Magnetic force on a current-carrying wire
Compute magnetic force on straight and differential current elements using vector form and right-hand-rule direction checks.
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Guided steps
- The directed wire length follows conventional current, and includes only the segment actually in the field. Take length cross field, in that order. The angle selects a perpendicular component; it is not a reason to reverse the direction again for electron carriers.
- Current density is carrier number density times signed charge times drift velocity. Multiplying it cross the external field gives magnetic force per volume. At fixed conventional current, changing carrier sign also changes drift direction, so the wire force does not reverse. Distinguish the field acting on the wire from the field the wire itself produces.
- Distribute the cross product over the wire length components. With field along positive z, the y length gives positive x force and the x length gives negative y force. Combine force components with the vector magnitude and test the force dot length is zero. A field parallel to one wire orientation remains invisible to that force measurement.
- Parallel current and field give zero force. Reversing one vector reverses a nonzero force; reversing both restores it. Count only the active wire length. A moving wire may gain mechanical energy from an electrical source, but magnetic force is not a new source of energy; each charge’s magnetic force remains perpendicular to its full velocity.
- Use the angle between conventional current and field for the sine factor. The example gives a magnitude; an absolute spatial direction requires actual current and field directions. Compare it with the perpendicular and parallel limits.
- A vector length supplies its direction and components, not separate scalar wire lengths to add. Resolve each contribution, then check both the resultant magnitude and perpendicularity before interpreting the result.
- Find the signed x and y force components and the positive force magnitude. Keep the current direction encoded in the given length vector. Explain signs and check perpendicularity separately.
- Explain both cross-product signs, why electron drift needs no extra reversal, and how the force dot length checks perpendicularity. Compare your argument with the public solution yourself; three checked numbers do not mark the reasoning. Paper work is fine.
- Ask what each force measurement can determine. A parallel field component has no force contribution, but that does not make the component zero. Changing wire orientation can turn that same unknown into a perpendicular component, while the assumed field and current remain fixed.
- Use the two measured force components to find the corresponding field components. Also choose what the first orientation alone establishes about the parallel component. The field’s absent y component is given, not inferred.
- Explain the information missing from the first measurement and how the second orientation supplies it. Compare the two cross products yourself; the checked field values and choice do not mark that inference argument.
- Predict the force magnitude for a parallel wire and uniform field. Test the angle before assuming a nonzero force.
- Find the force magnitude on the stated perpendicular segment, keeping force distinct from force per length.
- Predict what reversing only the current does to the direction of the existing nonzero force. The wire and field stay fixed.
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A current-carrying wire feels the combined magnetic forces on its moving charges. Choose the conventional-current direction, identify the part actually in the field, and take a vector cross product. A force magnitude alone cannot tell you the field’s full direction.
Use the magnetic force on a particle and signed current density. Here the current is steady and the external magnetic field is prescribed; finding the wire’s own field is a separate problem.
Orient the length along conventional current
For a straight segment in a uniform external field, F = IL × B, F = ILB sin θ. Here I > 0 is the current magnitude, L points along conventional current, and θ is the angle between L and B. Its magnitude L is the length of the segment within the field, not automatically the entire wire. The unit check is A m T = N.
Use the right-hand rule on length cross field: point your fingers along L, curl them towards B, and read the force direction from your thumb. Reversing the order gives the opposite result. In right-handed axes, x hat × z hat = -y hat and y hat × z hat = +x hat.
Connect carrier force to current density
Let A be cross-sectional area, n carrier number density, q signed carrier charge and vd drift velocity. A small volume A dℓ contains nA dℓ carriers. Its magnetic-force density is fB = nqvd × B = J × B, where J = nqvd is conventional current density. Since J A = Iℓ hat along a thin wire, dF = I dℓ × B. The element dℓ follows conventional current even when electrons drift oppositely. At the same conventional current, reversing both carrier charge and drift direction does not reverse the wire force. The conductor transmits the charge carriers’ magnetic loading to the wire through its internal interactions. This is the current-element model used in OpenStax’s treatment of wire forces.
Keep components, magnitudes and energy separate
For a straight wire and constant field, integrating the element gives F = IL × B. If B = Bzz hat and L = Lₓx hat + Lyy hat, distribute the cross product: Fₓ = IBzLy, Fy = -IBzLₓ, Fz = 0. The magnitude is square root of (Fₓ² + Fy²), not the sum of signed components. The result is perpendicular to both L and B. A component of field parallel to the wire contributes no force, so one wire orientation cannot reveal that field component.
If the wire moves, its mechanical power is F · u, where u is the wire’s velocity. This can be nonzero even though magnetic force does zero work on each charge at its full instantaneous velocity. For a driven motor, energy comes from the electrical source; conductor motion, electric fields and circuit energy transfer must be considered together. The force law is not an extra energy source.
