# Confused in one of the boundary conditions in Electrodynamics

I am confused about the derivation of the boundary condition relating the perpendicular components of magnetic field across the boundary.

In my optics course, the instructor used the fact that $$\epsilon$$ might vary spatially. And so he used this equation

$$\vec{\nabla} \cdot(\vec D)=\rho$$

And then took a closed surface to do the integral which resulted in $$\epsilon_{1}E_{1}^\perp = \epsilon_{2}E_{2}^\perp$$.

Now for the magnetic component he used similar approach and used this equation $$\vec{\nabla}\cdot{\vec B}=0~~~~~~~ ......(1)$$

And got $$B_{1}^\perp=B_{2}^{\perp}$$

But then I divided the equation (1) both sides by $$\mu$$ and then got the relation $$\mu_{1}B_{1}^{\perp}=\mu_{2}B_{2}^{\perp}$$

So can someone here point out why it's not true ?

• How the heck is this a homework question. This is something that I got confused into myself !! Commented Jan 24 at 13:41
• In what differential equation have you divided both sides by $\mu$ (which $\mu$?) to get this result? Commented Jan 24 at 13:44
• I don't think this question needs any edit. I just want to know why I can't apply the second method to reach the last relationship. Commented Jan 24 at 18:04
• I'm voting to reopen Commented Jan 24 at 18:09
• You can find a note here basics.altervista.org/jump-conditions-for-em-field Commented Jan 24 at 18:35

$$\mu^{-1}\vec\nabla \cdot \vec B = 0$$ does not imply $$\vec \nabla \cdot (\mu^{-1}\vec B)=0$$. In moving $$\mu$$ inside the divergence operator, you are assuming that $$\mu$$ is uniform. Note that $$\require{cancel} \vec\nabla \cdot(\mu^{-1}\vec B)=\vec\nabla (\mu^{-1})\cdot \vec B+\mu^{-1}\cancelto{0}{ \vec\nabla\cdot \vec B}=\vec\nabla (\mu^{-1})\cdot \vec B$$ The right-hand side is not zero in general because $$\mu$$ is a function of position.

At the interface of two media with different permeabilities, $$\mu$$ has a large discontinuity. This means it has a very large gradient here: in fact if we are assuming the two media are separated by an abrupt boundary, $$\vec \nabla \mu^{-1}$$ goes to infinity and must be described by a delta function. You can integrate and apply Gauss' theorem to $$\vec\nabla \cdot(\mu^{-1}\vec B)=\vec\nabla (\mu^{-1})\cdot \vec B$$ if you'd like, but carefully dealing with the integral of the right-hand side will still produce $$B_1^\perp=B_2^\perp$$.

From integral form of Maxwell's equations, you get

\begin{aligned} d_{n2} - d_{n1} & = \sigma \\ b_{n2} - b_{n1} & = 0 \ , \end{aligned} being $$\sigma$$ the surface charge density. If surface charge density is zero, and considering a surface between two linear isotropic non-dispersive media, described by the constitutive equations $$\mathbf{d}_k = \varepsilon_k \mathbf{e}_k$$ and $$\mathbf{b}_k = \mu_k \mathbf{h}_k$$, it's possible to write

\begin{aligned} d_{n2} & = d_{n1} \\ b_{n2} & = b_{n1} \ , \end{aligned}

or

\begin{aligned} \varepsilon_2 e_{n2} & = \varepsilon_1 e_{n1} \\ \mu_2 h_{n2} & = \mu_1 h_{n1} \ . \end{aligned}