I'm reading chapter 5.4.2 of Griffiths' Introduction to Electrodynamics, and I'm having trouble understanding some of the reasoning and some of the results he obtains.

He starts by considering a current sheet and a wafer-thin pillbox straddling the sheet (a current sheet with surface current $\mathbf{K}$ to be more precise); then he applies the integral form of $$\nabla \cdot \mathbf{B} = 0\quad \Leftrightarrow\quad \iint_\Sigma\mathbf{B}\cdot d\boldsymbol{\sigma} = 0,$$ where $\Sigma$ is any closed surface. He gets, by requiring that the flux through the closed surface equals zero, the condition on the perpendicular components of $\mathbf{B}$: $$B^{\perp}_{{\rm above}} = B^{\perp}_{{\rm below}}.$$ Now my problem is this (he omits any calculations, so I will try to give my explanation here): I first define a normal unit vector always pointing outward from the surface of the pillbox. Do we need to work under the assumption that the height of the pillbox $h \to 0$? Under this assumption, I still can only obtain his result if I assume the $B^{\perp}_{{\rm below}}$ component points upwards, the same as for the other component. Now, is there a reason behind this? I can't actually see why he assumes a priori that those components both point upwards, to be honest.

He then goes on to consider the tangential components of the magnetic field. He considers a closed loop (of length $l$ and a height which I guess is an infinitesimal of higher order with respect to $l$) perpendicular to $\mathbf{K}$ straddling the surface current: $$ \oint \mathbf{B} \cdot \mathbf{d}l= \left(B^{\parallel}_{{\rm above}}-B^{\parallel}_{{\rm below}}\right)l = \mu_0 Kl.$$ Again, why does he assume a priori that the two tangential components run in the same direction (otherwise, he would not get the minus sign from the dot product)? Didn't we explicitly (in an example really similar to this) obtain that the magnetic field due to a current sheet runs parallel to the surface, perpendicular to $\mathbf{K}$, but in opposite directions above and below the surface current?

Then he considers the same loop running parallel to $\mathbf{K}$ and derives the continuity of the tangential component parallel to the current. This is ok. The author goes on to consider the vector potential $\mathbf{A}$ and affirms that the vector potential is continuous across any boundary and that the condition $\nabla \cdot \mathbf{A} = 0$ guarantees this for the perpendicular component, by the same reasoning he used with $\mathbf{B}$, (I'm ok with this statement), and then finally he affirms that the following equalities: $$ \oint \mathbf{A} \cdot \mathbf{d}l = \Phi_{B\,{\rm surface}}$$ mean that the tangential components are continuous. I can't seem to grasp the correlation between the chain of equalities and this last result.

If someone could help me clarify these doubts, I'd be very grateful.


1 Answer 1


There is no such assumption that either the normal or the tangential components have any particular direction but you must assign a particular direction to the surface normals or the line tangents.

For example, in the case of the surface integral $\int \mathbf B d\boldsymbol{\sigma}$ all is needed that the dimension, "$h$", of the "pillbox" perpendicular to the local normal be infinitesimal, here the length of the cylinder. Then, assuming that $\mathbf B$ is finite everywhere, as $h\to 0$ the piece of the surface integral $\mathbf B d\boldsymbol{\sigma}$ around the cylinder goes to zero, and what remains of it is just the top and bottom flat pieces to which $\mathbf B$ is normal: $B_{n1} d\sigma_1+B_{n2} d\sigma_2$. But the normal is pointing outward, therefore $d\sigma_1=-d\sigma_2$ and all is left is that $B_{n1}-B_{n2}=0$. A similar argument goes for the line integral to show that the tangential components can be continuous if the contour integral is zero, or if the latter is not zero then their jump is the surface source density.

  • $\begingroup$ Wait, everything’s ok, except for one thing: the minus sign you wrote comes from the dot product between the field and the unit surface normal, that we defined at each point to be pointing outward, nevertheless I think we must take into consideration the nature of our magnetic field. I thought that maybe we could’ve had a field which flows outside the small pillbox at any given point, such as the electrical field generated by a point charge place at the center of the pillbox. In that case, the dot product would always be positive, but then again no magnetic monopoles exist so problem solved $\endgroup$ Aug 10, 2023 at 9:10
  • $\begingroup$ What I’m trying to say is that the minus sign you wrote doesn’t come from the direction of the normal unit vector which we assigned, but it also depends upon the direction of $\vec{B}$ $\endgroup$ Aug 10, 2023 at 9:11
  • 1
    $\begingroup$ $B_n$, the normal component of the vector $\mathbf B$ is not an "absolute" thing, it is defined relative to something, in this case it is defined relative to the normal vector that you can define any way you wish but it must be defined consistently. Here for the surface integral it must be continuous as you go around the closed surface of the pillbox. Then because the scalar product is also continuous and the the integral is zero, then $B_n$ must also be continuous. Same idea for the contour integral. A famous example is the Mobius strip where you cannot not do this. $\endgroup$
    – hyportnex
    Aug 10, 2023 at 13:37
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    $\begingroup$ The important physical view you should see is that if the field is not continuous at some surface then there is something to make it to jump. Smooth out the sudden transition at the interface, then everything is continuous. Now "slowly" make the transition shaper, the field is always continuous, but when it becomes sudden there must be something at the interface to sustain the jump in the field. If there is nothing there then there cannot be any jump. If the surface (or line) integral represents the field then this affects the normal (or tangential) components only. $\endgroup$
    – hyportnex
    Aug 10, 2023 at 13:46

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