Physics · Magnetism

Magnetic force on a current-carrying wire

Magnetic force on a straight current-carrying segment follows the cross product of conventional current direction and external field.

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Formula

Calculate

F = B ∣ I ∣ ℓ sin θ

F in N; B in T; I in A; ℓ in m; θ is an angle.

Variables & units

Variables & units
Symbol	Meaning	Unit
$F$	Magnitude of magnetic force	N
$B$	Magnitude of external magnetic flux density at the considered segment	T
$I$	Current	A
$ℓ$	Wire length in the field	m
$θ$	Angle between current and field	rad or °

LaTeX

F=B|I|\ell\sin\theta

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Bent conductor: current first travels 0.120 metres right, then 0.160 metres up. Field points out of plane. First force is 0.0900 newtons down, second 0.120 newtons right, resultant 0.150 newtons diagonally down-right. — Original example with I = 3.00 A and B = 0.250 T along +z. Wire geometry: 750 px/m; force arrows: 500 px/N. Forces sum to (+0.120, −0.0900) N. Adding wire lengths would incorrectly give 0.210 N. OpenStax §11.4.

Force on a current-carrying wire calculator

Find the force, current, wire length or flux density for a straight wire section in a uniform magnetic field.

Calculation model

Find

Current I

Current in the wire; its sign only changes the direction of the force, not its size, e.g. 4 A.

Straight segment length in the field ℓ

Length of the wire section that actually lies in the magnetic field, e.g. 5 cm.

Magnetic flux density B

Magnetic flux density at the wire, e.g. 0.1 T between the poles of a horseshoe magnet.

Angle between current and field θ

Angle between current direction and field: 90° if the wire lies across the field; parallel gives no force.

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

Result: Force on the wire |F|

|F| = 20 mN

With I = 4 A, ℓ = 5 cm, B = 0.1 T and θ = 90 °, |F| = 20 mN.

Working · rearrange for |F|, substitute, evaluate

1. Rearrange for |F|
$|F|=|I|\ell B\sin\theta$
2. Substitute values (in SI units)
- ℓ = 5 cm = 0.05 m
- θ = 90 ° = 1.5708 rad
$4\cdot 0.05\cdot 0.1\cdot\sin\left(1.57079632679\right)=0.02$
3. Result
|F| = 20 mN

Quick tips

Unit check: 1 T · 1 A · 1 m = 1 N. Convert the length to metres and mT to T first, otherwise the answer is off by a factor of 100 or 1000.
Only the part across the field counts: at 30° the force is half that at 90°, and a wire parallel to the field feels no force at all.

Assumptions and limits

Specified straight segment in a uniform external field. Length and field magnitude nonnegative. θ∈[0,π] is between the fixed path reference and B; I is signed against that reference. Result uses |I| and gives no force direction. Not self-field, total bent-wire length or total mechanical force. Inverse I gives magnitude only.

Derivation & background

For uniform field, $\vec F_B=I\vec\ell\times\vec B_{\mathrm{ext}}$ . Generally use $I\int\mathrm d\vec\ell\times\vec B$ . In uniform field, resultant force on a bent path depends on endpoint displacement rather than total wire length. A closed path has zero net force there but can have torque.

Exam blueprint

Validity range

Straight thin current path in uniform external magnetic field; specify current and active field length. Integrate for curved paths or spatially varying field. Distinguish magnitude, vector and total mechanical force.

Derivation steps

Sum magnetic Lorentz forces on drifting charges using conventional current, then integrate along the current path.

1Identify the conductor segment and its external magnetic field.
2Orient positive path element along conventional current.
3Local carrier force is $q\vec v_{\mathrm d}\times\vec B$ ; the short segment contains $nS\,\mathrm d\ell$ carriers.
4Use $\vec J=nq\vec v_{\mathrm d}$ to obtain $\mathrm d\vec F=I\,\mathrm d\vec\ell\times\vec B$ .
5For a straight segment, constant current and uniform field, this becomes the cross product using full active length.
6Magnitude is $F=|I|\ell B\sin\theta$ ; the cross product fixes direction.
7Check the parallel limit and separately consider weight, supports and return wires.

Rearrangements

Current magnitude

|I|=\frac{F}{B\ell\sin\theta}

B, active length and sine must be nonzero; determine direction separately.

Task variant

A straight 0.10 m segment carries 2.0 A perpendicular to uniform external field 0.50 T. Find magnetic force magnitude.

