magnetic field with no other forces, the perpendicular velocity component produces circular motion; the magnetic force supplies the centripetal force. The parallel component stays constant while the transverse component revolves. Together they form a helix with pitch h per revolution. Purely perpendicular entry gives zero pitch and a circle. Purely parallel motion has no transverse revolution; the radius expression must not be interpreted as a circle being traversed. For relativistic particles, transverse momentum replaces mass times transverse speed.[2]

Positive charge: downward quarter circle

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Original orbit for the third example, with equally scaled spatial axes. The positive charge travels from the origin to the lower-right endpoint; field points out of the xy-plane. This is a trajectory rather than field lines.[2]

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Tesla and a non-unique field measurement

\begin{aligned}1\,\mathrm T&=1\,\frac{\mathrm N\,\mathrm s}{\mathrm C\,\mathrm m}=1\,\frac{\mathrm N}{\mathrm A\,\mathrm m}\\{}B&=\frac{|F_B|}{|q|v\sin\alpha}\end{aligned}

Rearranging requires known nonzero charge and transverse speed.[1]

A single force measurement does not determine every field component: the field component parallel to velocity contributes nothing to the cross product. Zero force therefore does not prove zero field. Convert micro-, milli- and nano-values to SI before substitution. Elementary charge is exactly defined, but mass or speed supplied in an example may still be rounded. Results should respect input precision.[1][3]

Two charges with defined axes

Electron entering obliquely

Given: Electron with $q=-1.602176634\times10^{-19}\,\mathrm C$ , $\mathbf v=(3.00\times10^5,0,4.00\times10^5)\,\mathrm{m/s}$ and $\mathbf B=(0,0,0.200)\,\mathrm T$ . Other forces vanish.
Find: Calculate initial magnetic force and its power.

Approach: evaluate the cross product in the specified axes, then multiply by signed charge.
$\begin{aligned}F_y&=-qv_xB_z\\{}P_B&=F_yv_y\end{aligned}$
The z component parallel to the field supplies no force contribution.
$\begin{aligned}\mathbf v\times\mathbf B&=(0,-6.00\cdot10^4,0)\,\mathrm{m\,T\,s^{-1}}\\{}F_y&=-( -1.602176634\cdot10^{-19})(3.00\cdot10^5)(0.200)\,\mathrm N\\{}&=9.613059804\cdot10^{-15}\,\mathrm N\end{aligned}$

Result: Force points along +y with magnitude about 9.61·10⁻¹⁵ N; magnetic power is zero.

Check: Total speed is 5.00·10⁵ m/s but only 3.00·10⁵ m/s is transverse. Negative charge reverses direction.

A model helix

Given: Positive model charge $q=1.60\times10^{-19}\,\mathrm C$ , $m=1.67\times10^{-27}\,\mathrm{kg}$ , $B=0.250\,\mathrm T$ , $v_\perp=2.00\times10^5\,\mathrm{m/s}$ , $v_\parallel=1.50\times10^5\,\mathrm{m/s}$ . Uniform steady field without other forces; nonrelativistic.
Find: Find radius, period and pitch.

Approach: transverse magnetic force provides radial acceleration; parallel motion covers one pitch during a revolution.
$\begin{aligned}r&=\frac{mv_\perp}{|q|B}\\{}T&=\frac{2\pi m}{|q|B}\\{}h&=v_\parallel T\end{aligned}$
Substitute the rounded model values separately.
$\begin{aligned}r&=\frac{(1.67\cdot10^{-27})(2.00\cdot10^5)}{(1.60\cdot10^{-19})(0.250)}\,\mathrm m=0.00835\,\mathrm m\\{}T&=\frac{2\pi(1.67\cdot10^{-27})}{(1.60\cdot10^{-19})(0.250)}\,\mathrm s\approx2.62323\cdot10^{-7}\,\mathrm s\\{}h&=(1.50\cdot10^5)T\approx0.0393484\,\mathrm m\end{aligned}$

Result: $r=8.35\,\mathrm{mm}$ , $T\approx0.262\,\mathrm{\mu s}$ , $h\approx39.3\,\mathrm{mm}$ .

Check: Transverse circumference divided by $v_\perp$ gives the same period. Total speed stays 2.50·10⁵ m/s; pitch uses only $v_\parallel$ .

