Steady Electric Currents, Resistance, and Continuity

Unit 7: Inductance, energy and electromagnetic force

Four lessons: Self-inductance and flux linkage, Mutual inductance and coupled circuits, Magnetic energy and electromechanical force, Reluctance circuits and design limits.

Learning outcomes

By the end of this unit, explain the physical field variables and their SI units; solve a useful analytical reference geometry; reproduce a field-level numerical calculation; identify boundary assumptions; and explain an engineering decision that would be unjustified by the ideal model alone.

Unit investigation

Use a set of five measured inductance values L(x) for a movable armature, construct a smooth graph and estimate central-difference dL/dx. At constant current I, calculate the force F=0.5I²dL/dx. Explain why derivative noise can make force estimates unstable even when individual L estimates appear precise; compare central difference with a least-squares local slope. Show the change in estimated force when the position sampling step is halved. Do not claim actuator force can be inferred from energy at a single mechanical position.

As a separate magnetic-circuit approximation, estimate flux through a core dominated by an air gap. Explicitly include ampere-turns, gap length, effective area and the assumption of a high-permeability unsaturated core. Calculate the field pressure B²/(2μ0) as an order-of-magnitude estimate and compare force pressure times area with the energy-gradient estimate. A mismatch is an opportunity to examine fringe flux, nonlinearity and geometric modeling, not to tune a constant until two unrelated models agree.

Guided lesson path

Step 1: Self-inductance and flux linkage

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 1](../lessons/U07-L01-self-inductance-and-flux-linkage.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 2: Mutual inductance and coupled circuits

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 2](../lessons/U07-L02-mutual-inductance-and-coupled-circuits.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 3: Magnetic energy and electromechanical force

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 3](../lessons/U07-L03-magnetic-energy-and-electromechanical-force.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Step 4: Reluctance circuits and design limits

Review the governing relationship, reproduce the numerical variant with independent unit conversions, and explicitly note one observation that would falsify the model's assumptions. See [lesson 4](../lessons/U07-L04-reluctance-circuits-and-design-limits.html). The mathematical conclusion should follow the boundary and source assumptions, not merely the choice of familiar-looking formula.

Unit verification checklist

• Define at least one coordinate system and positive contour or normal orientation.
• State which source or boundary quantities are controlled and which are outputs.
• Compare analytical and computational predictions, including one limiting case.
• Report at least one sensitivity result rather than a single nominal value.
• Describe which material data would be needed for genuine engineering validation.
• Prepare a concise error budget identifying unit conversion, approximate geometry and material uncertainty.

Practice and project: [Unit assessment](../assessments/unit-07.html) · [Unit engineering project](../projects/unit-07.html) · [Interactive mathematical model](../simulations/model-07.html).

Unit 7, lesson 1 — Self-inductance and flux linkage

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

An inductor stores energy in its magnetic field. Flux linkage λ counts magnetic flux threaded by turns, and inductance connects λ to current under linear conditions. In a long ideal solenoid, increasing turns substantially increases inductance because current both creates the field and the resulting field links the turns. Real coils have winding resistance, parasitic capacitance, leakage, fringing and frequency-dependent cores; a single L is a low-frequency approximation, not an all-frequency device model.

Governing relationships and derivation pathway

For a linear inductor λ=LI and v=dλ/dt=L di/dt if L is constant. Ideal long solenoid L≈μ N² A/ell. Magnetic energy Wm=0.5 L I² in the linear regime.

Worked numerical illustration

For an air-core solenoid with N=200, A=1 cm² and ell=0.10 m, L≈μ0×200²×1e-4/0.10≈50.3 μH. At 1 A the stored energy is about 25.1 μJ.

Independent numerical variant

An ideal solenoid N=100, A=2 cm², length .2 m, air core has L≈12.6 μH and W≈6.28 μJ at 1 A.

Engineering context: Coil dimension sensitivity

For the ideal long air-core solenoid L scales with turns squared, cross-sectional area and inverse length. Vary turns by ±10% and observe that inductance changes approximately by +21% or -19%, not ±10%. This sensitivity can dominate geometry tolerances in a prototype. At high frequency, measure winding capacitance and resistance separately; relying on the ideal inductance relation beyond its range creates false resonance and loss predictions.