Approach: $F=B|I|\ell\sin\theta$ .
For $\theta=90^\circ$ , $F=0.10\,\mathrm N$ .

Common mistakes

Calculate force using the wire’s whole self-field instead of the external magnetic field.

Use the external field and oriented current direction; integrate contributions for a curved conductor.

Exam context

Oblique and curved conductors in an external field: check cross product, effective length and current sign.

Formula cluster

Magnetism

Wire force combines Lorentz forces and explains the torque on a current loop.

F = q(v × B)

Lorentz Force

B = μ₀|I|/(2πr)

Magnetic field of a long straight wire

ℰ(t) = NBAω sin(ωt)

Induced emf of a rotating generator

Worked example

For $B=0.50\,\mathrm T$ , $I=2.0\,\mathrm A$ , $\ell=0.10\,\mathrm m$ and $\theta=90^\circ$ , $F=B|I|\ell\sin\theta=0.10\,\mathrm N$ . The cross product additionally gives direction.

Applications

Explain electric motors, force measurements and motion of current-carrying conductors.

How do you calculate with Magnetic force on a current-carrying wire?

Here is how to work through a typical Magnetic force on a current-carrying wire (F = B|I|ℓ sinθ) task step by step:

1
Task
A straight 0.10 m segment carries 2.0 A perpendicular to uniform external field 0.50 T. Find magnetic force magnitude.
Solution path
1. Approach: $F=B|I|\ell\sin\theta$ .
2. For $\theta=90^\circ$ , $F=0.10\,\mathrm N$ .

Understand and apply the equation

A current-carrying conductor transfers magnetic action on its moving charges to the conductor segment. Direction follows conventional current; magnitude depends on length transverse to the field. A bent path and a closed circuit show why wire lengths cannot always simply be added.

When does the force formula apply to a straight current-carrying wire?

Imagine a straight wire between magnetic poles. Once current flows, the magnet can push it sideways. Current and field directions together determine the force direction. The wire needs moving charges and a field component transverse to their motion. We first calculate force on one segment. Whether the whole circuit moves also depends on its return wire and supports.[1]

\begin{aligned}\vec F_B&=I\vec\ell\times\vec B_{\mathrm{ext}}\\|\vec F_B|&=|I|\ell B\sin\theta\end{aligned}

Straight segment in uniform external field. Here $I\ge0$ is current magnitude and $\vec\ell$ follows actual conventional current. A signed current instead requires a fixed reference direction for $\vec\ell$ .[1]

Magnetic force is not automatically total mechanical force: weight, supports and electric forces can also act. Use the external field, not a wire’s own symmetric field as its driving force. In a real conductor, charge carriers and material transfer their interaction to the component. Its subsequent motion is determined by mechanical balance. A magnitude expression alone does not tell whether the wire is pushed upwards, downwards or sideways.[1]

From charge flow to a wire integral

\begin{aligned}\vec J&=nq\vec v_{\mathrm d}\\\mathrm d\vec F_B&=(nS\,\mathrm d\ell)q\vec v_{\mathrm d}\times\vec B\\&=I\,\mathrm d\vec\ell\times\vec B\\\vec F_B&=I\int_C\mathrm d\vec\ell\times\vec B\end{aligned}

Assume a thin current path and steady current; S is wire cross-section, not loop area.[1]

Carrier count in a short volume equals density times cross-section times length. Sum individual magnetic Lorentz forces and use the relation between drift and conventional current. Negative carriers drift opposite to positive current direction; their negative charge compensates that reversal. Do not combine electron direction and a positive current magnitude without accounting for the sign. Contributions of multiple carrier species add through current density. This reasoning uses the Lorentz law as its starting point rather than independently proving that experimental law.[1]

Two directions determine the third

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

Current along +y in a field along +x gives force along −z.[1]

Force is perpendicular to current direction and field. Reversing current reverses force, as does reversing field; reversing both preserves force. Parallel or antiparallel current path gives zero force even when neither current nor field vanishes. A field component along current contributes nothing to the cross product. In a planar drawing, specify whether a dot points out of the plane or a cross into it. A force arrow without that assignment cannot be checked.[1]

How do you calculate bent current paths and varying magnetic fields?

\begin{aligned}\vec F_B&=I\left(\int_C\mathrm d\vec\ell\right)\times\vec B\\&=I(\vec r_b-\vec r_a)\times\vec B\\\vec F_{\mathrm{closed}}&=0\end{aligned}

With constant current and spatially uniform external field, total force depends on endpoint displacement; a closed path gives zero net force.[1][2]