A quarter-circle orbit in a finite time

Given: A model charge $q=+1.60\times10^{-19}\,\mathrm C$ , $m=1.67\times10^{-27}\,\mathrm{kg}$ starts at the origin with $\mathbf v=(2.00\times10^5,0,0)\,\mathrm{m/s}$ . A uniform constant $B=0.250\,\mathrm T$ points along +z. Other forces vanish; use a nonrelativistic model.
Find: Find position and flight time after one quarter turn.

Approach: positive charge gives initial force along −y. The circle centre therefore lies below the starting point.
$\begin{aligned}r&=\frac{mv}{qB}\\x(t)&=r\sin(\omega_ct)\\y(t)&=r[\cos(\omega_ct)-1]\\\omega_c&=\frac{qB}{m}\end{aligned}$
Substitution: one quarter turn has phase π/2.
$\begin{aligned}r&=8.35\,\mathrm{mm}\\\Delta t&=\frac{\pi m}{2qB}\approx65.5807\,\mathrm{ns}\\(x,y)&=(8.35,-8.35)\,\mathrm{mm}\end{aligned}$

Result: After about 65.6 ns the charge is at (8.35, −8.35) mm; velocity then points along −y.

Check: Check: arc length $\frac{\pi r}{2}$ divided by v gives the same time. Distance from the circle centre stays r and kinetic energy remains constant. These coordinates do not describe straight displacement with the unchanged initial velocity.

Five exercises: magnitude, measurement and diagnosis

Calculate · Foundation

A +4.0 nC charge moves at 200 m/s at 30° to a 0.050 T field. Find force magnitude.

Show hint

The angle is between field and velocity.

Learning goal: Calculate angle-dependent force magnitude.

Show solution and calculation

Approach: $|F_B|=|q|vB\sin\alpha$ .
Use charge magnitude and transverse speed.
The result is 20 nN.

Check: 90° would give 40 nN; 30° gives half.

|F_B|=(4.0\cdot10^{-9})(200)(0.050)\sin30^\circ\,\mathrm N=2.0\cdot10^{-8}\,\mathrm N

[1]

Rearrange · Application

$q=2.0\,\mathrm{nC}$ , $v=80\,\mathrm{m/s}$ , $\alpha=90^\circ$ , measured magnetic force magnitude 48 nN. What is B?

Show hint

Do not use total force including other forces.

Learning goal: Infer field magnitude from known transverse motion.

Show solution and calculation

Approach: $B=\frac{|F_B|}{|q|v\sin\alpha}$ .
Divide magnetic force magnitude by absolute charge and speed.
$B=0.300\,\mathrm T$ .

Check: substitution gives 48 nN; this magnitude alone leaves field direction undetermined.

B=\frac{48\cdot10^{-9}}{(2.0\cdot10^{-9})(80)}\,\mathrm T=0.300\,\mathrm T

[1]

Interpret · Application

In right-handed axes v points along +x and B along +z. A sketch draws both force arrows along −y for equal-magnitude positive and negative charges. Evaluate it.

Show hint

First calculate the cross product without q.

Learning goal: Check force arrows and charge sign.

Show solution and calculation

Approach: $\mathbf F_B=q(\mathbf v\times\mathbf B)$ .
The cross product points along −y; multiply by q afterwards.
Positive: −y, negative: +y.
Magnitudes agree.

Check: both forces are perpendicular to v and B.

\hat{\mathbf x}\times\hat{\mathbf z}=-\hat{\mathbf y}

[1]

Error analysis · Application

A particle with $q\ne0$ and $v\ne0$ has zero magnetic force. A solution concludes $B=0$ . What other possibility exists? Suggest an additional measurement.

Show hint

One field component is invisible for the present motion.

Learning goal: Avoid interpreting zero magnetic force as zero field.

Show solution and calculation

Approach: $\mathbf v\times\mathbf B=0$ .
Examine when the cross product vanishes.
Field may be parallel or antiparallel to v.
A known nonparallel second velocity tests this: a nonzero field aligned with the first motion then gives a force.

Check: do not divide by $\sin0^\circ$ .

