I'm trying to figure out the reasoning of equation (5.74) and why it is stated that "this time the tangential component changes". Write the vectors as $\vec{B}^{\perp}_{above} = B^{\perp}_{above}\hat{z}$ and $\vec{B}^{\perp}_{below} = B^{\perp}_{below}\hat{z}$ then if $$\oint \vec{B} \cdot d\vec{a} = 0$$ we have $$\oint \vec{B} \cdot d\vec{a} = \int |\vec{B}^{\perp}_{above}||d\vec{a}| - \int|\vec{B}^{\perp}_{below}|d\vec{a}| = |\vec{B}^{\perp}_{above}|A - |\vec{B}^{\perp}_{below}|A = 0$$ hence $$|\vec{B}^{\perp}_{above}| = |\vec{B}^{\perp}_{below}|.$$ This is then not the same as $B^{\perp}_{above} = B^{\perp}_{below}$, since $B^{\perp}_{above} = B^{\perp}_{below} \implies |\vec{B}^{\perp}_{above}| = |\vec{B}^{\perp}_{below}|$ but the converse implication is not necessarily true. If in the text they meant that $B^{\perp}_{above} = |\vec{B}^{\perp}_{above}|$, then it isn't true as stated that only the tangetial component changes since we could still have $\vec{B}^{\perp}_{above} = -\vec{B}^{\perp}_{below}$ as the magnetic field moves accross the surface. Does anyone know where I am going wrong in my reasoning? Thanks.
2 Answers
You did it wrong because $$\oint \vec{B} \cdot d\vec{a} \ne \int |\vec{B}^{\perp}_{above}||d\vec{a}| - \int|\vec{B}^{\perp}_{below}|d\vec{a}|$$
Above the surface, we have $d\vec{a}=da\hat{z}$ and below we have $d\vec{a}=da(-\hat{z})$.
So $$\oint \vec{B} \cdot d\vec{a} = \int \vec{B}_{above}\cdot \hat{z}d\vec{a} - \int\vec{B}_{below}\cdot\hat{z}d\vec{a}$$ $$=B_{above}^\perp A-B_{below}^\perp A$$
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$\begingroup$ Thanks for your answer. I assume that the step you omitted is $$ \int B^{\perp}_{above}da - \int B^{\perp}_{below}da = B^{\perp}_{above}\int da - B^{\perp}_{below}\int da = (B^{\perp}_{above}-B^{\perp}_{below})A.$$ Is the reason we can take $B^{\perp}_{above}$ and $B^{\perp}_{below}$ outside the integral, because we can choose $A$ small enough so that $\vec{B}$ is uniform over the pillbox? $\endgroup$– user114445Commented Oct 27, 2016 at 17:50
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Let there be small area of transition layer through which the property of the medium is assumed to change rapidly but continuously amidst the discontinuous surface.
Now, consider a small cylinder (in op's case it is a pill box) bounded by small areas $\delta A_1$ and $\delta A_2$ along each side of the surface with height $\delta h\,.$
Now, by assumption, we can take that $\mathbf B$ is continuous throughout the cylinder.
So, $$\int\textrm{div}~\mathbf B~\mathrm dV = \int \mathbf B\cdot \mathbf n~\mathrm dS = 0 ~~~~~~\textrm{Gauss' Law} $$
where $\mathbf n$ is the unit vector normal to the infinitesimal surface $\mathrm dS$ in the cylinder.
Since, $\delta A_1$ and $\delta A_2$ are small, $\mathbf B$ can be considered to have constant values $\mathbf B^{(1)}$ and $\mathbf B^{(2)}$ on them.
So, the integral becomes
$$\mathbf B^{(1)}\cdot \mathbf n_1~\delta A_1 + \mathbf B^{(2)}\cdot \mathbf n_2~\delta A_2 + \textrm{contribution from the walls} = 0 $$
As the cylinder shrinks, so does $\delta h\to 0\,.$
So, in the limit $$\left(\mathbf B^{(1)}\cdot \mathbf n_1 + \mathbf B^{(2)}\cdot \mathbf n_2\right)\delta A = 0,$$ where $\delta A$ is the area of the surface intersected by the cylinder.
Now, we can take $\mathbf n_1 = -~\mathbf n_{12} $ and $\mathbf n_2 = \mathbf n_{12}\,.$
So, finally the integral can be written as
$$\mathbf n_{12}\cdot\left(-~\mathbf B^{(1)} + \mathbf B^{(2)} \right) = 0\,.\tag{I}$$
Does anyone know where I am going wrong in my reasoning?
What component did you work with? It is not clear from your work. The dot product of $\mathbf B$ with the normals $\mathbf n_1$ and $\mathbf n_2$ on the respective bounded areas above evidently points out that the integral is concerned with the normal component of the magnetic field $\mathbf B;$ this treatment is absent in your work.
References:
$\bullet$ Principles of Optics by Born, Wolf.