Local force contribution along the wire

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Original curve for the varying external field. This is magnitude per length; all contributions act along −y. Area under the curve gives force magnitude.[1]

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Zero net force does not mean zero local force or zero torque. Opposite forces can rotate or deform a closed rectangular frame. In a nonuniform field, $\mathbf B$ cannot be factored out of the integral, and even a closed current path can experience net force. Measuring a supported segment does not automatically measure the entire circuit including its return path. State which segment is mechanically free and which return paths lie outside the field or are separately supported. Torque on a current-carrying loop describes this turning effect in a uniform external field, referenced to its current-oriented area normal.[1][2]

Straight wire in a varying field

Given: A straight segment from $x=0$ to $L=0.250\,\mathrm m$ carries $I=2.00\,\mathrm A$ along +x. The prescribed external field is $\vec B(x)=(0.200+0.800\frac{x}{\mathrm m})\,\mathrm T\,\hat{\vec z}$ . The return wire is separately supported.
Find: Find force on the segment and identify the field value that permits a shortcut.

Approach: the cross product points along −y everywhere; integrate magnitude along the wire.
$F_y=-I\int_0^L(B_0+bx)\,\mathrm dx=-I\left(B_0L+\frac{bL^2}{2}\right)$
Substitution: field magnitude rises linearly from 0.200 T to 0.400 T.
$\begin{aligned}F_y&=-(2.00)\left[(0.200)(0.250)+\frac{(0.800)(0.250)^2}{2}\right]\,\mathrm N\\&=-0.150\,\mathrm N\end{aligned}$

Result: Force is 0.150 N along −y.

Check: Check: the length-averaged field magnitude is 0.300 T and $IL\overline B=0.150\,\mathrm N$ . Using only the initial or final field would give 0.100 N or 0.200 N. This average works because direction and current are fixed, not for arbitrary varying field vectors.

Example: one field component contributes nothing

Straight segment in an oblique field

Given: Straight segment $\ell=0.120\,\mathrm m$ along +y, $I=2.40\,\mathrm A$ ; uniform external field $\vec B=(0.300\hat{\vec x}+0.400\hat{\vec y})\,\mathrm T$ .
Find: Find force vector and verify magnitude using the angle.

Approach: only the x field component is transverse to current.
$\vec F=I\ell\hat{\vec y}\times(B_x\hat{\vec x}+B_y\hat{\vec y})=-I\ell B_x\hat{\vec z}$
Substitution: the parallel field component has zero cross product.
$\begin{aligned}F_z&=-(2.40)(0.120)(0.300)\,\mathrm N=-0.0864\,\mathrm N\\B&=\sqrt{0.300^2+0.400^2}\,\mathrm T=0.500\,\mathrm T\\\sin\theta&=\frac{0.300}{0.500}=0.600\end{aligned}$

Result: Force is 86.4 mN along −z.

Check: Using magnitude 0.500 T and sine 0.600 gives the same 86.4 mN. Omitting the angle correction would incorrectly give 144 mN.

Example: right-angle current path

Two segments, one resultant force

Given: $I=3.00\,\mathrm A$ first follows 0.120 m along +x, then 0.160 m along +y. Both segments lie in uniform $B=0.250\,\mathrm T$ along +z. Return wire is separately supported.
Find: Calculate resultant force and compare with total wire length.

Approach: the two cross products point along −y and +x. Add their vectors.
$\begin{aligned}\vec F_1&=-I\ell_1B\hat{\vec y}\\\vec F_2&=I\ell_2B\hat{\vec x}\\|\vec F|&=IB\sqrt{\ell_1^2+\ell_2^2}\end{aligned}$
Substitution: endpoint distance is 0.200 m, whereas wire length is 0.280 m.
$\begin{aligned}\vec F&=(0.120\hat{\vec x}-0.0900\hat{\vec y})\,\mathrm N\\|\vec F|&=\sqrt{0.120^2+0.0900^2}\,\mathrm N=0.150\,\mathrm N\end{aligned}$

Result: Resultant force has magnitude 0.150 N and components +0.120 N, −0.0900 N.

Check: Total length times IB would give 0.210 N, incorrectly adding magnitudes. The endpoint expression using displacement gives exactly the same components.