\mathbf v\times\mathbf B=0\ \not\Rightarrow\ \mathbf B=0

[1]

Transfer · Challenge

$\mathbf v=(600,0,0)\,\mathrm{m/s}$ , $\mathbf B=(0,0,0.080)\,\mathrm T$ . Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?

Show hint

Total electromagnetic force must vanish.

Learning goal: Balance electric and magnetic forces for straight motion.

Show solution and calculation

Approach: $\mathbf E+\mathbf v\times\mathbf B=0$ .
Set electric field equal to the negative cross product.
E points along +y with 48 V/m.

Check: reversing charge reverses both contributions together, preserving cancellation.

\begin{aligned}\mathbf v\times\mathbf B&=(0,-48,0)\,\mathrm{V\,m^{-1}}\\{}\mathbf E&=-\mathbf v\times\mathbf B=(0,48,0)\,\mathrm{V\,m^{-1}}\end{aligned}

[1]

What an observed trajectory actually shows

Deflection in a mass spectrometer can determine transverse momentum per charge magnitude when field and geometry are known. A unique mass also needs charge state and speed or an independent energy specification. Magnetic lenses and bending magnets shape beams; a compass responds to magnetic-dipole torque and is not fully described by force on a single moving test charge. The magnetic force on a current-carrying conductor instead describes the macroscopic effect on the moving charge carriers of a current.[2][4]

A spatially varying field must be evaluated at the particle position; a simple circle is not automatically valid. For very fast particles, force equals momentum rate rather than simply fixed rest mass times acceleration. Whether magnetic force itself does work remains determined by the scalar product. Scattering, radiation and accompanying electric fields must be stated as additional model ingredients.[2]

What to take away

Determine direction from velocity and field first, then account for charge sign. Only transverse velocity contributes to magnetic force. That force alone changes direction rather than kinetic energy. A circular orbit is a special case with definite modelling conditions.[1]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Separated magnetic and complete force; developed direction, energy and helix models with two original examples and five varied exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.
October 1, 2026Surfaced direction and work questions in headings, clarified orbit assumptions and linked electric field, centripetal force and conductor force in context.

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Quanta. Lorentz Force. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/lorentzkraft

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Frequently asked questions about Lorentz Force

How do you calculate the magnitude of the Lorentz force?+

The magnitude of the magnetic contribution is $|F_B|=|q|vB\sin\alpha$ , where the angle lies between velocity and magnetic field. Insert charge in coulombs, speed in metres per second and flux density in tesla. An electron with $v=10^6\,\mathrm{m/s}$ perpendicular to $B=0.1\,\mathrm T$ experiences about $1.60\cdot10^{-14}\,\mathrm N$ . Perpendicular motion gives maximum magnetic force; parallel motion gives zero. Charge sign determines direction: find the cross product first, then reverse it for negative charge. Include any electric field separately. The complete Lorentz force is $\vec F=q(\vec E+\vec v\times\vec B)$ ; its magnitude does not generally follow from the magnetic shortcut.

Why is there no Lorentz force when the charge moves parallel to the field?+

The magnetic force vanishes for parallel or antiparallel motion because the velocity–field cross product is zero. What matters is transverse velocity, not merely whether the charge moves. An additional electric field can nevertheless exert force, so “no Lorentz force” here means only the magnetic contribution. In a uniform time-independent field with no other forces, purely parallel motion stays straight. Combining parallel and transverse motion instead produces a helix, whereas a circle requires purely transverse entry. A single zero-force measurement therefore does not establish that the magnetic field is zero. A second known nonparallel velocity can test the previously invisible field component.

Why do charges move on circular paths in a magnetic field?+

A circle does not follow from the presence of an arbitrary magnetic field alone. It requires a uniform time-independent field, purely perpendicular entry and no other relevant forces. In the nonrelativistic model, magnetic force then provides radial acceleration: $r=\frac{mv}{|q|B}$ . Speed stays constant because magnetic force is perpendicular to velocity and does no work. Adding a parallel velocity component produces a helix, with only transverse velocity entering the radius. Relativistic particles require transverse momentum instead. Inferring mass from a measured trajectory also needs a known charge state and speed, or an appropriate independent energy specification. Field strength and observed geometry alone do not always identify the particle.

