WEBVTT - generated by VCS
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Have a wonderful day, ladies and gentlemen.
I warmly welcome you to another
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Article in the series Theoretical Electrical
Engineering. My name is Hans Georg Krauthäuser.
I am
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Holder of the professorship for Electromagnetic
Theory and Compatibility
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at the TU Dresden. Today we will deal with
magnetostatics.
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We will get into the topic, ie we will look
at the basics.
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There will then be a second video where we
actually look at the stomach statics
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in the matter a little closer, but as I said,
let's start with the basics.
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That means we have to get a clear picture of
the initial situation.
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The initial situation is that we are time-independent
in magnetostatics
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Fields will be considered - but now in a different
way than in electrostatics - of course
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magnetic fields ie time-independent magnetic
fields are the subject. and
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they are caused by direct currents. In any
case, we want to limit it to
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because - you know that of course - usually
there is another effect. It
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There are also permanent magnetic substances
and of course they can also be time-independent,
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create magnetic fields, but within the scope
of what we're doing here, that's what we want
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don't look at first. The main difference to
electrostatics is that in
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of magnetostatics do not have magnetic monopoles
- at least they are not observed.
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There are indications of relatively exotic
situations - substances, for example, which
can also be found in
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the class of spineis summarized. However ,
there are partially electrical structures
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which - if you take them as pseudoparticles
- can be interpreted in such a way that
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as if they were magnetic monopoles. However,
they always occur in pairs, so that they are
macroscopic
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the statement stands: There are no magnetic
monopoles. The leading term - if we think of
the
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Think multipole development - the leading term
here is not the monopole term, but the
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dipole term. Of course that has an impact.
Let's look at the equations again
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- as we have treated them so far. Here in front
are Maxwell's equations, again
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generally supplemented by the continuity equation.
In electrostatics we restricted this to
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the two Maxwell equations for the electric
fields, E-field and dielectric displacement.
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In the case of a steady flow field, in addition
to the constant rotation E equal to 0, we also
have divergence j
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considered equal to 0. And divergence j equal
to 0 follows immediately from rotation h
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same j. If you apply the difference here to
this Maxwell equation,
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yes you get it That's why I have up here -
even if we are in the stationary flow field
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have not considered magnetic phenomena - at
least write that down again in brackets.
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And in magentostatics we will now focus on
these two boxed equations,
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i.e. rotation h equals j and difference b equals
0 as basic equations.
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And here too, of course, from rotation haar
equals j immediately divergence j equals 0,
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that's why we wrote it again for the sake of
completeness - even if we are
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are not explicitly considered. Well, we can
do these basic equations of gastric statics
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of course in addition to the differential form
that we just had, yes of course also in an
integral form
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rewrite the form. It's just repeated here.
You know that by now. and
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of course we generally have to - we know that
from the axiomatic introduction,
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we know that from electrostatics and you probably
already know that
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of course - also add material equations. For
the magnetic phenomena is
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it so that the magnetic induction and the magnetic
field are linked in this way
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and way. And in general, this is a very complex
context that I am talking about here
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times indicated by a matrix m that in terms
of if it is anisotropic,
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the direction of b and the direction of h do
not have to agree either.
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And in addition, these material parameters
can of course also be location-dependent, then
it would be
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it just not homogeneous. It can depend on the
magnetic field itself, then it would not be
linear. It
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there can also be time dependencies and other
dependencies. Generally will
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However, if we restrict ourselves to the simplest
case, i.e. linear, homogeneous,
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isotropic media. And then we have a simple
relation b equals Mü_0 times Mü_r times h
or then
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simple - if I summarize the two Mü's again
- b equals Mü times h. And mu is
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then a simple scalar. But as I said, more on
that later. In principle, we stay in the first
place
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Vacuum, although there is no difference between
a vacuum and a simple medium, because they
are simply different
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constant is. In electrostatics, we very quickly
introduced a potential
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and found that it was very helpful. And in
electrostatics we could do that,
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because from rotation E equal to 0 it is clear
that E is a gradient field. we
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then introduced a scalar potential Phi with
E equal to minus gradient Phi . In the
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Magnetostatics is now rotation h not equal
to 0, but rotation h equals j,
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so this won't be a gradient field . But we
have a different relationship
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which we can exploit, namely divergence b equals
0 - so b is source free. And that helps
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introduce us to a magnetic vector potential.
Please remember when you think back to
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the prior knowledge block. There we introduced
the Helmholtz theorem - also known
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as a fundamental theorem of vector analysis,
which states that we can use any vector field
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can be divided into a divergence-free and a
rotation-free part, and then how
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these shares are also formed. And now we have
a field here, that's it
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divergence free. Then we can also write it
as a pure rotation field.