Units and practical force measurements

\begin{aligned}|I|&=\frac{F}{B\ell\sin\theta}\\B&=\frac{F}{|I|\ell\sin\theta}\\\mathrm T\,\mathrm A\,\mathrm m&=\mathrm N\end{aligned}

Denominators must be nonzero; a zero-force test along B cannot determine field magnitude.[1]

Force balances and loudspeakers place an active wire segment in an external field. A motor additionally involves torque, back emf and supply. A moving conductor can convert electrical power into mechanical work even though magnetic force on an individual charge is perpendicular to that charge’s instantaneous velocity. This requires the complete interaction of source, conductor and field. Contacts, heating and mechanical constraints additionally affect real measurements; an ideal force expression alone is not a complete motor model.[1][2]

Five exercises: balance and current paths

Calculate · Application

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. $B=0.500\,\mathrm T$ along −z; $g=9.81\,\mathrm{m/s^2}$ along −y. Leads are separately supported. What current balances it without other vertical forces?

Show hint

First choose current direction for upward +y force.

Learning goal: Balance weight with directed conductor force.

Show solution and calculation

Approach: $\sum F_y=I\ell B-mg=0$ .
Current along +x produces required +y force.
Set magnetic force equal to mg.
Result 0.981 A.

Check: both magnetic force and weight equal 0.11772 N.

|I|=\frac{mg}{B\ell}=\frac{(0.0120)(9.81)}{(0.500)(0.240)}\,\mathrm A=0.981\,\mathrm A

[1]

Rearrange · Foundation

Straight segment: $I=1.60\,\mathrm A$ , $\ell=0.100\,\mathrm m$ , angle 30° to field, magnetic force magnitude 48.0 mN. Find B.

Show hint

Divide by sine of the angle, not by the angle itself.

Learning goal: Infer oblique field magnitude from force data.

Show solution and calculation

Approach: $B=\frac{F}{|I|\ell\sin\theta}$ .
Isolate B using the known geometric factor.
Result 0.600 T.

Check: perpendicular alignment would double force at the same field.

B=\frac{0.0480}{(1.60)(0.100)\sin30^\circ}\,\mathrm T=0.600\,\mathrm T

[1]

Interpret · Application

A segment carries current along +y while field points along +x. State force direction initially, after current reversal, and after reversing both current and field.

Show hint

The cross product is linear in both factors.

Learning goal: Distinguish current reversal from field reversal.

Show solution and calculation

Approach: $\vec F=I\vec\ell\times\vec B$ .
Initially force points along −z.
Reversing only current gives +z; reversing both gives −z again.

Check: equal input magnitudes give equal force magnitudes.

(-\vec\ell)\times(-\vec B)=\vec\ell\times\vec B

[1]

Error analysis · Application

Only 0.0800 m of a straight 0.300 m cable lies in uniform 0.400 T field perpendicular to its 2.00 A current. External field is zero elsewhere. A calculation uses full cable length. Correct it.

Show hint

Split the force integral into regions with and without field.

Learning goal: Use active field length rather than cable length.

Show solution and calculation

Approach: $F=I\int B_\perp\,\mathrm d\ell$ .
Only the segment within the field contributes.
Correct force is 0.0640 N rather than 0.240 N.

Check: extending cable outside the field cannot change magnetic force here.

F=IB\ell_{\mathrm{active}}=(2.00)(0.400)(0.0800)\,\mathrm N=0.0640\,\mathrm N

[1]

Transfer · Challenge

Two long parallel thin wires in air carry 6.00 A and 4.00 A in the same direction, axis separation 25.0 mm. Find magnetic interaction on a central 0.400 m segment far from the ends.

Show hint

The first wire’s field is external field for the second.

Learning goal: Combine generated field with force on a second wire.

Show solution and calculation

Approach: $F=\frac{\mu_0I_1I_2\ell}{2\pi d}$ .
Calculate wire field and insert it into perpendicular conductor force.
Magnitude 76.8 µN, attractive.

Check: force on the corresponding equal-length segment of the other wire is equal and opposite.

\begin{aligned}B_1&=\frac{\mu_0 I_1}{2\pi d}\\F_2&=I_2\ell B_1=\frac{\mu_0 I_1I_2\ell}{2\pi d}\approx76.8\,\mathrm{\mu N}\end{aligned}

[4][3][1]

What to take away

Specify the segment, use conventional current direction and the external field. A straight segment in a uniform field permits the shortcut; a varying field requires summing local contributions. Add forces as vectors and do not infer zero turning effect from zero net force.[1]

Sources for this explanation

Quanta · Updated October 2, 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.