How do you account for the sign of the charge when finding the direction?+

The right-hand rule and the three-finger rule apply directly to positive charges: thumb in the direction of motion, index finger in the field direction, middle finger shows the force. For negative charges such as electrons the force direction reverses, so you flip the result by 180°. Alternatively use the left hand for electrons. The magnitude stays unchanged, only the orientation turns. A typical exam mistake is to apply the right-hand rule to an electron and forget the reversal; then the numerical value is right, but the predicted deflection direction is exactly wrong.

What is the difference between the electric force and the Lorentz force?+

Electric force $\vec F_E=q\vec E$ acts on a charge even when it is stationary in the chosen reference frame. For positive charge it points along the electric field, and for negative charge in the opposite direction. The magnetic contribution $\vec F_B=q\vec v\times\vec B$ requires motion transverse to the magnetic field and is perpendicular to velocity. It does no work itself and, acting alone, does not change speed. Both contributions form the complete Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$ . Use the same frame for fields and velocity. A problem without relevant electric field may calculate only the magnetic part; electric and magnetic contributions need not remain the same when changing frames.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §11.2 Magnetic Fields and Lines

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Lorentz force

The upper equation gives the magnetic contribution to the Lorentz force on a moving charge; an electric field adds electric force.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

F⃗

Magnetic contribution to the Lorentz force (vector)

q

Electric charge

v⃗

Velocity of the charge

m/s

B⃗

Magnetic flux density

Derivation & background

Exam blueprint

Validity range

Derivation steps

Evaluate the empirical magnetic force expression through cross product, energy check and orbit model.

1Separate magnetic contribution from electric force; the upper equation contains only the magnetic contribution.
2Specify velocity and field in the same frame; find their mutual angle.
3The cross product gives magnitude $|F_B|=|q|vB\sin\alpha$ .
4Determine direction first for positive charge, then reverse for negative charge.
5The scalar product $\mathbf F_B\cdot\mathbf v=0$ gives zero magnetic work.
6Uniform steady field with no other forces gives a circle for transverse motion, or a helix with additional parallel motion.

How do you calculate with Lorentz Force?

Here is how to work through a typical Lorentz Force (F = q(v × B)) task step by step:

Task

An electron moves parallel to the magnetic field. What is the Lorentz force?

Solution path

$\alpha=0^\circ,\quad F_{\mathrm{mag}}=|q|vB\sin0^\circ=0$ Parallel motion produces no magnetic force; no additional electric field is specified here.

Magnetic contribution and complete force

\begin{aligned}\mathbf F_B&=q\,\mathbf v\times\mathbf B\\{}\mathbf F_{\mathrm{em}}&=q(\mathbf E+\mathbf v\times\mathbf B)\end{aligned}

The upper formula describes only the magnetic contribution. All fields and velocity belong to the same reference frame.[1]

How do you find the Lorentz force direction for an electron?

\begin{aligned}|\mathbf F_B|&=|q|vB\sin\alpha=|q|Bv_\perp\\{}\mathbf v\times\mathbf B&=(v_yB_z-v_zB_y,\ v_zB_x-v_xB_z,\ v_xB_y-v_yB_x)\end{aligned}

The angle is between velocity and field; the magnitude equation uses absolute charge.[1]

Why does the magnetic Lorentz force do no work?

\begin{aligned}P_B&=\mathbf F_B\cdot\mathbf v=0\\{}\frac{dK}{dt}&=q\mathbf E\cdot\mathbf v\end{aligned}

The cross product has zero scalar product with velocity. This follows from the force expression; it does not derive the empirical force law.[2]

Circle or helix?

\begin{aligned}|q|v_\perp B&=\frac{mv_\perp^2}{r}\\{}r&=\frac{mv_\perp}{|q|B}\\{}T&=\frac{2\pi m}{|q|B}\\{}h&=v_\parallel T\end{aligned}

These nonrelativistic relations apply to nonzero transverse motion in a uniform, time-independent field with no electric or other forces.[2]

Positive charge: downward quarter circle

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Tesla and a non-unique field measurement

\begin{aligned}1\,\mathrm T&=1\,\frac{\mathrm N\,\mathrm s}{\mathrm C\,\mathrm m}=1\,\frac{\mathrm N}{\mathrm A\,\mathrm m}\\{}B&=\frac{|F_B|}{|q|v\sin\alpha}\end{aligned}

Rearranging requires known nonzero charge and transverse speed.[1]

Two charges with defined axes

Electron entering obliquely

Given: Electron with $q=-1.602176634\times10^{-19}\,\mathrm C$ , $\mathbf v=(3.00\times10^5,0,4.00\times10^5)\,\mathrm{m/s}$ and $\mathbf B=(0,0,0.200)\,\mathrm T$ . Other forces vanish.
Find: Calculate initial magnetic force and its power.

