A field can push a current-carrying wire
You will be able to: Calculate force on a straight current-carrying segment and determine its direction.
Which part of a wire contributes to magnetic force?
A horizontal wire carrying current to the right lies in an upward magnetic field. Its magnetic force points out of the page. Only the portion inside the field region contributes to the simple uniform-field calculation.
A useful starting point: Add fields at the same observation point →
Words and symbols before equations
- Active length L
- Length of straight wire within the uniform field, in m.
- Current direction
- Conventional current direction; do not substitute electron drift direction.
- Wire force
- Combined magnetic force associated with moving charge carriers, transferred to the conductor.
What this picture assumes
Uniform external field 0.50 T upward; straight horizontal segment perpendicular to B. Only magnetic force is shown; supports and gravity are not included.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- F_z=0.3 N: out of page. Magnitude=ILB=0.3 N.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For a straight segment, |F|=ILB sinθ, with θ between conventional current and B. Perpendicular current gives ILB; parallel current gives zero. Find direction from L×B, where L points with current.
Doubling I, the active length or B doubles the perpendicular force if other quantities stay fixed. Increasing total wire length outside the field does not increase the force from that field region.
A supported wire may also experience tension, gravity and contact forces. The model shows magnetic force only; equilibrium requires adding any other forces separately. This provides a direct way to connect current and field to a measurable force.
A worked example, step by step
A 0.20 m segment carries 3 A rightward through a 0.50 T upward field. Find magnetic force.
- The current is perpendicular to the field.
- F=ILB=(3)(0.20)(0.50)=0.30 N.
- Right crossed with up gives out of the page.
- Reversing current reverses the force but retains the 0.30 N magnitude.
Use the length inside the field region, not automatically the total length of wire in the circuit.
Add wire outside the field without changing current. Does this field’s force increase?
Compare with an explanation
No. The active length inside the field is unchanged.
Predict. Change one thing. Explain.
Vary current and active length in the fixed 0.50 T upward field. Reverse current and compare force sign with magnitude.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
F_z=0.3 N: out of page. Magnitude=ILB=0.3 N.
Uniform external field 0.50 T upward; straight horizontal segment perpendicular to B. Only magnetic force is shown; supports and gravity are not included.
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 0.40 m segment carries 2 A perpendicular to 0.25 T. (a) Find force magnitude. (b) Predict the effect of doubling active length. (c) Predict parallel orientation. (d) State the current convention.
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Compare with the answer and four-point rubric
- 1 point: F=0.20 N.
- 1 point: Force doubles to 0.40 N.
- 1 point: Parallel orientation gives zero magnetic force.
- 1 point: Use conventional current, not electron drift direction.
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 length belongs in ILB?
The straight segment inside the relevant uniform field.
RECALL 2What direction is wire force relative to I and B?
Perpendicular to both when nonzero.
RECALL 3Does the magnetic-force diagram alone prove equilibrium?
No. Other forces must also be considered.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A field can push a current-carrying wire
- Straight wire: |F|=ILB sinθ.
- Perpendicular: F=ILB; parallel: F=0.
- Direction uses conventional current.
Remember: Use the length inside the field region, not automatically the total length of wire in the circuit.
Conditions: Uniform external field 0.50 T upward; straight horizontal segment perpendicular to B. Only magnetic force is shown; supports and gravity are not included.
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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