0
$\begingroup$

I'm failing to understand the derivation of the interface conditions for the tangential components of the magnetic field given her (based on d.j,griffiths)

Ampere's law in integral form is given as $$\oint_C \mathbf{H}\cdot d\mathbf{l}=\int_S(\mathbf{j_f}+\frac{\partial \mathbf{D}}{\partial t})\cdot d\mathbf{a}$$ where $\mathbf{j}_f$ is the current-density of free charge carriers. Now considering the gaussian rectangle below, in the limit $h\rightarrow 0$ gives $$H^\parallel_1 l-H^\parallel_2 l=I_f$$,where $I_f$ represents the enclosed current of free charges $I_f=\int_S \mathbf{j}_f\cdot d\mathbf{a}$ gaussian rectangle

fine.

Then the author says

...The free surface current is the product of a surface current density $\mathbf{K}_f$ and the width of the loop;...$H^\parallel_1 l-H^\parallel_2 l=K_f l$

This is what confuses me:Where is $\mathbf{K}_f$ coming from? We already have a free surface current density- and it was called $\mathbf{j}_f$. What's the difference between these two? And how can a surface current density $\mathbf{K}_f$ have the same units as a magnetic field (A/m)? This doens't look like a density to me.

$\endgroup$
  • $\begingroup$ $J_f$ is volume current, ie. the free charges if they were to stop for an instant would have nonzero volume density. $K_f$ is surface current ie., so the corresponding moving free charges would have zero volume density but not zero surface density. $\endgroup$ – hyportnex Feb 12 at 16:51
  • $\begingroup$ Are you sure? S.I. units of D are C/m^2 so the time-derivative of D has units of A/m^2 and J_f has to match these units..isn't that a surface-current-density? $\endgroup$ – OD IUM Feb 12 at 16:58
  • $\begingroup$ $J_f$ represents current (charges per second) passing through a unti surface, but if the charges that are moving at speed $v$ stopped for an instant ($\delta t$ seconds) then they would represent a volume of charges $v\delta t dA$ passing through that unit surface $dA$. Similarly for $K_f$ but now charges confined to a surface passing through a unit length of a line $\endgroup$ – hyportnex Feb 12 at 17:10
  • $\begingroup$ ok,that helps me $\endgroup$ – OD IUM Feb 12 at 17:13
  • $\begingroup$ Note: because of the skin effect this concept of surface current is especially useful for high frequency currents in metals. $\endgroup$ – hyportnex Feb 12 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.