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Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Explained conductor force through drift current, separated direction and path-integral models; developed two original examples and five individual DE/EN exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Magnetic force on a current-carrying wire. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetische-kraft-leiter

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Frequently asked questions about Magnetic force on a current-carrying wire

What does Magnetic force on a current-carrying wire describe physically?+

The relation describes magnetic force on a current-carrying conductor segment in an external field. For a straight segment in uniform field, $\vec F_B=I\vec\ell\times\vec B$ . Length is oriented along positive conventional current. Only the part transverse to field produces force; parallel or antiparallel alignment gives zero. Direction is perpendicular to current and field. Reversing current reverses force, while additionally reversing field restores it. The magnitude expression provides no direction and includes neither weight nor supports. The same conductor’s own field contribution must not be blindly inserted as an external driving field. State the mechanically considered segment separately from its return circuit.

How can I check a calculation involving Magnetic force on a current-carrying wire?+

First draw conventional current and external field in the same axes. Find the force arrow through their cross product, then calculate magnitude. In the existing example, 0.50 tesla, two amperes and 0.10 metres in perpendicular geometry give $0.10\,\mathrm N$ . At thirty degrees between current path and field it would be half. Also check that the full length used actually lies in the field. For a bent segment, add partial forces vectorially. Adding wire lengths times field magnitude can then be wrong even if every individual force was calculated correctly. Treat return conductors and supports in a separate mechanical balance.

What assumptions and limits apply to Magnetic force on a current-carrying wire?+

The product expression requires a straight active segment, specified current and external field uniform across that segment. For curvature or varying field, use $I\int\mathrm d\vec\ell\times\vec B$ when current is constant along the thin path considered. Uniform field reduces this integral to the endpoint displacement vector. A closed circuit consequently has zero net force but can experience torque or local stresses. That zero-force conclusion does not generally hold in a nonuniform field. Real contact forces, heating and supports additionally influence measured motion. Ideal magnetic force therefore describes one contribution to the balance, not automatically the complete mechanical behaviour.

How do I check units and scale for Magnetic force on a current-carrying wire?+

One tesla equals one newton per ampere and metre. Tesla times amperes times active length in metres therefore gives newtons; sine is dimensionless. Convert milliamperes to amperes and centimetres to metres. In the existing example, maximum force magnitude is $B|I|\ell=0.10\,\mathrm N$ . At a different angle the magnitude cannot exceed this maximum. Correct units determine neither the force arrow nor whether active field length was selected incorrectly. Also check whether a force measurement isolates the magnetic contribution or includes weight and contact forces. Round only after substitution, particularly for small millinewton values and after completing the geometrical projection.

When should I combine Magnetic force on a current-carrying wire with another relation?+

For equilibrium or acceleration, supplement conductor force with the mechanical force balance. A perpendicular segment can, for example, balance its weight with suitable current direction, using $B|I|\ell=mg$ . For two parallel conductors, first calculate one wire’s field, then its force on the other. A loop additionally needs torque balance even when its net force in uniform field is zero. Motion under load also brings induced emf and electrical supply into the description. Each additional relation has its own assumptions. A single magnetic force proves neither continuous motor rotation nor a complete energy balance of the actual circuit and its moving components.

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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, §11.4: Magnetic Force on a Current-Carrying Conductor

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\tau=N|I|AB\sin\theta

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\frac{F}{\ell}=\frac{\mu_0 I_1I_2}{2\pi d}

Force Between Parallel Currents

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\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

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\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

U = R \cdot I

Ohm's Law

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Magnetic force on a current-carrying wire

Magnetic force on a straight current-carrying segment follows the cross product of conventional current direction and external field.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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

Symbol

Meaning

Unit

F

Magnitude of magnetic force

B

Magnitude of external magnetic flux density at the considered segment

I

Current

ℓ

Wire length in the field

θ

Angle between current and field

rad or °

Derivation & background

Exam blueprint

Validity range

Derivation steps

Sum magnetic Lorentz forces on drifting charges using conventional current, then integrate along the current path.

1Identify the conductor segment and its external magnetic field.
2Orient positive path element along conventional current.
3Local carrier force is $q\vec v_{\mathrm d}\times\vec B$ ; the short segment contains $nS\,\mathrm d\ell$ carriers.
4Use $\vec J=nq\vec v_{\mathrm d}$ to obtain $\mathrm d\vec F=I\,\mathrm d\vec\ell\times\vec B$ .
5For a straight segment, constant current and uniform field, this becomes the cross product using full active length.
6Magnitude is $F=|I|\ell B\sin\theta$ ; the cross product fixes direction.
7Check the parallel limit and separately consider weight, supports and return wires.

