WEBVTT - generated by VCS
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Have a wonderful good day, ladies and gentlemen.
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Today we want to use this little video from
the Electromagnetic Theory series to
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think about the behavior of the fields at the
interfaces, as it is
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obtained directly from Maxwell's equations.
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My name is Hans Georg Krauthäuser, I am the
holder of the professorship for
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Electromagnetic Theory and Compatibility at
the TU Dresden.
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Well, when we start we can start right away
with the Coulomb-Gaussian
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Law that is written down here again.
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Divergence D equals rho_v.
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And we apply these laws to a geometry that
is also popular in the
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Literature is referred to as Gaussian box or
Gaussian box or also as
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Gaussian box or something like that and yes
the geometry is very simple.
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We have here an interface between two media.
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Medium 1 is characterized by an epsilon_1
and a mu_1.
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Medium 2 is characterized by an epsilon_2
and a mu_2.
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And now we imagine a volume that is just such
a small box or can
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that begins in one volume on one side and
extends down to the other
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Page extends and for the sake of simplicity
we take one here
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cylindrical area.
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The interfaces are parallel to the interface.
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The interface itself is characterized by its
normal vector n and the
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The front sides, i.e. the top and bottom of
the box, so to speak, are characterized by
the
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Normal vectors n_1 on the one hand and n_2
on the other.
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Yes, and we actually look at dielectric displacement
first
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D, which now has some values on both
sides.
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And we will assume that these surfaces of
the cylinder - top and bottom
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of the cylinder - are small enough to view
D as constant over this area.
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Well so what can we do?
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We have a volume, we have the differential
relationship - the local one
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Relationship - and yes then I think it's obvious
that we should just get on with it
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skip an integral relationship.
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We do that on both sides and apply the volume
integral as we already do
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know, we can use Gauss’s integral theorem
from the volume integral over the
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Make divergence a surface integral.
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And now we make a limit h equal to 0, h is
the height of this cylinder.
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That means we let this whole structure shrink
, the volume is then
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no longer a real volume, of course.
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That means at the border crossing the volume
to 0 and then it is clear - there we even
have
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no more lateral surface of the cylinder -
then becomes the surface integral
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actually only the top and bottom of this cylinder
contribute.
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Because we said that D should be constant
over this area A, that is
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Then quickly written down integral.
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That is simply the contribution of one area
- the first area - n_1 times D_1
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scalar multiplied and the whole thing then
multiplied by the area A itself.
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And to this comes the contribution of the
soil, area 2, n_2 times D_2 as well
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again multiplied by the size of the area.
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Very simple surface integral in the case.
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On the other hand we have the volume integral
where we are now
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must take into account that the volume is
compressed onto the surface at
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The limit transition h towards 0 and, accordingly
, the volume charge density is also none
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real volume charge density more, but something
that only has one when x equals 0
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Has value. We can express that by being here
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actually replaced the bulk charge density
with a surface charge density
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rho_F, which then has the unit coulomb per
square meter multiplied by the
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Delta function of x.
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Here you have to note that x is a dimensional
quantity, a length,
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accordingly, the delta of x is also dimensional
and has the unit 1
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by length, 1 by meter.
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If you don't remember the relationship, please
look again
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Video with the previous knowledge.
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That's explained in there.
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Yes, and in that case this volume integral
will also be relatively simple, we can
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execute the x integration immediately, there
then only remains an area integral
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left. And with the area integral we again
assume that the area is like this
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small is - which we can of course achieve
again and again - that roh_F on the
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Area is also constant again, then it is simply
rho_F times A.
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Here we can of course shorten the A on both
sides and n times
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an n times D - that is now the normal component
D.
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We just have to be a little careful in which
direction we do it.
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We actually had n_1 and n_2 here before ,
we have now replaced,
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n_2 is directly n, but n_1 is equal to minus
n, so here comes straight
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also this sign and therefore there is minus
n times D_1 plus n times D_2.
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These are normal components, so that we can
summarize them now in
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this clear form: normal component D_2 minus
normal component
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D_1 is equal to rho_F, i.e. the normal component
of D is at the transition
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from medium 1 to medium 2 discontinuously and
we can also specify how big this is
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Discontinuity, how big the jump of the normal
component is.
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Attention: even for simple media, linear isotropic
homogeneous media, this can
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cannot be transferred to the E-field because
they are one on both sides
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have different epsilons , which of course
have to be factored in here
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so that they get a slightly different formulation
here .
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Similarly, we can of course do the whole thing
for B, for the B-field, for them
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magnetic induction.
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Here it becomes even easier because the divergence
B is equal to 0, that is, if
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we carry out the same calculation, the 0 and
are always on the right-hand side
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accordingly we get here that the normal component
of B is continuous,
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this means that the normal component does
not jump at the transition from medium 1 to
medium 2.
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Here again, attention: what we said here for
the B field
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does not apply to the H field.
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With the H-field you would still have different
mus on both sides , so that
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they would actually get them in one go.
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But first of all, these two relationships
are important.
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Well, these are the statements for the normal
components of D and B.
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And in fact we can make statements for them
in a similar way
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Tangential components of E and H.
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And for that we consider a slightly different
geometry and we do that first
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here on this slide actually only for the E-field.
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What we are looking at is a so-called Stokes
surface.
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So, here we have our interface between
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the two media 1 and 2.
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n is again the normal vector on this interface.
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And now we put a loop, a loop, perpendicular
to this interface
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rectangular loop as shown here
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which spans the area A.
