When designing coupled magnetics, flyback transformers, or wireless power links, the equation of mutual inductance is the mathematical bridge between physical coil geometry and observable circuit behavior. At its core, mutual inductance (M) defines how a changing current in one coil induces a voltage in a nearby coil. The two governing formulas you will use on the bench are the physical coupling form, M = k√(L1L2), and the circuit behavior form, v2 = M(di1/dt). Understanding how to manipulate these equations, track the units, and recognize where the math diverges from reality is what separates a textbook student from a practicing engineer.

The Core Equation of Mutual Inductance and Symbol Definitions

Mutual inductance is fundamentally a measure of magnetic flux linkage between two distinct circuits. According to Georgia State University's HyperPhysics, it can be defined by the flux passing through the secondary coil per unit of current in the primary coil, or by the coupling coefficient relating their self-inductances.

Symbol Parameter Standard SI Unit Practical Bench Unit
M Mutual Inductance Henry (H) μH, mH, nH
k Coupling Coefficient Dimensionless (0 to 1) Percentage (%)
L1, L2 Self-Inductance of Coil 1 and 2 Henry (H) μH, mH
v2 Induced Voltage in Coil 2 Volt (V) mV, V
i1 Current in Coil 1 Ampere (A) mA, A
t Time Second (s) μs, ns
N2 Number of Turns in Coil 2 Dimensionless Turns
Φ12 Magnetic Flux from Coil 1 linking Coil 2 Weber (Wb) μWb, mWb

Rearranged Forms: Solving for Every Variable

On the bench, you rarely need to solve for M directly. Usually, you have a target induced voltage and need to find the required switching speed (di/dt), or you have measured M and need to determine the physical coupling coefficient (k). Here are the algebraic rearrangements of the primary formulas:

From the Coupling Equation: M = k√(L1L2)

  • Solve for Coupling Coefficient: k = M / √(L1L2)
  • Solve for Primary Inductance: L1 = M2 / (k2L2)
  • Solve for Secondary Inductance: L2 = M2 / (k2L1)

From the Circuit Equation: v2 = M(di1/dt)

  • Solve for Mutual Inductance: M = v2 / (di1/dt)
  • Solve for Current Slew Rate: di1/dt = v2 / M
  • Solve for Induced Voltage: v2 = M × (di1/dt)

Assumptions, Limits, and Unit Mistakes That Break the Math

The equation of mutual inductance is elegant, but it relies on strict physical assumptions. If you violate these, your spreadsheet math will not match your oscilloscope traces.

When the Formula Applies (and When It Doesn't)

  1. Linear Magnetic Materials: The formula assumes the core permeability (μ) is constant. If you drive a ferrite core into saturation, L1 and L2 collapse, and M drops non-linearly. The math only holds in the linear region of the B-H curve.
  2. Stationary Coils: The equation accounts for transformer EMF (changing current). It does not account for motional EMF. If the coils are physically moving relative to one another (like in a generator), you must add the v × B Lorentz force terms.
  3. Sinusoidal or Piecewise Linear Waveforms: For complex, noisy switching waveforms (like a hard-switching MOSFET with parasitic ringing), you must use the derivative di/dt at the exact instant in time you are measuring, not an average over the switching period.

Unit Mistakes That Break the Calculation

The most common bench error is mixing SI prefixes. According to standard magnetics design guides from Coilcraft, inductance is usually in μH and time in μs.

The Prefix Trap: If your inductance is in μH (10-6) and your time is in μs (10-6), the prefixes cancel out perfectly when calculating di/dt, leaving you with Volts. However, if you measure current in milliAmps (mA) but time in microseconds (μs), your di/dt is off by a factor of 1,000. Always convert everything to base SI units (Henries, Amperes, Seconds, Volts) before plugging numbers into the formula, then convert back.

What a Realistic Answer Magnitude Looks Like

  • Power Grid Transformers: M is in the range of Henries (H). High inductance, low frequency (50/60Hz).
  • Switch-Mode Power Supplies (SMPS): M is typically in microHenries (μH). Fast switching (100kHz - 1MHz) compensates for low inductance.
  • PCB Traces and RF: Parasitic mutual inductance is in nanoHenries (nH) or picoHenries (pH). At GHz frequencies, even 50pH of mutual inductance can cause massive crosstalk.

Solved Problems with Strict Unit Tracking

Let's walk through two distinct problems, tracking every unit to ensure the math holds up.

