One wire’s field acts on the other wire
You will be able to: Combine long-wire field and wire-force reasoning to explain attraction and repulsion.
Why do parallel currents attract?
Two long side-by-side wires both carry upward conventional current. Each creates a field at the other, and each feels a force toward its neighbor. Reversing just one current makes the forces repulsive.
A useful starting point: A field can push a current-carrying wire →
Words and symbols before equations
- Separation d
- Perpendicular distance between the wire axes, in m.
- Force per length F/L
- Useful comparison for long wires, in N/m.
- External field
- When finding force on wire 2, use the field produced by wire 1, not wire 2’s own field.
What this picture assumes
Long parallel wires in vacuum; left current 3 A up, right magnitude 4 A. Equal-opposite forces shown schematically, not to scale; computed force is per length.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Attraction; each force per length has magnitude 24 μN/m. Pair forces equal and opposite; arrows schematic.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
First find wire 1’s field at wire 2 using the current right-hand rule. Then apply the force right-hand rule to wire 2’s current in that external field. Keeping these two steps separate prevents sign mistakes.
Combining B₁=μ₀I₁/(2πd) with F₂=I₂LB₁ gives F/L=μ₀|I₁I₂|/(2πd) for long parallel wires in vacuum. Same-direction currents attract; opposite-direction currents repel.
The forces on the two wires have equal magnitudes and opposite directions, even if the currents differ. They act on different objects, so they do not cancel in either wire’s individual force diagram.
A worked example, step by step
Two long parallel wires separated by 0.10 m carry 3 A and 4 A in the same direction. Find force per length.
- Field from wire 1 at wire 2: B₁=2×10⁻⁷(3)/0.10=6 μT.
- F₂/L=I₂B₁=(4)(6×10⁻⁶)=24 μN/m.
- Same-direction currents attract.
- The other wire has the same force-per-length magnitude toward its neighbor.
Do not use a wire’s own field as the external field in the simple force-on-wire calculation.
If one current doubles, does only that wire’s force change?
Compare with an explanation
No. The interaction-force magnitudes on both wires double.
Predict. Change one thing. Explain.
Reverse one current and vary separation. Predict whether attraction becomes repulsion and how force per length changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Attraction; each force per length has magnitude 24 μN/m. Pair forces equal and opposite; arrows schematic.
Long parallel wires in vacuum; left current 3 A up, right magnitude 4 A. Equal-opposite forces shown schematically, not to scale; computed force is per length.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant field, force, flux or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTwo long wires carry 2 A and 5 A with 0.20 m separation. (a) Find F/L. (b) State behavior for equal current directions. (c) Reverse one current. (d) Compare the two force magnitudes.
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Compare with the answer and four-point rubric
- 1 point: F/L=2×10⁻⁷(10)/0.20=10 μN/m.
- 1 point: Attraction.
- 1 point: Repulsion with unchanged magnitude.
- 1 point: Equal magnitudes on the two wires, opposite directions.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which field acts on the second wire?
The external field produced by the first wire.
RECALL 2What law helps check the pair forces?
Newton’s third law.
RECALL 3Why use force per length?
For long uniform parallel segments, total force scales with active length.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
One wire’s field acts on the other wire
- F/L=μ₀|I₁I₂|/(2πd), long parallel wires.
- Same-direction currents attract; opposite-direction currents repel.
- The pair forces are equal and opposite on different wires.
Remember: Do not use a wire’s own field as the external field in the simple force-on-wire calculation.
Conditions: Long parallel wires in vacuum; left current 3 A up, right magnitude 4 A. Equal-opposite forces shown schematically, not to scale; computed force is per length.
Refresh Kid · AP Physics 2 Unit 4 (official Unit 12) · Objectives 12.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 12.3, objectives 12.3.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the fourth AP Physics 2 unit; College Board numbers it Unit 12; the first unit in this course is official Unit 9. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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