How do you calculate with Magnetic force on a current-carrying wire?

Here is how to work through a typical Magnetic force on a current-carrying wire (F = B|I|ℓ sinθ) task step by step:

Task

A straight 0.10 m segment carries 2.0 A perpendicular to uniform external field 0.50 T. Find magnetic force magnitude.

Solution path

Approach: $F=B|I|\ell\sin\theta$ .
For $\theta=90^\circ$ , $F=0.10\,\mathrm N$ .

When does the force formula apply to a straight current-carrying wire?

\begin{aligned}\vec F_B&=I\vec\ell\times\vec B_{\mathrm{ext}}\\|\vec F_B|&=|I|\ell B\sin\theta\end{aligned}

From charge flow to a wire integral

\begin{aligned}\vec J&=nq\vec v_{\mathrm d}\\\mathrm d\vec F_B&=(nS\,\mathrm d\ell)q\vec v_{\mathrm d}\times\vec B\\&=I\,\mathrm d\vec\ell\times\vec B\\\vec F_B&=I\int_C\mathrm d\vec\ell\times\vec B\end{aligned}

Assume a thin current path and steady current; S is wire cross-section, not loop area.[1]

Two directions determine the third

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

Current along +y in a field along +x gives force along −z.[1]

How do you calculate bent current paths and varying magnetic fields?

\begin{aligned}\vec F_B&=I\left(\int_C\mathrm d\vec\ell\right)\times\vec B\\&=I(\vec r_b-\vec r_a)\times\vec B\\\vec F_{\mathrm{closed}}&=0\end{aligned}

With constant current and spatially uniform external field, total force depends on endpoint displacement; a closed path gives zero net force.[1][2]

Local force contribution along the wire

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Original curve for the varying external field. This is magnitude per length; all contributions act along −y. Area under the curve gives force magnitude.[1]

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Straight wire in a varying field

Given: A straight segment from $x=0$ to $L=0.250\,\mathrm m$ carries $I=2.00\,\mathrm A$ along +x. The prescribed external field is $\vec B(x)=(0.200+0.800\frac{x}{\mathrm m})\,\mathrm T\,\hat{\vec z}$ . The return wire is separately supported.
Find: Find force on the segment and identify the field value that permits a shortcut.

Approach: the cross product points along −y everywhere; integrate magnitude along the wire.
$F_y=-I\int_0^L(B_0+bx)\,\mathrm dx=-I\left(B_0L+\frac{bL^2}{2}\right)$
Substitution: field magnitude rises linearly from 0.200 T to 0.400 T.
$\begin{aligned}F_y&=-(2.00)\left[(0.200)(0.250)+\frac{(0.800)(0.250)^2}{2}\right]\,\mathrm N\\&=-0.150\,\mathrm N\end{aligned}$

Result: Force is 0.150 N along −y.

Example: one field component contributes nothing

Straight segment in an oblique field

Given: Straight segment $\ell=0.120\,\mathrm m$ along +y, $I=2.40\,\mathrm A$ ; uniform external field $\vec B=(0.300\hat{\vec x}+0.400\hat{\vec y})\,\mathrm T$ .
Find: Find force vector and verify magnitude using the angle.

Approach: only the x field component is transverse to current.
$\vec F=I\ell\hat{\vec y}\times(B_x\hat{\vec x}+B_y\hat{\vec y})=-I\ell B_x\hat{\vec z}$
Substitution: the parallel field component has zero cross product.
$\begin{aligned}F_z&=-(2.40)(0.120)(0.300)\,\mathrm N=-0.0864\,\mathrm N\\B&=\sqrt{0.300^2+0.400^2}\,\mathrm T=0.500\,\mathrm T\\\sin\theta&=\frac{0.300}{0.500}=0.600\end{aligned}$

Result: Force is 86.4 mN along −z.

Check: Using magnitude 0.500 T and sine 0.600 gives the same 86.4 mN. Omitting the angle correction would incorrectly give 144 mN.

Example: right-angle current path

Two segments, one resultant force

Given: $I=3.00\,\mathrm A$ first follows 0.120 m along +x, then 0.160 m along +y. Both segments lie in uniform $B=0.250\,\mathrm T$ along +z. Return wire is separately supported.
Find: Calculate resultant force and compare with total wire length.