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That means we can introduce a field A, the
so-called magnetic vector potential,
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and then express b as rotation A. The A then
has units of volt-seconds per meter
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and with this introduction, which, as I said,
follows from the Helmholtz theorem that we
do that
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allowed, it is then clear - and this is an
advantage - that divergence b equals 0 is automatically
fulfilled,
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because divergence of rotation of any field
vanishes. It's the same benefit
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as with the scalar potential. One of the two
basic equations is automatically fulfilled
and then
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we just need to concern ourselves with the
other. Then the other one
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gets a bit more complicated, we'll see that
in the next few steps. First of all, we assume
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that we would only have a linear and isotropic,
but still inhomogeneous medium,
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so that the mu would still be a mu of r. If
I then form rotation h, then that is rotation
1
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by mu of r times b. And I can now write b as
a rotation A, ie that would be a rotation of
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1 divided by Mü of r times the rotation of
A. And that would equal the current density.
But you see
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here you don't really get any further, because
you could only resolve it further if
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one would have statements about the location
dependence of Mü . Therefore one goes over
here to a medium,
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which is at least regionally homogeneous, so
that this area
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next to it can be regarded as a homogeneous
area. Then the mu is just a scalar. And have
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then of course we could pull this out, then
pull it in front of the rotation. And I can
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For the sake of simplicity , multiply the whole
equation by Mü to get the fractions
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get rid of and then get such an expression
Mü times rotation h. That's it then
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Rotation b, if I put b rotation A, that's rotation
rotation A. And rotation rotation,
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you can also write that - that 's a vector
identity - that can
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we write as gradiv minus Laplace, so we get
total gradient
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Divergence A minus Laplace A equals Mü times
j. It now says Mü mal j, because I yes
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multiplied by Mü. Well, and that's actually
a relatively promising one
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Equation, because here in the back part, there
is what we already know from the Poisson equation,
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but now as a vectorial equation. In other words,
that would be in Cartesian coordinates
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just three equations. But this here in front,
that is of course an expression of us still
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a bit disturbing. If we could get it down to
a pure Poisson shape then of course it would
have
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the advantage that we can adopt the solution
methods. And we will see now
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how is this actually done. This leads us to
the notion of gauge transformation. and
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as follows: we can assume we have an arbitrary
scalar field Psi. if
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we form a gradient field from this Gradient
Psi and then apply the rotation again,
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then it disappears because the rotation gradient
of something is always 0.
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Rotation is interesting because the vector
potential and the field,
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the magnetic induction, are related via the
rotation operator. So
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let's assume we create a new field B', that's
supposed to be a rotation of A'. And write
A'
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we now A plus gradient with any Psi, so A plus
gradient Psi. Then when we
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now apply the rotation operator, we see that
actually only rotation A is left here,
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because Rotation Gradient Psi vanishes for
any target i.e. when I transition from A
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to an A' in which I add a gradient field, any
gradient field, then this becomes
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do not change the magnetic induction. We see
the same thing again when we meet
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actually look directly at the equation we just
had: rotation rotation equals mu times
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j. Here, too, I can of course go over to an
A' . And rotation A' is the same though
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like rotation A, ie the equation does not change.
With the same j is then also A'
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a solution if A is a solution. In other words,
the vector potential is only - except for one
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additive constant -gradient Phi determined.
And that is extremely helpful. First of all,
give
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we give the whole thing a name. In this transition
from A to A' I just use a gradient field
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add - this is called a gauge transformation.
And the fact that the magnetic induction
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remains unaffected - ie that B' equals B -
we call these properties
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the gauge invariance, i.e. B is gauge invariant.
And look to see what that's good for
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let's just look at the divergence of A'. The
divergence of A' is then the divergence of
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A plus gradient Psi. And that's - I'll just
put the operator in the brackets - divergence
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A' plus divergence gradient Psi. And divergence
gradient that's not 0 now ie back here
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there is actually a summand. I can, therefore,
the difference of A' becomes another
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be than the divergence of A, although I don't
change the field in this operation. and
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if we now look at our differential equations
that we have to solve, then we see
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that's where the divergence comes in. And because
I can actually change the difference, through
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this transformation, it ultimately means that
I can choose the difference of A arbitrarily.
I
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can always find a psi so I have a hum through
this term divergence gradient phi,
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which allows me to ultimately set Difference
A to any value or difference
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A' to any value. Say this property allows me
for
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Divergence A also just to make an assumption
that I like right now. And something calibration.
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Yes, and where is this equation look at, what
would you particularly like. The first,
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what comes to mind, of course, is what freedom
I have to choose divergence A arbitrarily.
Then choose
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I just divergence A equals 0, then this whole
term is dropped. And that's something
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is actually also very successful in statics
for static problems. And this shape
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of the calibration is called the Coulomb calibration.
When I use coulom measurement, that means I'm
betting
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Divergence A just equals 0 and I'm allowed
to, as we just saw. This simplifies
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our differential equations for A are of course
very strong.