Approach: evaluate the cross product in the specified axes, then multiply by signed charge.
$\begin{aligned}F_y&=-qv_xB_z\\{}P_B&=F_yv_y\end{aligned}$
The z component parallel to the field supplies no force contribution.
$\begin{aligned}\mathbf v\times\mathbf B&=(0,-6.00\cdot10^4,0)\,\mathrm{m\,T\,s^{-1}}\\{}F_y&=-( -1.602176634\cdot10^{-19})(3.00\cdot10^5)(0.200)\,\mathrm N\\{}&=9.613059804\cdot10^{-15}\,\mathrm N\end{aligned}$

Result: Force points along +y with magnitude about 9.61·10⁻¹⁵ N; magnetic power is zero.

Check: Total speed is 5.00·10⁵ m/s but only 3.00·10⁵ m/s is transverse. Negative charge reverses direction.

A model helix

Given: Positive model charge $q=1.60\times10^{-19}\,\mathrm C$ , $m=1.67\times10^{-27}\,\mathrm{kg}$ , $B=0.250\,\mathrm T$ , $v_\perp=2.00\times10^5\,\mathrm{m/s}$ , $v_\parallel=1.50\times10^5\,\mathrm{m/s}$ . Uniform steady field without other forces; nonrelativistic.
Find: Find radius, period and pitch.

Approach: transverse magnetic force provides radial acceleration; parallel motion covers one pitch during a revolution.
$\begin{aligned}r&=\frac{mv_\perp}{|q|B}\\{}T&=\frac{2\pi m}{|q|B}\\{}h&=v_\parallel T\end{aligned}$
Substitute the rounded model values separately.
$\begin{aligned}r&=\frac{(1.67\cdot10^{-27})(2.00\cdot10^5)}{(1.60\cdot10^{-19})(0.250)}\,\mathrm m=0.00835\,\mathrm m\\{}T&=\frac{2\pi(1.67\cdot10^{-27})}{(1.60\cdot10^{-19})(0.250)}\,\mathrm s\approx2.62323\cdot10^{-7}\,\mathrm s\\{}h&=(1.50\cdot10^5)T\approx0.0393484\,\mathrm m\end{aligned}$

Result: $r=8.35\,\mathrm{mm}$ , $T\approx0.262\,\mathrm{\mu s}$ , $h\approx39.3\,\mathrm{mm}$ .

Check: Transverse circumference divided by $v_\perp$ gives the same period. Total speed stays 2.50·10⁵ m/s; pitch uses only $v_\parallel$ .

A quarter-circle orbit in a finite time

Given: A model charge $q=+1.60\times10^{-19}\,\mathrm C$ , $m=1.67\times10^{-27}\,\mathrm{kg}$ starts at the origin with $\mathbf v=(2.00\times10^5,0,0)\,\mathrm{m/s}$ . A uniform constant $B=0.250\,\mathrm T$ points along +z. Other forces vanish; use a nonrelativistic model.
Find: Find position and flight time after one quarter turn.

Approach: positive charge gives initial force along −y. The circle centre therefore lies below the starting point.
$\begin{aligned}r&=\frac{mv}{qB}\\x(t)&=r\sin(\omega_ct)\\y(t)&=r[\cos(\omega_ct)-1]\\\omega_c&=\frac{qB}{m}\end{aligned}$
Substitution: one quarter turn has phase π/2.
$\begin{aligned}r&=8.35\,\mathrm{mm}\\\Delta t&=\frac{\pi m}{2qB}\approx65.5807\,\mathrm{ns}\\(x,y)&=(8.35,-8.35)\,\mathrm{mm}\end{aligned}$

Result: After about 65.6 ns the charge is at (8.35, −8.35) mm; velocity then points along −y.

