Mutual Inductance and the Coupling Coefficient
How much flux do they share?
Sweep the coupling coefficient k from 0 to 1. Of the flux lines the left coil makes, how many reach all the way through the right coil?
From self to mutual
A lone coil links the flux made by its own current. That is its self-inductance L. Put a second coil nearby and part of that flux threads the second coil too. How much links the second coil is the mutual inductance M.
k is the linking fraction
Only part of one coil’s flux passes through the other. The rest leaks away to the side as leakage flux. Define the linking fraction as the coupling coefficient k, and 0 ≤ k ≤ 1. At k = 0 the coils are magnetic strangers; at k = 1 every flux line of one coil threads the other — the ideal transformer.
Why the ceiling is √(L₁L₂)
The flux reaching the second coil cannot exceed what the first coil holds, and the same the other way around. Account for both directions of coupling and M stops at the geometric mean of the two inductances, √(L₁L₂). Hence M = k√(L₁L₂), with k telling you, between 0 and 1, how far up to that ceiling you have climbed.
Coupling seen in the voltage
In sinusoidal steady state the two coil voltages bind together as V₁ = jωL₁I₁ + jωMI₂ and V₂ = jωMI₁ + jωL₂I₂. The M term is the bridge between the two circuits — the larger M is, the more one current rides into the other’s voltage.
Back to the first screen
As you raised k from 0 to 1, leakage lines turned one by one into linking lines that wrap both coils. That fraction is exactly k, and M climbed only by k times the ceiling √(L₁L₂). At k = 1 every flux line threads both coils and M touches the ceiling. Coupling is, in the end, a question of how much flux is shared.