I notice that the equations of Permittivity and Permeability in a linear medium are exactly opposite of each other. One is $$\mathbf{D} \equiv \varepsilon \mathbf{E}$$ while the other one is $$\mathbf{B} \equiv \mu \mathbf{H}$$ For the electrostatic, the $$\varepsilon$$ multiples directly to the electric field in free space, but the $$ \mu$$ multiples directly to the magnetic field in a medium. Why is it the case? My current guess is that it is because the electric field is weaker in the present of a medium(because the induced charge partly cancelled out the external electric field), but the magnetic field, depending on the medium, may become stronger than the external magnetic field

  • $\begingroup$ H is not the magnetic field in a medium. It is just the magnetic field. In vacuum too. $\endgroup$
    – nasu
    Oct 31, 2021 at 14:57
  • $\begingroup$ But like D, H is something that we cannot measure directly as field and stay the same even if I change the medium. Am I right? $\endgroup$
    – galoischan
    Oct 31, 2021 at 16:40

1 Answer 1


This is really a matter of convention. We could have defined $\mu' = 1/\mu$ and defined $\mathbf{H} = \mu' \mathbf{B}$. It turns out that although $\mathbf{E}$ and $\mathbf{B}$ are the "fundamental" electromagnetic quantities, $\mathbf{H}$ is really more like the magnetic counterpart to $\mathbf{E}$, and $\mathbf{B}$ is more like the magnetic counterpart to $\mathbf{D}$. Consequently, it makes sense to keep the definitions in parallel with the $\mathbf{E}$ field.

One way to see this parallel is to notice the symmetry between $\mathbf{D}$ and $\mathbf{B}$ and between $\mathbf{E}$ and $\mathbf{H}$ in the macroscopic Maxwell's equations with magnetic charges and currents:

$\nabla \cdot \mathbf{D} = \rho_{e}$

$\nabla \cdot \mathbf{B} = \rho_{m}$

$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mathbf{J}_m$

$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J}_e$

In an experiment, you also tend control to control $\mathbf{E}$ and $\mathbf{H}$ and not $\mathbf{D}$ and $\mathbf{B}$. e.g. if you use voltages and currents to generate electric and magnetic fields. Voltage directly determines $\mathbf{E}$ and the free current that you supply determines $\mathbf{H}$, not $\mathbf{B}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.