Check direction and the active length
Parallel or antiparallel current and field give zero force. Reversing only current or only field reverses a nonzero force; reversing both restores it. Count only the wire inside the stated field region. For a curved wire or a field varying along it, retain I∫ dℓ × B(r) rather than substituting the total wire length into the straight-wire formula.
Force on a wire at an angle
Worked example 1
The perpendicular component matters
Problem
A straight wire segment of length L = 0.30 m carries I = 5.0 A in a uniform field B = 0.40 T. The angle between the conventional current and the field is 30°. Find the magnetic-force magnitude.
Show full solution
Include the angle factor
Method
Use F = ILB sin θ with θ = 30°.Reason
The cross product selects the component perpendicular to the wire; using ILB alone would assume a right angle.Working
Answer
F = ILB sin θ; = (5.0)(0.30)(0.40) sin 30°; = (5.0)(0.30)(0.40)(0.5) = 0.30 N.State the magnitude
Working
Answer
F = 0.30 N, half the perpendicular-case force. A specific spatial direction requires the current and field directions, then follows vector L × vector B.
Quick check
If the wire lay along the field, θ = 0, there would be no force at all.
Resolve a wire force into components
Guided practice 1
A wire with two length components
Problem
A straight segment has directed length L = (0.120x hat + 0.160y hat) m along conventional current I = 4.00 A. The entire segment lies in uniform external field B = (0.500 T)z hat. Find Fₓ, Fy and the force magnitude. Explain the signs and verify that the force is perpendicular to the wire. Use right-handed axes.
Keep the two cross products separate
Hints
Hint 1
Distribute the cross product
Treat the x and y length components separately before combining their forces.Hint 2
Check perpendicularity
For this in-plane wire, evaluate FₓLₓ + FyLy.Show solution step by step
Resolve the direction
Method
Use y hat × z hat = x hat and x hat × z hat = -y hat.Reason
Conventional current already defines L; do not reverse it again for electron drift.Working
Answer
Fₓ = (4.00)(0.500)(0.160) = +0.320 N and Fy = -(4.00)(0.500)(0.120) = -0.240 N.Combine and test the vector
Method
Find F = square root of (Fₓ² + Fy²) and calculate F · L.Reason
The components are perpendicular; the magnetic force must be normal to the wire.Working
Answer
F = 0.400 N. Also (0.320)(0.120) + (-0.240)(0.160) = 0, as required. The wire length is 0.200 m, giving the same magnitude ILB = 0.400 N. Compare the direction and dot-product argument yourself; the checker verifies the three values only.
Quick check
Test the field with a different wire direction
Check your understanding 1 Use two force measurements
A straight 0.200 m segment carries 5.00 A. A uniform field is known to have the form B = Bₓx hat + Bzz hat, so By = 0. With conventional current along +x, the measured force is -(0.400 N)y hat. What value of Bz follows, and what can this first measurement tell you about Bₓ?
The same segment is then rotated so its conventional current points along +z. Current magnitude and the field are unchanged; the measured force is now + (0.300 N)y hat. Find Bₓ. Explain why the second orientation supplies information the first lacks.
Show hint
Show answer
The first orientation gives Fy = -ILBz, so Bz = +0.400 T. It leaves Bₓ undetermined: any parallel component gives the same first force. The absence of By was supplied in the question, not inferred from that single force component.
For the second orientation, Fy = +ILBₓ because z hat × x hat = +y hat. Hence Bₓ = +0.300 T. A field component invisible in one orientation becomes perpendicular in the other. Compare this inference argument yourself; the checker verifies the two field values and the first-measurement choice only.
Check your understanding 2
A straight current-carrying wire is parallel to a uniform magnetic field. What is its magnetic force?
Show hint
Evaluate the sine factor for parallel directions.
Show answer
F = ILB sin 0° = 0 N. The field has no component perpendicular to the wire.
Check your understanding 3
A 0.20 m straight wire carries 8.0 A perpendicular to a uniform 0.50 T field. Find the force magnitude.
Show hint
For perpendicular directions the sine factor is one.
Show answer
F = ILB = (8.0)(0.20)(0.50) = 0.80 N. Here the entire field is perpendicular to the wire.
Check your understanding 4
For a wire experiencing a nonzero magnetic force, what happens to the force direction if the current reverses while the wire and field stay fixed?
Show hint
Reverse vector L, which points along conventional current, in I vector L × vector B.
Show answer
The force reverses because reversing the current reverses the cross product. If the current magnitude is unchanged, the force magnitude is unchanged too.
Keep the vector law and its assumptions together
Choose conventional current, the active length and the external field before calculating. Use the cross product for direction, combine force components as vectors, and test perpendicularity. A field component parallel to one wire direction is not determined by that wire’s force measurement.
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