Sensitivity and design interpretation

For an ideal long solenoid at fixed geometry, doubling N quadruples L and quadruples energy at fixed I.

Why the model can fail

The inductance formula neglects end fringing, and using a constant high μ for a saturating core can overstate stored energy. A coil with an air gap may store much of its energy in the gap, even though the iron carries the flux.

Two resolved conceptual checkpoints

Question A. If turns N double at fixed ideal geometry, by what factor does L change?

Reasoned answer. Inductance scales as N², so it becomes four times larger in the ideal linear solenoid model.

Question B. Why can a core with an air gap store most energy in its gap?

Reasoned answer. Gap H can be large for a given B because μ is small; field energy density B²/(2μ) is correspondingly high.

Next steps

[Unit 7 interactive model](../simulations/model-07.html) · [Assessment](../assessments/unit-07.html) · [Project](../projects/unit-07.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 7, lesson 2 — Mutual inductance and coupled circuits

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

Two coils can share magnetic flux so that changing current in one produces voltage in the other. Mutual inductance describes this linkage and reciprocity makes M12=M21 under ordinary linear passive conditions. The sign of induced voltages depends on dot convention and winding orientation, not simply whether a displayed current is positive. Coupling coefficient k expresses the fraction of perfect coupling and has magnitude at most one for passive linear two-coil models.

Governing relationships and derivation pathway

λ1=L1 i1+M i2 and λ2=M i1+L2 i2 under aligned dot references; M=k sqrt(L1 L2), 0≤|k|≤1. Induced terms v1=L1 di1/dt+M di2/dt and v2=M di1/dt+L2 di2/dt are signed by chosen terminal conventions.

Worked numerical illustration

With L1=4 mH, L2=9 mH and k=0.5, M=0.5 sqrt(36) mH=3 mH. A primary current slope of +200 A/s creates magnitude 0.6 V in the secondary, with polarity decided by dot convention.

Independent numerical variant

For L1=1 mH, L2=16 mH and k=.25, M=1 mH. A current ramp of 1000 A/s induces 1 V magnitude in the second winding.

Engineering context: Polarity verification

Mark the terminals of two coupled coils using a dot convention. Increase current entering the dotted primary terminal and determine which secondary terminal becomes positive under the chosen voltage reference. Check M ≤ sqrt(L1 L2) to protect the energy model from an unphysical coupling claim. In a real instrument, phase delay and winding capacitance can complicate measurements; the textbook ideal is intentionally low-frequency.

Sensitivity and design interpretation

Doubling coupling k doubles M when L1 and L2 are fixed; physical |k| cannot exceed one in the passive model.

Why the model can fail

A coupling coefficient from one operating point may vary with mechanical position or saturating material. Do not ignore core loss and leakage when interpreting an actual transformer at high frequency.

Two resolved conceptual checkpoints

Question A. What is the bound for mutual inductance of passive linear two-coil models?

Reasoned answer. Its magnitude is no larger than sqrt(L1L2), corresponding to

Question B. Does a positive numerical M alone establish actual secondary voltage polarity?

Reasoned answer. No. Terminal and winding dot convention determines the sign of mutual voltage terms.

Next steps

[Unit 7 interactive model](../simulations/model-07.html) · [Assessment](../assessments/unit-07.html) · [Project](../projects/unit-07.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 7, lesson 3 — Magnetic energy and electromechanical force

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

In magnetic actuators the field energy depends on current and mechanical geometry. The force that moves an armature can be obtained from energy or co-energy derivatives, but the choice of held-fixed electrical variables matters. For a linear coil with position-dependent inductance L(x), magnetic co-energy at constant current is 0.5L(x)I²; the force along x is its derivative. This provides a bridge between field calculations and practical relays, speakers and solenoids.

Governing relationships and derivation pathway

For an ideal linear actuator, F_x|I = 0.5 I² dL/dx. Magnetic field energy density is u_m=0.5 B·H for a linear nondispersive material. Across a simple air gap the field pressure scale is p≈B²/(2μ0).