Approach: the two cross products point along −y and +x. Add their vectors.
$\begin{aligned}\vec F_1&=-I\ell_1B\hat{\vec y}\\\vec F_2&=I\ell_2B\hat{\vec x}\\|\vec F|&=IB\sqrt{\ell_1^2+\ell_2^2}\end{aligned}$
Substitution: endpoint distance is 0.200 m, whereas wire length is 0.280 m.
$\begin{aligned}\vec F&=(0.120\hat{\vec x}-0.0900\hat{\vec y})\,\mathrm N\\|\vec F|&=\sqrt{0.120^2+0.0900^2}\,\mathrm N=0.150\,\mathrm N\end{aligned}$

Result: Resultant force has magnitude 0.150 N and components +0.120 N, −0.0900 N.

Check: Total length times IB would give 0.210 N, incorrectly adding magnitudes. The endpoint expression using displacement gives exactly the same components.

Units and practical force measurements

\begin{aligned}|I|&=\frac{F}{B\ell\sin\theta}\\B&=\frac{F}{|I|\ell\sin\theta}\\\mathrm T\,\mathrm A\,\mathrm m&=\mathrm N\end{aligned}

Denominators must be nonzero; a zero-force test along B cannot determine field magnitude.[1]

Five exercises: balance and current paths

Calculate · Application

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. $B=0.500\,\mathrm T$ along −z; $g=9.81\,\mathrm{m/s^2}$ along −y. Leads are separately supported. What current balances it without other vertical forces?

Show hint

First choose current direction for upward +y force.

Learning goal: Balance weight with directed conductor force.

Show solution and calculation

Approach: $\sum F_y=I\ell B-mg=0$ .
Current along +x produces required +y force.
Set magnetic force equal to mg.
Result 0.981 A.

Check: both magnetic force and weight equal 0.11772 N.

|I|=\frac{mg}{B\ell}=\frac{(0.0120)(9.81)}{(0.500)(0.240)}\,\mathrm A=0.981\,\mathrm A

[1]

Rearrange · Foundation

Straight segment: $I=1.60\,\mathrm A$ , $\ell=0.100\,\mathrm m$ , angle 30° to field, magnetic force magnitude 48.0 mN. Find B.

Show hint

Divide by sine of the angle, not by the angle itself.

Learning goal: Infer oblique field magnitude from force data.

Show solution and calculation

Approach: $B=\frac{F}{|I|\ell\sin\theta}$ .
Isolate B using the known geometric factor.
Result 0.600 T.

Check: perpendicular alignment would double force at the same field.

B=\frac{0.0480}{(1.60)(0.100)\sin30^\circ}\,\mathrm T=0.600\,\mathrm T

[1]

Interpret · Application

A segment carries current along +y while field points along +x. State force direction initially, after current reversal, and after reversing both current and field.

Show hint

The cross product is linear in both factors.

Learning goal: Distinguish current reversal from field reversal.

Show solution and calculation

Approach: $\vec F=I\vec\ell\times\vec B$ .
Initially force points along −z.
Reversing only current gives +z; reversing both gives −z again.

Check: equal input magnitudes give equal force magnitudes.

(-\vec\ell)\times(-\vec B)=\vec\ell\times\vec B

[1]

Error analysis · Application

Only 0.0800 m of a straight 0.300 m cable lies in uniform 0.400 T field perpendicular to its 2.00 A current. External field is zero elsewhere. A calculation uses full cable length. Correct it.

Show hint

Split the force integral into regions with and without field.

Learning goal: Use active field length rather than cable length.

Show solution and calculation

Approach: $F=I\int B_\perp\,\mathrm d\ell$ .
Only the segment within the field contributes.
Correct force is 0.0640 N rather than 0.240 N.

Check: extending cable outside the field cannot change magnetic force here.

F=IB\ell_{\mathrm{active}}=(2.00)(0.400)(0.0800)\,\mathrm N=0.0640\,\mathrm N

[1]

Transfer · Challenge

Two long parallel thin wires in air carry 6.00 A and 4.00 A in the same direction, axis separation 25.0 mm. Find magnetic interaction on a central 0.400 m segment far from the ends.

Show hint

The first wire’s field is external field for the second.

Learning goal: Combine generated field with force on a second wire.

Show solution and calculation

Approach: $F=\frac{\mu_0I_1I_2\ell}{2\pi d}$ .
Calculate wire field and insert it into perpendicular conductor force.
Magnitude 76.8 µN, attractive.