Five exercises: magnitude, measurement and diagnosis

Calculate · Foundation

A +4.0 nC charge moves at 200 m/s at 30° to a 0.050 T field. Find force magnitude.

Show hint

The angle is between field and velocity.

Learning goal: Calculate angle-dependent force magnitude.

Show solution and calculation

Approach: $|F_B|=|q|vB\sin\alpha$ .
Use charge magnitude and transverse speed.
The result is 20 nN.

Check: 90° would give 40 nN; 30° gives half.

|F_B|=(4.0\cdot10^{-9})(200)(0.050)\sin30^\circ\,\mathrm N=2.0\cdot10^{-8}\,\mathrm N

[1]

Rearrange · Application

$q=2.0\,\mathrm{nC}$ , $v=80\,\mathrm{m/s}$ , $\alpha=90^\circ$ , measured magnetic force magnitude 48 nN. What is B?

Show hint

Do not use total force including other forces.

Learning goal: Infer field magnitude from known transverse motion.

Show solution and calculation

Approach: $B=\frac{|F_B|}{|q|v\sin\alpha}$ .
Divide magnetic force magnitude by absolute charge and speed.
$B=0.300\,\mathrm T$ .

Check: substitution gives 48 nN; this magnitude alone leaves field direction undetermined.

B=\frac{48\cdot10^{-9}}{(2.0\cdot10^{-9})(80)}\,\mathrm T=0.300\,\mathrm T

[1]

Interpret · Application

In right-handed axes v points along +x and B along +z. A sketch draws both force arrows along −y for equal-magnitude positive and negative charges. Evaluate it.

Show hint

First calculate the cross product without q.

Learning goal: Check force arrows and charge sign.

Show solution and calculation

Approach: $\mathbf F_B=q(\mathbf v\times\mathbf B)$ .
The cross product points along −y; multiply by q afterwards.
Positive: −y, negative: +y.
Magnitudes agree.

Check: both forces are perpendicular to v and B.

\hat{\mathbf x}\times\hat{\mathbf z}=-\hat{\mathbf y}

[1]

Error analysis · Application

A particle with $q\ne0$ and $v\ne0$ has zero magnetic force. A solution concludes $B=0$ . What other possibility exists? Suggest an additional measurement.

Show hint

One field component is invisible for the present motion.

Learning goal: Avoid interpreting zero magnetic force as zero field.

Show solution and calculation

Approach: $\mathbf v\times\mathbf B=0$ .
Examine when the cross product vanishes.
Field may be parallel or antiparallel to v.
A known nonparallel second velocity tests this: a nonzero field aligned with the first motion then gives a force.

Check: do not divide by $\sin0^\circ$ .

\mathbf v\times\mathbf B=0\ \not\Rightarrow\ \mathbf B=0

[1]

Transfer · Challenge

$\mathbf v=(600,0,0)\,\mathrm{m/s}$ , $\mathbf B=(0,0,0.080)\,\mathrm T$ . Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?

Show hint

Total electromagnetic force must vanish.

Learning goal: Balance electric and magnetic forces for straight motion.

Show solution and calculation

Approach: $\mathbf E+\mathbf v\times\mathbf B=0$ .
Set electric field equal to the negative cross product.
E points along +y with 48 V/m.

Check: reversing charge reverses both contributions together, preserving cancellation.

\begin{aligned}\mathbf v\times\mathbf B&=(0,-48,0)\,\mathrm{V\,m^{-1}}\\{}\mathbf E&=-\mathbf v\times\mathbf B=(0,48,0)\,\mathrm{V\,m^{-1}}\end{aligned}

[1]

What an observed trajectory actually shows

What to take away

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Separated magnetic and complete force; developed direction, energy and helix models with two original examples and five varied exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.
October 1, 2026Surfaced direction and work questions in headings, clarified orbit assumptions and linked electric field, centripetal force and conductor force in context.

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Quanta. Lorentz Force. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/lorentzkraft

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Frequently asked questions about Lorentz Force

How do you calculate the magnitude of the Lorentz force?+

Why is there no Lorentz force when the charge moves parallel to the field?+

Why do charges move on circular paths in a magnetic field?+

How do you account for the sign of the charge when finding the direction?+

What is the difference between the electric force and the Lorentz force?+

Lorentz force

Variables & units

Lorentz force calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrodynamics

Worked example

Applications

How do you calculate with Lorentz Force?