Worked numerical illustration

An actuator has dL/dx=0.8 mH/m and constant current I=2 A. The ideal force is 0.5×4×0.0008=0.0016 N along increasing inductance. This small magnitude is a reminder that geometry sensitivity matters as much as current.

Independent numerical variant

For a linear actuator dL/dx=.002 H/m and I=3 A, the ideal constant-current force is .009 N in the increasing-L direction.

Engineering context: Force derivative model

For a movable coil with L(x) estimated at three equally spaced positions, approximate dL/dx using a central difference. Insert current into F=0.5 I² dL/dx to estimate force direction. Repeat with current doubled and confirm a factor-of-four scaling in the linear model. If a measured force departs from that prediction, inspect saturation, mechanical friction, end effects and current-source constraints before blaming the numerical derivative.

Sensitivity and design interpretation

Doubling I at fixed geometry multiplies magnetic force by four if the linear co-energy model remains valid.

Why the model can fail

The energy method gives net generalized force, not necessarily local stress concentration. Friction, springs, saturation and eddy currents can dominate in real devices. Always state what “held constant” means.

Two resolved conceptual checkpoints

Question A. If I doubles at fixed L(x), how does ideal magnetic force change?

Reasoned answer. Force at constant current is proportional to I², so it becomes four times as large.

Question B. What must be held fixed in the co-energy force derivative?

Reasoned answer. Current, rather than flux linkage, for F=(∂Wco/∂x)

Next steps

[Unit 7 interactive model](../simulations/model-07.html) · [Assessment](../assessments/unit-07.html) · [Project](../projects/unit-07.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 7, lesson 4 — Reluctance circuits and design limits

Book 21 — Electromagnetic Field Theory I · Prerequisites: multivariable calculus, introductory circuits and SI units. Quality note: Original educational draft; physical equations and instructional solutions require independent review before publication.

Physical idea and meaning

A magnetic circuit approximates flux as current through a reluctance network. Magnetomotive force NI drives flux; reluctance grows with length and shrinks with permeability and cross-sectional area. The analogy to resistive networks is useful but imperfect: flux leakage, fringing and saturation make permeability nonconstant and routes nonunique. Air gaps often dominate total reluctance, allowing intentionally controlled inductance but also introducing strong local fields and mechanical force.

Governing relationships and derivation pathway

Reluctance Rm=ell/(μ A), magnetic flux Φ≈NI/Rm and B≈Φ/A. For series sections with a common flux, reluctances add. In an air-gap-dominated core of gap g and cross section A, Rgap≈g/(μ0 A) ignoring fringe.

Worked numerical illustration

Take a gap 1 mm long, area 4 cm², N=500 turns and I=0.2 A. Approximate Bgap≈μ0 NI/g=0.126 T when iron reluctance is negligible. Estimated gap pressure is B²/(2μ0)≈6.3 kPa.

Independent numerical variant

A 0.5 mm air gap, area 2 cm², N=200, I=.1 A gives ideal B≈50.3 mT neglecting iron reluctance and gap fringing.

Engineering context: Reluctance and fringing

Model a core gap as g/(μ0A) and then compare two approximations for the effective gap area: geometric A and a slightly enlarged effective area accounting for fringing. The second gives lower reluctance and somewhat larger flux. No fixed fringing correction is universal; geometry sets the correction. A useful student report shows both cases and identifies which physical data would be needed to justify either approximation.

Sensitivity and design interpretation

Doubling air-gap length doubles dominant reluctance and roughly halves flux at fixed NI.

Why the model can fail

A magnetic network is not a substitute for a field solver when air gaps are wide or dimensions change sharply. A nominally one-dimensional gap includes fringing, and gap B at saturation can differ drastically.

Two resolved conceptual checkpoints

Question A. If an ideal dominant gap doubles, what happens to its reluctance?

Reasoned answer. Reluctance doubles at fixed effective area. The predicted flux halves at fixed NI if the gap truly dominates.

Question B. Why is a large-fringing gap hard to model with one area?

Reasoned answer. The flux spreads in space, making effective area and local B nonuniform. A field solver or measured correction is needed.

