Electromagnetic Coupling Simulator Back
Electromagnetic Simulator

Electromagnetic Coupling Simulator

Visualize two-coil electromagnetic coupling via mutual inductance $M = k\sqrt{L_1 L_2}$. Switch between transformer, wireless power transfer, and induction heating modes to see current and voltage waveforms in real time.

Mode
Coil Parameters
L₁ (Primary Inductance)
mH
L₂ (Secondary Inductance)
mH
k (Coupling Coefficient)
R₁ (Primary Resistance)
Ω
R₂ (Secondary Resistance)
Ω
V₀ (Source Voltage)
V
Frequency
Hz
Results
Results
0.90 mH
Mutual Inductance M
81.0%
Transfer Efficiency η
Secondary Power P₂
Resonant Freq. f_res
Coupling
Wave
Theory & Key Formulas
$$M = k\sqrt{L_1 L_2}$$ $$V_2 = -M \frac{dI_1}{dt}$$ $$\eta \approx \frac{k^2 \omega^2 L_1 L_2}{R_1 Z_2 + k^2 \omega^2 L_1 L_2}$$

What is Electromagnetic Coupling?

🙋
What exactly is "mutual inductance"? I see it's the main output of this simulator, but what does it physically represent?
🎓
Basically, it's a measure of how well two coils can "talk" to each other magnetically. When the current in the primary coil (L₁) changes, it creates a changing magnetic field. Mutual inductance (M) quantifies how much of that changing field cuts through the secondary coil (L₂) to induce a voltage. In practice, a higher M means a stronger induced signal. Try moving the k slider above from 0 to 1 and watch how M changes instantly.
🙋
Wait, really? So the coupling coefficient k is like the efficiency of this magnetic link? What happens if I set L₁ and L₂ to very different values?
🎓
Exactly! k is the magnetic link efficiency, ranging from 0 (no link) to 1 (perfect link). The formula $M = k\sqrt{L_1 L_2}$ shows M depends on both k AND the geometric mean of the two inductances. For instance, if you set L₁ very high and L₂ very low using the sliders, the simulator will show a moderate M, but the energy transfer might still be poor because of the impedance mismatch. A common case is a wireless phone charger (high L₁) powering a small receiver coil (lower L₂).
🙋
That makes sense. So what's the role of the frequency parameter? And why do we have resistances R₁ and R₂ in the simulator if they're not in the M formula?
🎓
Great question! Frequency is crucial because the induced voltage $V_2 = -M \frac{dI_1}{dt}$ depends on how fast the current changes. Higher frequency means a faster change, leading to a higher induced voltage V₂. The resistances R₁ and R₂ represent real-world losses—copper wire isn't perfect. They don't affect M, but they drastically affect the efficiency (η) of power transfer. When you change R₁ or R₂, you'll see how they drain useful energy as heat, which is a major design challenge.

Physical Model & Key Equations

The core relationship defining the strength of magnetic linkage between two coils. The mutual inductance M is proportional to the geometric mean of their self-inductances, scaled by the coupling coefficient.

$$M = k\sqrt{L_1 L_2}$$

M: Mutual Inductance (Henries). k: Coupling Coefficient (0 to 1). L₁, L₂: Self-Inductance of primary and secondary coils (Henries).

This equation describes the induced voltage in the secondary coil. It is the fundamental law of electromagnetic induction (Faraday's Law) applied to two coupled circuits.

$$V_2 = -M \frac{dI_1}{dt}$$

V₂: Induced voltage in the secondary coil (Volts). dI₁/dt: Rate of change of current in the primary coil (Amps/second). The negative sign indicates Lenz's Law (the induced voltage opposes the change in current).

An approximate formula for the power transfer efficiency (η). It shows how efficiency depends on the square of k and the operating frequency (ω = 2πf), but is reduced by resistive losses.

$$\eta \approx \frac{k^2 \omega^2 L_1 L_2}{R_1 Z_2 + k^2 \omega^2 L_1 L_2}$$

η: Efficiency. ω: Angular frequency (rad/s) = 2π × Frequency. Z₂: Impedance of the secondary circuit, heavily influenced by R₂. This shows why high k and high frequency are sought for wireless power, but resistances are the enemy.

