Two coils placed near each other will exchange energy through their shared magnetic field — but how much? That depends on their individual inductances, their geometry, and how tightly their flux links together. Use this Mutual Inductance Interactive Calculator to calculate mutual inductance (M), coupling coefficient (k), induced voltage, mutual flux, secondary inductance, or coaxial solenoid M using inputs like self-inductance values, turns counts, current change rate, and core geometry. Getting this right matters in transformer design, wireless EV charging, and RFID antenna engineering — undershoot the coupling and your system wastes power or fails to start; overshoot and you're dealing with voltage transients you didn't plan for. This page includes the key formulas, a worked transformer example, engineering theory, real-world application scenarios, and an FAQ.
What is Mutual Inductance?
Mutual inductance is a measure of how strongly two coils are electromagnetically linked — specifically, how much voltage one coil induces in the other when its current changes. It is measured in henries (H). The stronger the magnetic coupling between the coils, the higher the mutual inductance.
Simple Explanation
Think of 2 coils sitting near each other. When current flows through the first coil, it creates a magnetic field — like an invisible bubble of energy around the wire. If that bubble reaches the second coil, it pushes a voltage into it, even though nothing is physically touching. Mutual inductance is just a number that tells you how effective that invisible push is: a high number means strong coupling, a low number means most of the magnetic energy misses the second coil entirely.
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Table of Contents
Diagram
Mutual Inductance Calculator
How to Use This Calculator
- Select your calculation mode from the dropdown — choose what you want to solve for (M, k, induced voltage, flux, secondary inductance, or coaxial solenoid M).
- Enter the required input values that appear for your selected mode — these may include self-inductance values (H), coupling coefficient, turns counts, current change rate (A/s), or solenoid geometry.
- Check your units — inductance in henries, current in amps, area in m², length in metres.
- Click Calculate to see your result.
📹 Video Walkthrough — Mutual Inductance Interactive Calculator
Mutual Inductance Interactive Visualizer
Watch magnetic flux lines couple between two coils as you adjust their inductances, coupling coefficient, and geometry. See real-time calculations for mutual inductance, induced voltage, and coupling status.
MUTUAL INDUCTANCE
10.95 mH
INDUCED VOLTAGE
-1.10 V
COUPLING STATUS
MODERATE
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Equations & Formulas
Use the formula below to calculate mutual inductance from the coupling coefficient and coil self-inductances.
Mutual Inductance from Coupling Coefficient
M = k√(L₁L₂)
M = mutual inductance (henries, H)
k = coupling coefficient (dimensionless, 0 ≤ k ≤ 1)
L₁ = primary coil self-inductance (H)
L₂ = secondary coil self-inductance (H)
Use the formula below to calculate induced voltage using Faraday's Law.
Induced Voltage (Faraday's Law)
V₂ = -M(dI₁/dt)
V₂ = voltage induced in secondary coil (volts, V)
M = mutual inductance (H)
dI₁/dt = rate of change of primary current (amperes per second, A/s)
The negative sign indicates Lenz's law (induced EMF opposes the change)
Use the formula below to calculate mutual flux linkage through the secondary coil.
Mutual Flux Linkage
Φ₂₁ = MI₁/N₂
Φ₂₁ = magnetic flux through secondary coil from primary current (webers, Wb)
M = mutual inductance (H)
I₁ = primary current (A)
N₂ = number of turns in secondary coil (dimensionless)
Use the formula below to calculate mutual inductance for coaxial solenoids.
Mutual Inductance for Coaxial Solenoids
M = (μN₁N₂A)/l
M = mutual inductance (H)
μ = permeability of core material (henries per meter, H/m)
N₁ = number of turns in primary solenoid (dimensionless)
N₂ = number of turns in secondary solenoid (dimensionless)
A = cross-sectional area (square meters, m²)
l = length of solenoid (meters, m)
Use the formula below to calculate coupling coefficient from measured inductances.
Coupling Coefficient
k = M/√(L₁L₂)
k = coupling coefficient (dimensionless, 0 ≤ k ≤ 1)
M = mutual inductance (H)
L₁, L₂ = self-inductances of the two coils (H)
Perfect coupling (k = 1) means all flux from one coil links the other; k = 0 means no coupling
Simple Example
Two coils with k = 0.5, L₁ = 0.04 H, and L₂ = 0.04 H.
M = k × √(L₁ × L₂) = 0.5 × √(0.04 × 0.04) = 0.5 × 0.04 = 0.02 H
If primary current then changes at 100 A/s, the induced secondary voltage is V₂ = −0.02 × 100 = −2 V.