Next steps

[Unit 7 interactive model](../simulations/model-07.html) · [Assessment](../assessments/unit-07.html) · [Project](../projects/unit-07.html) · [Shared study guidance](../STUDY-GUIDE.html).

Unit 7 engineering project: Solenoid and actuator trade-offs

Problem statement and intended decision

Design an idealized moving-armature coil with L(x) specified at five positions and constant drive current. Estimate force from numerical dL/dx, plot stored energy and show the effect of changing current by a factor of two. Then construct an air-gap-reluctance approximation and compare qualitative force direction. Flag which quantities require physical characterization: B–H curve, friction, eddy-current loss, mechanical limits and coil heating. The result should distinguish an ideal electromechanical feasibility calculation from hardware validation.

Detailed investigation plan

Use a set of five measured inductance values L(x) for a movable armature, construct a smooth graph and estimate central-difference dL/dx. At constant current I, calculate the force F=0.5I²dL/dx. Explain why derivative noise can make force estimates unstable even when individual L estimates appear precise; compare central difference with a least-squares local slope. Show the change in estimated force when the position sampling step is halved. Do not claim actuator force can be inferred from energy at a single mechanical position.

As a separate magnetic-circuit approximation, estimate flux through a core dominated by an air gap. Explicitly include ampere-turns, gap length, effective area and the assumption of a high-permeability unsaturated core. Calculate the field pressure B²/(2μ0) as an order-of-magnitude estimate and compare force pressure times area with the energy-gradient estimate. A mismatch is an opportunity to examine fringe flux, nonlinearity and geometric modeling, not to tune a constant until two unrelated models agree.

Submission requirements

Follow the seven evidence deliverables and the 100-point project rubric in the [Shared study guidance](../STUDY-GUIDE.html). The unit-specific source model, investigation plan, parameter choices and physical limitations above determine whether the report is adequate.

Submission safeguard

All activities are simulation and analysis only. If a project might ordinarily be performed with mains current, charged capacitors, electromagnets, high-voltage insulation or RF power, replace hardware with the ideal mathematical model and a synthetic dataset. Do not treat this project brief as authorization or guidance to perform dangerous laboratory measurements.

Unit 7 assessment — Inductance, energy and electromagnetic force

Instructions: Work independently before expanding answers. These twelve questions cover all four lessons. State your coordinate reference, units, sign convention and physical assumptions. The answers below explain the expected reasoning; identical numerical output alone is insufficient.

Question 1 — Self-inductance and flux linkage

Prompt: If turns N double at fixed ideal geometry, by what factor does L change?

Reveal worked reasoning

Inductance scales as N², so it becomes four times larger in the ideal linear solenoid model. The matching lesson and shared study guide provide additional verification questions.

Question 2 — Self-inductance and flux linkage

Prompt: Why can a core with an air gap store most energy in its gap?

Reveal worked reasoning

Gap H can be large for a given B because μ is small; field energy density B²/(2μ) is correspondingly high. The matching lesson and shared study guide provide additional verification questions.

Question 3 — Self-inductance and flux linkage numerical and decision challenge

Prompt: Independently derive and check this scenario: An ideal solenoid N=100, A=2 cm², length .2 m, air core has L≈12.6 μH and W≈6.28 μJ at 1 A. Then choose a parameter to change, predict its effect from for an ideal long solenoid at fixed geometry, doubling N quadruples L and quadruples energy at fixed I. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: An ideal solenoid N=100, A=2 cm², length .2 m, air core has L≈12.6 μH and W≈6.28 μJ at 1 A.

Sensitivity logic: For an ideal long solenoid at fixed geometry, doubling N quadruples L and quadruples energy at fixed I.

Engineering qualification: The inductance formula neglects end fringing, and using a constant high μ for a saturating core can overstate stored energy. A coil with an air gap may store much of its energy in the gap, even though the iron carries the flux. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 4 — Mutual inductance and coupled circuits

Prompt: What is the bound for mutual inductance of passive linear two-coil models?

Reveal worked reasoning

Its magnitude is no larger than sqrt(L1L2), corresponding to |k|≤1 from nonnegative stored magnetic energy. The matching lesson and shared study guide provide additional verification questions.

Question 5 — Mutual inductance and coupled circuits

Prompt: Does a positive numerical M alone establish actual secondary voltage polarity?

