Relate generator coil positions to voltage peaks and zeros
Use the angle between coil normal and field to distinguish maximum linking from maximum rate of change.
Section 4.5.2 · The a.c. generator
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Explain induced e.m.f. as a changing-linking effect.
- Use parallel and perpendicular geometric directions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is a coil normal?
What you will learn
- Use the angle between coil normal and field to distinguish maximum linking from maximum rate of change.
- Connect coil normal orientation to linking and its rate of change before locating voltage peaks and zeros.
Define orientation without ambiguous plane language
The coil normal is a line perpendicular to its plane. Define angle θ between that normal and the magnetic field. At θ = 0° or 180°, the plane is perpendicular to the field and magnetic linking is at a positive or negative extreme. Its instantaneous rate of change is zero, so induced e.m.f. is zero.
At θ = 90° or 270°, the plane is parallel to the field. The linking passes through zero but changes most rapidly in the ideal constant-speed model, so e.m.f. reaches a peak or trough. Maximum linking and maximum induced e.m.f. therefore occur at different orientations.
Keep rotation and terminal conventions stated
For the supplied convention with θ increasing from 0°, the output sequence is 0, +6, 0, −6 and 0 V at 0°, 90°, 180°, 270° and 360°. Another terminal convention can invert signs while retaining the zero and peak positions. The key geometric distinction is extreme linking versus fastest change.
The plane’s appearance in a drawing depends on the viewing direction. Use the normal and field angle, not whether a sketch happens to look upright. Coil position alone gives magnitude class; positive versus negative also needs the rotation and output terminal conventions.
Pause and explain
Linking is at its maximum magnitude during ideal uniform rotation. What is the instantaneous e.m.f.?
Worked example
In the supplied ideal model, the coil normal is parallel to the field. Is induced e.m.f. at a peak?
Show the worked solution
- Normal parallel means plane perpendicular to field.
- Linking is at an extreme.
- The instantaneous linking change rate is zero.
Answer No: e.m.f. is zero at maximum-magnitude linking.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Classify e.m.f. magnitude for coil normal parallel and perpendicular to the field.
Give me a hint
Convert normal orientation to linking change rate.
Compare my reasoning
- Normal parallel gives linking extreme.
- Its instantaneous change rate and e.m.f. are zero.
- Normal perpendicular gives fastest linking change and peak magnitude.
Normal parallel: zero; normal perpendicular: peak magnitude.
Look for these in your work
- Use the normal convention.
- Distinguish linking and change rate.
Use the supplied increasing-θ convention to give e.m.f. at 0°, 90°, 180°, 270° and 360°.
Give me a hint
Follow a complete zero–peak–zero–trough cycle.
Compare my reasoning
- At 0° and 180° it is zero.
- At 90° it is +6 V and at 270° it is −6 V.
- At 360° it returns to zero.
0, +6, 0, −6, 0 V.
Look for these in your work
- Retain the supplied signs.
- Complete the revolution.
A diagram says plane parallel to field means maximum linking and zero e.m.f. Correct it for the ideal uniform-rotation model.
Give me a hint
Define the normal before deciding the linking state.
Compare my reasoning
- Plane parallel means normal perpendicular.
- Linking is zero and changes most rapidly.
- E.m.f. magnitude is at a peak, with sign needing the convention.
Plane parallel gives fastest linking change and peak e.m.f. magnitude, not maximum linking.
Look for these in your work
- Correct the plane/normal relation.
- State the sign limitation.
Common mistakes
- Angle to normal differs from angle to plane.
- Maximum linking is not maximum e.m.f.
- Peak sign needs a convention.
What can you explain now?
At which stated orientation is ideal e.m.f. magnitude greatest?
Compare with the explanation
Coil normal perpendicular to the field
The plane is parallel to the field and linking changes most rapidly during constant-speed rotation.
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Make it stick
Connect coil normal orientation to linking and its rate of change before locating voltage peaks and zeros.
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