Theory & Engineering Applications
Fundamental Electromagnetic Coupling Principles
Mutual inductance quantifies the magnetic coupling between two circuits through their shared magnetic flux. When alternating or time-varying current flows through a primary coil, it generates a changing magnetic field that induces electromotive force (EMF) in any nearby secondary coil. This phenomenon, governed by Faraday's law of electromagnetic induction, forms the operational basis for transformers, inductive coupling systems, and proximity sensors across electrical engineering disciplines.
The coupling coefficient k represents the fraction of magnetic flux from the primary coil that successfully links the secondary coil. In practice, k rarely equals unity due to geometric constraints, air gaps, core material limitations, and flux leakage paths. A non-obvious limitation arises in high-frequency applications: above certain frequencies determined by coil geometry and core material properties, skin effect and proximity effect redistribute current density within conductors, effectively reducing the mutual inductance from its low-frequency value by 15-30% in typical power electronics transformers operating above 100 kHz.
Core Material Selection and Permeability Effects
The permeability μ of the magnetic core material dramatically influences mutual inductance magnitude. Air-core transformers use μ₀ = 4π × 10⁻⁷ H/m (permeability of free space), yielding relatively weak coupling suitable for radio-frequency tuned circuits. Ferrite cores with relative permeabilities between 1,500 and 15,000 increase mutual inductance proportionally, enabling compact high-frequency transformer designs. Silicon steel laminations (μᵣ ≈ 4,000) dominate power-frequency transformers where eddy current losses must be minimized through lamination thickness optimization.
A critical non-textbook consideration: core saturation fundamentally limits mutual inductance validity. When magnetic flux density exceeds the core's saturation threshold (typically 1.2-1.8 Tesla for silicon steel, 0.3-0.5 Tesla for ferrites), permeability collapses nonlinearly, and mutual inductance becomes current-dependent rather than constant. This nonlinearity generates harmonic distortion in transformer secondary voltages and reduces power transfer efficiency by 5-20% depending on operating point.
Wireless Power Transfer Engineering
Modern inductive charging systems for electric vehicles and consumer electronics exploit resonant mutual inductance where both primary and secondary circuits are tuned to the same frequency. The quality factor Q and coupling coefficient k determine maximum achievable efficiency through the figure of merit k²Q. Automotive wireless charging pads operating at 85 kHz typically achieve k = 0.15-0.25 due to ground clearance constraints (150-250 mm air gaps), necessitating Q values above 200 to maintain 90% efficiency. Precise coil alignment mechanisms compensate for lateral misalignment, which degrades coupling coefficient by approximately 8% per centimeter of horizontal displacement in typical pad geometries.
Worked Example: Transformer Design Calculation
Problem: An engineer designs a step-down transformer for a switched-mode power supply. The primary coil has 847 turns with measured self-inductance L₁ = 3.76 mH. The secondary coil has 124 turns with L₂ = 82.3 μH. Measurements using an LCR meter show mutual inductance M = 467 μH. Calculate the coupling coefficient, determine coupling quality, predict the secondary voltage when primary current changes at 2,350 A/s, and calculate the flux through each secondary turn when primary current is 1.83 A.
Solution:
Step 1: Calculate coupling coefficient
Using k = M / √(L₁L₂):
L₁ = 3.76 mH = 3.76 × 10⁻³ H
L₂ = 82.3 μH = 82.3 × 10⁻⁶ H
M = 467 μH = 467 × 10⁻⁶ H
√(L₁L₂) = √(3.76 × 10⁻³ × 82.3 × 10⁻⁶) = √(3.094 × 10⁻⁷) = 5.562 × 10⁻⁴ H
k = 467 × 10⁻⁶ / 5.562 × 10⁻⁴ = 0.840
Step 2: Assess coupling quality
With k = 0.840, this transformer exhibits strong coupling (k greater than 0.7). This indicates minimal flux leakage and efficient magnetic circuit design, typical of ferrite-core high-frequency transformers with close winding coupling.
Step 3: Calculate induced voltage
Using V₂ = -M(dI₁/dt):
dI₁/dt = 2,350 A/s
V₂ = -(467 × 10⁻⁶) × 2,350 = -1.097 V
The magnitude of induced voltage is 1.10 V (negative sign indicates polarity per Lenz's law)
Step 4: Calculate flux per secondary turn
Using Φ₂₁ = MI₁/N₂:
I₁ = 1.83 A
N₂ = 124 turns
Φ₂₁ = (467 × 10⁻⁶ × 1.83) / 124 = 6.891 × 10⁻⁶ Wb = 6.89 μWb
Engineering Interpretation: The coupling coefficient of 0.840 confirms excellent magnetic coupling, justifying the transformer design for high-efficiency operation. The induced voltage calculation demonstrates that during rapid current transitions (as occur in switching converters), significant voltage transients appear across the secondary, requiring appropriate snubber or clamping circuits. The flux calculation reveals that each secondary turn intercepts 6.89 μWb, which can be used to verify core saturation margins by comparing against the core's cross-sectional area and maximum flux density specification.