Check: force on the corresponding equal-length segment of the other wire is equal and opposite.

\begin{aligned}B_1&=\frac{\mu_0 I_1}{2\pi d}\\F_2&=I_2\ell B_1=\frac{\mu_0 I_1I_2\ell}{2\pi d}\approx76.8\,\mathrm{\mu N}\end{aligned}

[4][3][1]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Explained conductor force through drift current, separated direction and path-integral models; developed two original examples and five individual DE/EN exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Magnetic force on a current-carrying wire. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetische-kraft-leiter

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Magnetic force on a current-carrying wire

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Force on a current-carrying wire calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetism

Worked example

Applications

How do you calculate with Magnetic force on a current-carrying wire?

When does the force formula apply to a straight current-carrying wire?

From charge flow to a wire integral

Two directions determine the third

How do you calculate bent current paths and varying magnetic fields?

Local force contribution along the wire

Straight wire in a varying field

Example: one field component contributes nothing

Straight segment in an oblique field

Example: right-angle current path

Two segments, one resultant force

Units and practical force measurements

Five exercises: balance and current paths

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. B=0.500 TB=0.500\,\mathrm TB=0.500T along −z; g=9.81 m/s2g=9.81\,\mathrm{m/s^2}g=9.81m/s2 along −y. Leads are separately supported. What current balances it without other vertical forces?

Straight segment: I=1.60 AI=1.60\,\mathrm AI=1.60A, ℓ=0.100 m\ell=0.100\,\mathrm mℓ=0.100m, angle 30° to field, magnetic force magnitude 48.0 mN. Find B.

A segment carries current along +y while field points along +x. State force direction initially, after current reversal, and after reversing both current and field.

Only 0.0800 m of a straight 0.300 m cable lies in uniform 0.400 T field perpendicular to its 2.00 A current. External field is zero elsewhere. A calculation uses full cable length. Correct it.

Two long parallel thin wires in air carry 6.00 A and 4.00 A in the same direction, axis separation 25.0 mm. Find magnetic interaction on a central 0.400 m segment far from the ends.

What to take away

Sources for this explanation

Frequently asked questions about Magnetic force on a current-carrying wire

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Magnetic force on a current-carrying wire for exams

Magnetic force on a current-carrying wire

Variables & units

Force on a current-carrying wire calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetism

Worked example

Applications

How do you calculate with Magnetic force on a current-carrying wire?

When does the force formula apply to a straight current-carrying wire?

From charge flow to a wire integral

Two directions determine the third

How do you calculate bent current paths and varying magnetic fields?

Local force contribution along the wire

Straight wire in a varying field

Example: one field component contributes nothing

Straight segment in an oblique field

Example: right-angle current path

Two segments, one resultant force

Units and practical force measurements

Five exercises: balance and current paths

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. B=0.500 TB=0.500\,\mathrm TB=0.500T along −z; g=9.81 m/s2g=9.81\,\mathrm{m/s^2}g=9.81m/s2 along −y. Leads are separately supported. What current balances it without other vertical forces?

Straight segment: I=1.60 AI=1.60\,\mathrm AI=1.60A, ℓ=0.100 m\ell=0.100\,\mathrm mℓ=0.100m, angle 30° to field, magnetic force magnitude 48.0 mN. Find B.

A segment carries current along +y while field points along +x. State force direction initially, after current reversal, and after reversing both current and field.

Only 0.0800 m of a straight 0.300 m cable lies in uniform 0.400 T field perpendicular to its 2.00 A current. External field is zero elsewhere. A calculation uses full cable length. Correct it.

Two long parallel thin wires in air carry 6.00 A and 4.00 A in the same direction, axis separation 25.0 mm. Find magnetic interaction on a central 0.400 m segment far from the ends.

What to take away

Sources for this explanation

Frequently asked questions about Magnetic force on a current-carrying wire

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Magnetic force on a current-carrying wire for exams

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. $B=0.500\,\mathrm T$ along −z; $g=9.81\,\mathrm{m/s^2}$ along −y. Leads are separately supported. What current balances it without other vertical forces?

Straight segment: $I=1.60\,\mathrm A$ , $\ell=0.100\,\mathrm m$ , angle 30° to field, magnetic force magnitude 48.0 mN. Find B.

A mechanically free segment has mass 12.0 g and length 0.240 m along +x. $B=0.500\,\mathrm T$ along −z; $g=9.81\,\mathrm{m/s^2}$ along −y. Leads are separately supported. What current balances it without other vertical forces?

Straight segment: $I=1.60\,\mathrm A$ , $\ell=0.100\,\mathrm m$ , angle 30° to field, magnetic force magnitude 48.0 mN. Find B.