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The contour of A would therefore run around
accordingly if the
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The normal vector of this surface would now
come out of the film plane.
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Otherwise it has the length l on one side
and the width h on the other side.
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The E field on both sides is again labeled
E_1 and E_2.
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Here, too, we would first do an integration.
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Now, however, an area integration, namely
the integration via this right now
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pasted surface A.
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On the one hand we get an area integral over
the rotation of E, da
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we know again, we can convert that using Stokes'
integral theorem
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into a contour integral, i.e. a line integral
that is closed
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Way around the area. That's this integral
here E ds.
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And on the other hand there is simply the
area integral, that is
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minus integral A dB after dt times dA.
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What is written here is ultimately the change
in flow through this area.
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If we now let the border crossing point h
go towards 0, then it is clear that the
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Flux change through this area, when this area
becomes 0, will also be the
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Flux changes through this area disappear,
so that on this side on
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in any case, a 0 comes up.
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And with the orbital integral it is so that
in the border crossing h against 0, of course,
this
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Cross pieces of length h no longer play a role,
only these two
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Long sides still make a contribution.
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This post is written here.
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The whole thing is now written down in terms
of the tangent vector
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t introduced here above.
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And we see that this one path here in medium
1 runs in the direction of minus t,
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therefore this minus sign appears here.
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The other way is in the direction of t, accordingly
there is a positive one here
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Character. I can then shorten the length l
so that I can see immediately that
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the tangential component of E is continuous
at this transition.
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Again, please be careful again: you could
use it for simple media now
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of course replace E by 1 by epsilon times D,
but there the epsilon on both
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Pages were different for simple media , they
would get a discontinuity.
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So you can not refer to the continuity of
the tangential component of E
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Infer the continuity of the tangential component
of D.
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The tangential vector t is of course not unique.
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You can access this area, which is characterized
by the
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Normal vector n define any number of tangent
vectors in which you
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simply rotate this tangential vector.
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And you could of course make a corresponding
consideration again.
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You can then introduce a surface again, introduce
a Stoke surface,
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which is also spanned again by t and n, so
that ultimately it is this
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Continuity condition for any tangential vector
of this boundary surface
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to get. Yes, and we can now also consider
that in a very analogous way
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Make the H-field, but there is the equation,
our Maxwell equation, that
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Flooding law a little more complicated.
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Therefore, the result is also a little more
complicated.
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But first of all we would lay a surface just
as flat again , vertically
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for the interface between media 1 and 2 and
the steps
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are initially the same.
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First of all I do an area integral, use the
again
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Stokes' integral theorem um from the area integral
over the rotation
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To make circulation integral.
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And look at my orientation again and see again
that one way goes
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toward minus t and the other way is toward
t, so I'm on side 1
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get a negative sign again.
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The positive sign remains on page 2.
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And the change in the displacement current
will also disappear again when the
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The area becomes small, so that this term,
the integral over this term, falls into the
limit
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h also disappears towards 0.
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What remains is then the integral over the
current density J dA.
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And that is a volume flow density, first of
all, that in this one
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Area would prevail.
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If I let the area shrink, let h go towards
0, then I have
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of course, again only the contribution on the
interface itself.
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I do that in a very similar way to what I have
just done by introducing a delta function
and
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now have a surface current density distribution
J_A, i.e. J_A times the delta of x
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my normal current density again.
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And so these are the same integrals, only
I can now do the integral back here
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easy to evaluate, of course, because the delta
function is included.
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And if we again assume that the loop is small
enough that this
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Surface current density is the same over the
whole area , then I get it here
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accordingly, simply just multiply the tangential
component
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out with the length.
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Now I have to be careful which tangential component.
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The current density would be the current density
that penetrates this area.
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And if I let that shrink to the interface now
, that's it
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00:18:00.080 --> 00:18:07.080
Tangential component which is given here by
this tangential vector t_2, which is here
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plays a role. t_2 is nothing more than the
normal vector on this loop that i
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I've defined here and well, obviously these
vectors also form
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a legal system and that is how I get this
simple relationship.
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And here too, of course, you can immediately
see that the length l is shortened out again
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can, so that overall I get a simple statement
again .
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Namely the one that the tangential component
of H is discontinuous and the discontinuity
of the
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Tangential component is given by a tangential
component of
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00:18:59.080 --> 00:19:08.040
Surface current density. But we have to be
careful, t and t_2 are different
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00:19:08.040 --> 00:19:16.040
Tangential components. So if you have the
discontinuity in relation to the
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00:19:16.040 --> 00:19:26.040
Want to write down the tangent vector t, then
you have to see that you get from the
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Surface current density get the tangential
component with respect to t_2.
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So they are perpendicular to each other.
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00:19:34.040 --> 00:19:44.040
First of all, of course, this applies again
to any tangential components t that would
be
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yes again freely selectable.
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But I also have to turn t_2 accordingly so
that it works, so
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t_2 must be the normal vector on this loop.
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Or to put it another way: there is a fixed
relationship between t_2, n and t that always
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must be adhered to. Otherwise t is arbitrary.
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Well. Yes, that's what we say about the interfaces,
the components of the fields
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wanted to testify at the interfaces.
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The relationships are fundamental.
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You can see that we can get that from the Maxwell
equations only with pure
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Have derived mathematics, did not need anything
else.
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In other words, you can actually always fall
back on these border relationships,
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these are not approximations or approximations,
they actually always apply.
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Well. Yes, thank you for your attention.
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As always, you can find more information on
the website.
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Until next time.