Reveal worked reasoning

No. Terminal and winding dot convention determines the sign of mutual voltage terms. The matching lesson and shared study guide provide additional verification questions.

Question 6 — Mutual inductance and coupled circuits numerical and decision challenge

Prompt: Independently derive and check this scenario: For L1=1 mH, L2=16 mH and k=.25, M=1 mH. A current ramp of 1000 A/s induces 1 V magnitude in the second winding. Then choose a parameter to change, predict its effect from doubling coupling k doubles M when L1 and L2 are fixed; physical |k| cannot exceed one in the passive model. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: For L1=1 mH, L2=16 mH and k=.25, M=1 mH. A current ramp of 1000 A/s induces 1 V magnitude in the second winding.

Sensitivity logic: Doubling coupling k doubles M when L1 and L2 are fixed; physical |k| cannot exceed one in the passive model.

Engineering qualification: A coupling coefficient from one operating point may vary with mechanical position or saturating material. Do not ignore core loss and leakage when interpreting an actual transformer at high frequency. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 7 — Magnetic energy and electromechanical force

Prompt: If I doubles at fixed L(x), how does ideal magnetic force change?

Reveal worked reasoning

Force at constant current is proportional to I², so it becomes four times as large. The matching lesson and shared study guide provide additional verification questions.

Question 8 — Magnetic energy and electromechanical force

Prompt: What must be held fixed in the co-energy force derivative?

Reveal worked reasoning

Current, rather than flux linkage, for F=(∂Wco/∂x)|I. Mixing constraints gives incorrect signs or energy balances. The matching lesson and shared study guide provide additional verification questions.

Question 9 — Magnetic energy and electromechanical force numerical and decision challenge

Prompt: Independently derive and check this scenario: For a linear actuator dL/dx=.002 H/m and I=3 A, the ideal constant-current force is .009 N in the increasing-L direction. Then choose a parameter to change, predict its effect from doubling I at fixed geometry multiplies magnetic force by four if the linear co-energy model remains valid. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: For a linear actuator dL/dx=.002 H/m and I=3 A, the ideal constant-current force is .009 N in the increasing-L direction.

Sensitivity logic: Doubling I at fixed geometry multiplies magnetic force by four if the linear co-energy model remains valid.

Engineering qualification: The energy method gives net generalized force, not necessarily local stress concentration. Friction, springs, saturation and eddy currents can dominate in real devices. Always state what “held constant” means. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.

Question 10 — Reluctance circuits and design limits

Prompt: If an ideal dominant gap doubles, what happens to its reluctance?

Reveal worked reasoning

Reluctance doubles at fixed effective area. The predicted flux halves at fixed NI if the gap truly dominates. The matching lesson and shared study guide provide additional verification questions.

Question 11 — Reluctance circuits and design limits

Prompt: Why is a large-fringing gap hard to model with one area?

Reveal worked reasoning

The flux spreads in space, making effective area and local B nonuniform. A field solver or measured correction is needed. The matching lesson and shared study guide provide additional verification questions.

Question 12 — Reluctance circuits and design limits numerical and decision challenge

Prompt: Independently derive and check this scenario: A 0.5 mm air gap, area 2 cm², N=200, I=.1 A gives ideal B≈50.3 mT neglecting iron reluctance and gap fringing. Then choose a parameter to change, predict its effect from doubling air-gap length doubles dominant reluctance and roughly halves flux at fixed NI. State which quantities remain fixed and identify one important excluded physical effect.

Reveal worked reasoning and review criteria

Reference calculation: A 0.5 mm air gap, area 2 cm², N=200, I=.1 A gives ideal B≈50.3 mT neglecting iron reluctance and gap fringing.

Sensitivity logic: Doubling air-gap length doubles dominant reluctance and roughly halves flux at fixed NI.

Engineering qualification: A magnetic network is not a substitute for a field solver when air gaps are wide or dimensions change sharply. A nominally one-dimensional gap includes fringing, and gap B at saturation can differ drastically. The final report must include a quantity with correct SI units, an analytical limit and a specific reason the ideal calculation is not sufficient for hardware sign-off.