Q:

Does inductance depend on a bias field?

Considering that the energy $U$ stored in a magnetic field $B$ is scales as $U \propto B^2$.
It is clear that if we have two identical coils (same number of turns, area and length), but with opposing currents of equal magnitude in a strong bias field:

one will have a more energetic configuration than the other.
On the other hand, the energy stored in an inductor is given by $U = LI^2 / 2$.
This suggests that that $L_{A} \neq L_{B}$. But from experience, I have never seen this effect (in resonant circuits and so), and I am lead to believe that $L_{A} = L_{B}$.
So that's the question, is $L_{A}$ equal to $L_{B}$? What am I missing?

A:

The work done to energize an inductor from its state of zero current $i=o$  to having current $i=I$ is the integral $W_m=\int_0^I Lidi = \frac{1}{2}LI^2$. If there were no other fields present while changing the current this then is equal to the inductor's magnetic energy: $\mathcal{E}_m =W_m$. This formula says nothing about other possible sources of energy that may be present, i.e., kinetic, potential, or even other magnetic field energies and their interactions. If, for example, what you call bias field $B_0$ is generated by another coil and its current then during the increase of the current from $i=0$ to $i=I$ you are also inducing a current opposing this field (Lenz) through the mutual inductance of the coils. To maintain the biasing field work has to be put in the "biasing" coil and its current source, etc.