Frequently Asked Questions

k represents the shared proportion of magnetic flux and can be set in the range of 0 to 1. When k=1, it is an ideal transformer (full magnetic flux coupling). The smaller k is, the more leakage flux increases, reducing power transmission efficiency to the secondary side. In wireless power transfer, k is typically around 0.1 to 0.5.
Transformer mode uses tight coupling (k≈1) for high-efficiency power conversion, wireless power transfer mode uses loose coupling (k<0.5) for contactless transmission, and induction heating mode simulates heat generation from eddy currents by using the secondary coil as a short-circuit load. The load resistance and frequency range are automatically adjusted depending on the mode.
Divergence mainly occurs when the coupling coefficient k is too large or when the load resistance is extremely small. First, reduce k to 0.5 or lower, set the load resistance to 1Ω or higher, confirm stable operation, and then gradually adjust the parameters toward the target values. The automatic time-step adjustment function is also available.
For general applications, both L₁ and L₂ are recommended to be in the range of 1μH to 10mH. In transformer mode, L₁=L₂≈1mH is typical; in wireless power transfer, 10 to 100μH; and in induction heating, 1 to 10μH is a guideline. If the values are too small, current becomes excessive, and if too large, the response becomes slow.

Real-World Applications

Wireless Smartphone Charging: This uses loose coupling (k ~ 0.2-0.5). The transmitter pad (L₁) and the receiver coil inside the phone (L₂) are separated by air and a plastic case. Designers optimize the frequency and coil geometry to maximize M and efficiency despite the low k.

Power Transformers: The core of the electrical grid. Here, coils are wound on a shared iron core to achieve very high coupling (k > 0.99). This allows efficient voltage stepping (e.g., 11kV down to 240V) with minimal energy loss, as predicted by the high M value.

Inductive Heating: Used in industrial melting and kitchen induction cooktops. A high-frequency alternating current in L₁ induces large eddy currents in a metal workpiece (which acts as a lossy L₂). The key is a moderate M coupled with high frequency to generate heat via the secondary's effective resistance.

Implantable Medical Devices: Charging pacemakers or neural implants wirelessly through the skin. This is an extreme challenge because k is very low (coils are small and far apart). Engineers use resonant tuning at specific frequencies to boost effective power transfer, as suggested by the efficiency equation.

Common Misconceptions and Points to Note

First, understand that "the coupling coefficient k is not a fixed value." While you can easily change it with a slider in the simulator, in actual design it fluctuates significantly based on coil shape, orientation, distance, and surrounding metal (like shields or cases). For example, simply shifting a smartphone slightly on a wireless charging pad can drop k from 0.3 to 0.15, potentially halving the efficiency. In practice, it's crucial to evaluate this "misalignment tolerance" beforehand through simulation.

Next, beware of the pitfall that "the resonant frequency does not behave as calculated." The tool uses the simple resonant frequency determined by L and C, $$f_r = \frac{1}{2\pi\sqrt{LC}}$$, but real coils have parasitic capacitance (distributed capacitance) between windings. Especially at high frequencies, resonance occurs at a lower frequency than calculated. For instance, a design calculated for 1MHz might peak at 800kHz in actual measurement. Consider simulation results as a first approximation only; actual measurement with a prototype is essential.

Finally, do not finalize your design based on efficiency (η) alone. 90% efficiency is excellent, but the remaining 10% is mostly heat loss (copper loss) from coil resistance. This heat can degrade the coil's insulation or cause the device temperature to rise. For example, reducing R1 from 0.1Ω to 0.05Ω might increase efficiency by 2%, but requires thicker, more expensive copper wire. Judging the trade-offs between cost, size, and heat generation holistically is where engineering skill comes in.

How to Use

  1. Enter primary coil inductance L₁ (mH) in nL1 and coupling coefficient k (0–1) in nk, where k=1 represents perfect coupling
  2. Input secondary coil inductance L₂ (mH) in nL2 and operating frequency (kHz) in sk to establish the two-coil system
  3. Click simulate to compute mutual inductance M = k√(L₁L₂), transfer efficiency η, secondary power P₂, and resonant frequency f_res = 1/(2π√(L₁L₂C))

Worked Example

For a wireless power transfer system: L₁ = 15 mH (primary coil), L₂ = 12 mH (secondary coil), k = 0.85 (air-gap coupling), operating frequency = 125 kHz. Calculated mutual inductance M = 0.85√(15×12) = 10.74 mH. At 50W input with η = 78%, secondary power output P₂ = 39 W. Resonant frequency f_res ≈ 145 kHz for matched impedance.

Practical Notes

  1. Ferrite core transformers achieve k > 0.95; air-core wireless chargers typically k = 0.6–0.8 depending on coil separation and alignment
  2. Secondary power P₂ degrades nonlinearly with misalignment; verify k value experimentally before production design
  3. For inductive heating and RFID systems, operate 10–15% above resonant frequency to avoid Q-factor peaking and component stress