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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 $$\mu_{1}B_{1}^{\perp}=\mu_{2}B_{2}^{\perp}$$

Now when I asked about this derivation to my instructor he himself got confused and had no answers why the above result is not true.

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

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  • $\begingroup$ How the heck is this a homework question. This is something that I got confused into myself !! $\endgroup$
    – Ankit
    Commented Jan 24 at 13:41
  • 2
    $\begingroup$ In what differential equation have you divided both sides by $\mu$ (which $\mu$?) to get this result? $\endgroup$ Commented Jan 24 at 13:44
  • $\begingroup$ 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. $\endgroup$
    – Ankit
    Commented Jan 24 at 18:04
  • $\begingroup$ I'm voting to reopen $\endgroup$
    – basics
    Commented Jan 24 at 18:09
  • $\begingroup$ You can find a note here basics.altervista.org/jump-conditions-for-em-field $\endgroup$
    – basics
    Commented Jan 24 at 18:35

2 Answers 2

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$\mu^{-1}\vec\nabla \cdot \vec B = 0$ 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 $B_1^\perp=B_2^\perp$.

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From integral form of Maxwell's equations, you get

\begin{equation} \begin{aligned} d_{n2} - d_{n1} & = \sigma \\ b_{n2} - b_{n1} & = 0 \ , \end{aligned}\end{equation} 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{equation} \begin{aligned} d_{n2} & = d_{n1} \\ b_{n2} & = b_{n1} \ , \end{aligned} \end{equation}

or

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

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