Current makes a field around a wire
You will be able to: Use the right-hand rule and inverse-distance relation for a long straight wire.
How does field strength change with distance from a straight wire?
A compass beside a current-carrying wire turns when the current changes. The field circles the wire. At twice the perpendicular distance, the ideal long-wire field is half as strong.
A useful starting point: Sideways charge separation creates a voltage →
Words and symbols before equations
- Conventional current I
- Positive-charge flow direction, in A; use this direction for the wire right-hand rule.
- Perpendicular distance r
- Shortest distance from the wire’s central axis to the observation point, in m.
- Long-wire approximation
- Wire much longer than the observation distance, away from its ends.
- μT
- Microtesla: 10⁻⁶ T.
What this picture assumes
Long straight wire in vacuum, away from ends. Probe is right of wire. B=μ₀I/(2πr); μ₀≈4π×10⁻⁷ T·m/A. Cross-section drawing is schematic; graph uses a fixed scale.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- |B|=10 μT. At the right-side probe, B points up; circulation counterclockwise.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Point your right thumb along conventional current. Curled fingers give the circulating B direction. In a cross section, current out of the page produces counterclockwise field arrows; current into the page produces clockwise arrows.
For a long straight wire in vacuum or approximately air, B=μ₀I/(2πr). Field vectors are tangent to circles centered on the wire, not radial and not along the wire.
For an experiment, keep I fixed and measure B at several r values; plot B against 1/r to seek a straight line. Account for background fields and probe orientation, and measure r from the wire axis. The simulation is ideal data, not experimental evidence.
A worked example, step by step
A long wire carries 5 A. Find B at r=0.10 m, using μ₀≈4π×10⁻⁷ T·m/A.
- Use B=μ₀I/(2πr).
- B=(4π×10⁻⁷)(5)/(2π×0.10)=1.0×10⁻⁵ T.
- B=10 μT. At 0.20 m it would be 5 μT.
- Use current direction and the right-hand rule to supply direction at the chosen point.
The distance dependence is 1/r for a long straight wire, not the point-charge electric field’s 1/r².
Double both I and r. What happens to B magnitude?
Compare with an explanation
It remains unchanged because I/r stays constant.
Predict. Change one thing. Explain.
Change current and probe distance. Predict each factor change and inspect the common-scale field-versus-distance graph. Reverse current to check the direction.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
|B|=10 μT. At the right-side probe, B points up; circulation counterclockwise.
Long straight wire in vacuum, away from ends. Probe is right of wire. B=μ₀I/(2πr); μ₀≈4π×10⁻⁷ T·m/A. Cross-section drawing is schematic; graph uses a fixed scale.
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 questionA long wire carries 3 A. (a) Find B at 0.060 m. (b) Find B at 0.120 m. (c) State the field direction at the right side for outward current. (d) Name one experimental control.
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Compare with the answer and four-point rubric
- 1 point: B=2×10⁻⁷×3/0.060=10 μT.
- 1 point: B=5 μT.
- 1 point: Upward at the wire’s right side.
- 1 point: Hold current fixed, account for background B, and keep probe orientation consistent.
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 1What distance appears in the wire formula?
Perpendicular distance from the wire axis.
RECALL 2What does the field direction look like?
Tangent to concentric circles.
RECALL 3Which graph linearizes the ideal fixed-current relation?
B versus 1/r.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Current makes a field around a wire
- B=μ₀I/(2πr), long straight wire in vacuum.
- Right thumb along conventional I; fingers follow B.
- B is tangent to circles around the wire.
Remember: The distance dependence is 1/r for a long straight wire, not the point-charge electric field’s 1/r².
Conditions: Long straight wire in vacuum, away from ends. Probe is right of wire. B=μ₀I/(2πr); μ₀≈4π×10⁻⁷ T·m/A. Cross-section drawing is schematic; graph uses a fixed scale.
Refresh Kid · AP Physics 2 Unit 4 (official Unit 12) · Objectives 12.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 12.3, objectives 12.3.A. 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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