Problem 1: Flux Linkage Definition

Given: A primary coil carries a steady current I1 = 0.5 A. This current generates a magnetic flux that links a secondary coil. The secondary coil has N2 = 200 turns, and the flux passing through each turn of the secondary is Φ12 = 15 μWb.

Find: The mutual inductance M.

Step-by-Step Solution:

  1. Convert flux to base units: Φ12 = 15 × 10-6 Wb (or V·s).
  2. Apply the flux linkage equation: M = (N2 × Φ12) / I1
  3. Substitute values: M = (200 × 15 × 10-6 V·s) / 0.5 A
  4. Calculate numerator: 3000 × 10-6 V·s
  5. Divide by current: (3000 × 10-6 V·s) / 0.5 A = 6000 × 10-6 (V·s)/A
  6. Recognize that 1 (V·s)/A = 1 Henry. Therefore, M = 6000 μH, or 6 mH.

Problem 2: Coupling and Induced Voltage

Given: Two RF coils on a test board. L1 = 40 μH, L2 = 60 μH. The measured coupling coefficient is k = 0.75. The primary current switches from 0 A to 5 A in 2 μs.

Find: The mutual inductance M and the peak induced voltage v2.

Step-by-Step Solution:

  1. Convert inductances: L1 = 40 × 10-6 H, L2 = 60 × 10-6 H.
  2. Calculate M: M = 0.75 × √(40 × 10-6 × 60 × 10-6)
  3. Multiply inside the root: 2400 × 10-12 H2
  4. Take the square root: 48.99 × 10-6 H
  5. Multiply by k: M = 0.75 × 48.99 μH = 36.74 μH (or 36.74 × 10-6 H).
  6. Calculate di1/dt: Δi = 5 A - 0 A = 5 A. Δt = 2 × 10-6 s.
  7. di1/dt = 5 A / (2 × 10-6 s) = 2.5 × 106 A/s.
  8. Calculate v2: v2 = M × (di1/dt)
  9. v2 = (36.74 × 10-6 H) × (2.5 × 106 A/s)
  10. The 10-6 and 106 cancel out perfectly. v2 = 36.74 × 2.5 = 91.85 V.

Real-World Bench Scenario: Designing a Wireless Power Transfer Coil

Textbook equations assume ideal conditions. Here is what happens when you take the equation of mutual inductance to the workbench to design a 100 kHz Qi-style wireless power transfer (WPT) prototype.

The Setup

We needed to design a loosely coupled transformer to step a 12V TX inverter output down to a 5V RX rectified output. We wound two identical planar spiral coils on TDK PC40 ferrite sheets.

  • Measured L1 = 12 μH
  • Measured L2 = 12 μH
  • Target coupling coefficient at 10mm air gap: k = 0.4

The Numbers

Using the coupling equation, our target mutual inductance was:
M = 0.4 × √(12 μH × 12 μH) = 4.8 μH.

To get a peak induced voltage of 5V on the secondary (ignoring rectifier drops for a moment), we rearranged the circuit equation to find the required primary current slew rate:
di1/dt = 5V / 4.8 μH = 1.04 × 106 A/s (or roughly 1 A/μs). Our half-bridge inverter was easily capable of this switching speed.

The Outcome

We hooked up the oscilloscope, powered the TX coil, and probed the open-circuit RX coil. Instead of the calculated 5V peak, the scope showed a pathetic 1.2V peak. The math was correct, but the physical reality had diverged from the model.

What Went Wrong

We had violated the assumption of a linear, isolated magnetic environment. The RX coil was mounted directly on top of a solid copper ground plane on the PCB. The changing magnetic flux from the TX coil didn't just link the RX coil; it also passed through the solid copper plane.

By Faraday's Law, this induced massive eddy currents in the copper pour. These eddy currents generated their own opposing magnetic fields, effectively acting as a shorted tertiary coil. This destroyed the effective coupling coefficient (k dropped from 0.4 to roughly 0.08) and lowered the apparent L2 due to the reflected impedance of the shorted turn.

The Fix: We took a Dremel tool and slotted the ground plane beneath the RX coil in a star pattern, breaking the continuous conductive loops while maintaining DC grounding. We re-measured the open-circuit voltage, and it jumped to 4.6V—close enough to our 5V target once accounting for minor misalignment losses. As noted in Electronics Tutorials, mutual inductance is highly sensitive to the physical medium between and around the coils; always account for nearby conductive structures in high-frequency designs.