Automotive Sensor Applications
Inductive position sensors in automotive throttle bodies, camshaft positioning systems, and ride height measurement employ mutual inductance variation with target proximity. A primary coil driven at 50-500 kHz creates an alternating magnetic field. Eddy currents induced in nearby metallic targets alter the effective mutual inductance between primary and secondary coils, producing amplitude or phase changes in the secondary voltage proportional to target distance. These sensors operate reliably across -40°C to +150°C, immune to contamination from oil, dirt, and moisture that would compromise optical or capacitive technologies.
For further exploration of electromagnetic design principles and related engineering calculations, visit the complete calculator library covering inductance, magnetic circuits, and electromagnetic compatibility topics.
Practical Applications
Scenario: Electric Vehicle Charging System Design
Rebecca, an automotive electrical engineer at an EV manufacturer, designs an inductive charging pad system that must deliver 11 kW across a 200 mm air gap. Her transmitter coil specifications show L₁ = 87.3 μH, and the receiver coil measures L₂ = 91.6 μH. Through electromagnetic simulation and prototype testing, she measures M = 14.2 μH between aligned coils. Using this calculator's coupling coefficient mode, Rebecca calculates k = 14.2 μH / ���(87.3 μH × 91.6 μH) = 0.159. This relatively low coupling (due to the large air gap required for ground clearance) tells her she must implement resonant compensation circuits with quality factors above 180 to achieve the target 92% efficiency. The calculation directly influences her component selection for compensation capacitors and informs tolerance requirements for coil positioning mechanisms that maintain alignment within ±25 mm to prevent efficiency degradation below specification.
Scenario: Industrial Transformer Verification
Marcus, a quality assurance technician at a transformer manufacturing facility, tests production units for a 480V-to-120V isolation transformer rated at 3.5 kVA. The design specifications call for primary inductance L₁ = 2.14 H and secondary inductance L₂ = 134 mH, with minimum coupling coefficient k ≥ 0.95 to meet efficiency requirements. Using an LCR bridge at 1 kHz with one winding open, he measures mutual inductance M = 506 mH. Entering these values into the calculator's coupling mode yields k = 0.946, which falls just below the specification threshold. This alerts Marcus to a potential core assembly defect—possibly incomplete seating of the E-I laminations creating a small air gap. He flags the unit for core inspection, discovering a 0.12 mm gap caused by burrs on lamination edges. After reassembly with deburred laminations, remeasurement shows M = 535 mH and k = 0.999, confirming the transformer now meets production requirements. This calculator enables rapid go/no-go testing without complex test equipment or lengthy analysis.
Scenario: RFID Antenna Optimization
Dr. Chen, a senior RF engineer developing contactless payment terminals, optimizes the reader antenna for maximum read range with ISO 14443 Type A cards operating at 13.56 MHz. Her reader coil design has N₁ = 4 turns wound on a ferrite sheet with measured L₁ = 1.82 μH. Payment cards contain standardized coils with approximately N₂ = 5 turns and L₂ = 2.1 μH. At the target reading distance of 35 mm, near-field electromagnetic coupling yields measured M = 127 nH between reader and card coils. Using this calculator, she determines k = 127 nH / √(1.82 μH × 2.1 μH) = 0.065, indicating weak coupling as expected for this operating distance. She then uses the induced voltage mode: with reader coil current changing at dI/dt = 2.8 × 10⁷ A/s (corresponding to the 13.56 MHz carrier modulation), the induced voltage in the card coil is V₂ = M × dI/dt = 127 nH × 2.8 × 10⁷ A/s = 3.56 V. This calculation confirms her design provides sufficient voltage to power the card's integrated circuit (minimum 1.8V required), validating the antenna geometry before committing to tooling for the injection-molded enclosure.
Frequently Asked Questions
What is the difference between mutual inductance and self-inductance? +
Why can't the coupling coefficient k exceed 1.0? +
How does coil separation distance affect mutual inductance? +
What causes the negative sign in the induced voltage equation V₂ = -M(dI₁/dt)? +
How does operating frequency affect mutual inductance measurement and behavior? +
Can mutual inductance be negative, and what does that mean? +
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About the Author
Robbie Dickson — Chief Engineer & Founder, FIRGELLI Automations
Robbie Dickson brings over two decades of engineering expertise to FIRGELLI Automations. With a distinguished career at Rolls-Royce, BMW, and Ford, he has deep expertise in mechanical systems, actuator technology, and precision engineering.
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