Magnetic contribution and complete force

How do you find the Lorentz force direction for an electron?

Why does the magnetic Lorentz force do no work?

Circle or helix?

Positive charge: downward quarter circle

Tesla and a non-unique field measurement

Two charges with defined axes

Electron entering obliquely

A model helix

A quarter-circle orbit in a finite time

Five exercises: magnitude, measurement and diagnosis

A +4.0 nC charge moves at 200 m/s at 30° to a 0.050 T field. Find force magnitude.

q=2.0 nCq=2.0\,\mathrm{nC}q=2.0nC, v=80 m/sv=80\,\mathrm{m/s}v=80m/s, α=90∘\alpha=90^\circα=90∘, measured magnetic force magnitude 48 nN. What is B?

In right-handed axes v points along +x and B along +z. A sketch draws both force arrows along −y for equal-magnitude positive and negative charges. Evaluate it.

A particle with q≠0q\ne0q=0 and v≠0v\ne0v=0 has zero magnetic force. A solution concludes B=0B=0B=0. What other possibility exists? Suggest an additional measurement.

v=(600,0,0) m/s\mathbf v=(600,0,0)\,\mathrm{m/s}v=(600,0,0)m/s, B=(0,0,0.080) T\mathbf B=(0,0,0.080)\,\mathrm TB=(0,0,0.080)T. Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?

What an observed trajectory actually shows

What to take away

Sources for this explanation

Frequently asked questions about Lorentz Force

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Lorentz Force for exams

Lorentz force

Variables & units

Lorentz force calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrodynamics

Worked example

Applications

How do you calculate with Lorentz Force?

Magnetic contribution and complete force

How do you find the Lorentz force direction for an electron?

Why does the magnetic Lorentz force do no work?

Circle or helix?

Positive charge: downward quarter circle

Tesla and a non-unique field measurement

Two charges with defined axes

Electron entering obliquely

A model helix

A quarter-circle orbit in a finite time

Five exercises: magnitude, measurement and diagnosis

A +4.0 nC charge moves at 200 m/s at 30° to a 0.050 T field. Find force magnitude.

q=2.0 nCq=2.0\,\mathrm{nC}q=2.0nC, v=80 m/sv=80\,\mathrm{m/s}v=80m/s, α=90∘\alpha=90^\circα=90∘, measured magnetic force magnitude 48 nN. What is B?

In right-handed axes v points along +x and B along +z. A sketch draws both force arrows along −y for equal-magnitude positive and negative charges. Evaluate it.

A particle with q≠0q\ne0q=0 and v≠0v\ne0v=0 has zero magnetic force. A solution concludes B=0B=0B=0. What other possibility exists? Suggest an additional measurement.

v=(600,0,0) m/s\mathbf v=(600,0,0)\,\mathrm{m/s}v=(600,0,0)m/s, B=(0,0,0.080) T\mathbf B=(0,0,0.080)\,\mathrm TB=(0,0,0.080)T. Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?

What an observed trajectory actually shows

What to take away

Sources for this explanation

Frequently asked questions about Lorentz Force

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Lorentz Force for exams

$q=2.0\,\mathrm{nC}$ , $v=80\,\mathrm{m/s}$ , $\alpha=90^\circ$ , measured magnetic force magnitude 48 nN. What is B?

A particle with $q\ne0$ and $v\ne0$ has zero magnetic force. A solution concludes $B=0$ . What other possibility exists? Suggest an additional measurement.

$\mathbf v=(600,0,0)\,\mathrm{m/s}$ , $\mathbf B=(0,0,0.080)\,\mathrm T$ . Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?

$q=2.0\,\mathrm{nC}$ , $v=80\,\mathrm{m/s}$ , $\alpha=90^\circ$ , measured magnetic force magnitude 48 nN. What is B?

A particle with $q\ne0$ and $v\ne0$ has zero magnetic force. A solution concludes $B=0$ . What other possibility exists? Suggest an additional measurement.

$\mathbf v=(600,0,0)\,\mathrm{m/s}$ , $\mathbf B=(0,0,0.080)\,\mathrm T$ . Which uniform electric field permits straight constant-speed motion of a nonzero charge with no other forces?