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It takes on a Poisson form, of the form Laplace
A equals minus mu times j. We have to do a
little
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be careful, because this Laplace operator,
which we are about to apply to a vector field,
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this is actually a different operator than
what we see for scalar fields. So
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Laplacian applied to vector fields is not the
same operator as Laplacian does
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applied to scalar fields. If you know what
is the Laplace operator,
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then eventually I have to go back into the
relationship we just had. We had
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yes exploited that Rod Rod equals GradDiv minus
Laplace. Of course I can turn that around
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and say Laplace equals GraDiv minus Rod Rod.
And in a general coordinate system
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I will actually calculate the Laplacian operator
via this relation, GradDiv minus Rod Rod
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must. And that becomes a problem and that becomes
a bit more complex. Even in curvilinear coordinates,
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because then the unit vectors are no longer
location-independent and that leads
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through these operations that the components
mix. This means,
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that no longer simply breaks down into xyz
proportions - it does in Cartesian coordinates
- but
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this is not the case in general curved line
coordinates . You have to
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calculate explicitly and you really have to
be a bit careful, not that
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Laplacian applied to vector fields easily confused
with the scalar Laplacian.
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For Cartesian coordinates, but also only for
Cartesian coordinates, but then it breaks down
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as usual. This means that we can actually use
this vectorial equation
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just write it in three scalar equations and
then you have the Laplace operator
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actually the scalar Laplacian for Cartesian
coordinates as we know it.
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And I just wrote A and j again in Cartesian
coordinates.
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So it works in Cartesian coordinates, but for
example - to name just one example - in
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Cylindrical coordinates, the same equation
looks much more complex. Here you can see how
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the components mix, ie I have an equation here
in which the A_Rho and
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j_Rho appear, but not alone , there is also
a derivation after Phi and one
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Derivation to Rho. And that is also the case
here in the second equation. Only the third
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looks normal again. That's actually what you
expect from the cylinder coordinates
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so. And the Laplace operator that now appears
here is actually that again
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scalar Laplacian - written in cylindrical coordinates.
Good, so you see, that's it
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actually generally much more complex - but
only simple in Cartesian coordinates.
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But that will also help us. We see that here,
for example. And now we are interested
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of course what does the solution then look
like for the vector potential. And that's where
it's helpful
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actually go in first in Cartesian coordinates
. Then we - like we just did
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have seen - three scalar Poisson equations
and if we want to solve them, then we know
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yes how we solve a Poisson equation - if we
can solve one, we can also solve 3. and
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all the solution methods that we looked at
in electrostatics,
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we can actually take care of that one-to-one
. And also means that we do something like
that
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how to find an analogue to the Coulomb integral.
That means we can do A_x, A_y and A_z easily
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write down as a volume integral j. The corresponding
component through r minus r' and
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then of course you need a prefactor that we
have to adjust accordingly. But that
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we can in principle take directly from what
we have learned about electrostatics.
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Except that we now have it three times for
the three scalar components. Now could
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you say: and what good does that do? That which
has now been solved enough in Cartesian coordinates,
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Of course that helps, because we can use what
we found here in Cartesian coordinates
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have of course also summarized in coordinate-free
. You can read that directly
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that it takes this form. And now , of course,
we can easily argue:
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Well, we used Cartesian coordinates to calculate
it, because then it's special
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was easy, but the result is of course true
for any coordinate system,
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because the result cannot depend on the arbitrary
choice of the coordinate system
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be dependent. It is therefore quite helpful
that the differential equations for the
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Vector potential is at least simply written
in Cartesian coordinates . That is enough
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to actually write down the overall solution
. That means that would be the
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Vector potential solution. Now there is an
important special case, namely that the
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Current density is limited to linear paths.
So speak on wires - everyone will see that
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that this is an important special case in electrical
engineering - and then of course I can
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split this volume integral here. On the one
hand, I have the current density that is caused
by the
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cross-sectional area this conductor passes
through and then I still have as further integration
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the integration along the stream thread. And
what goes through the cross-sectional area
that
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00:22:55.020 --> 00:23:01.080
is simply the current by definition. We have
already defined it in this way for the flow
field. speak
196
00:23:01.080 --> 00:23:10.080
for a streamlined arrangement, this integral
becomes much simpler. We have the power
197
00:23:10.080 --> 00:23:18.080
and from the volume integral that actually
remains only a line integral
198
00:23:18.080 --> 00:23:25.080
along the current thread, which of course then
has to be closed. this stream
199
00:23:25.080 --> 00:23:30.080
it will be location-independent because of
the continuity equation, we can do that now
200
00:23:30.080 --> 00:23:38.080
actually pull ahead of the integral. And then
this is a very simple expression,
201
00:23:38.080 --> 00:23:45.080
dm we can then also calculate many cases .
And ultimately what we are now
202
00:23:45.080 --> 00:23:50.080
have calculated here for the vector potential
- either in general or then just for the
203
00:23:50.080 --> 00:23:57.080
Stream threads - that then also leads us to
the law of Biot and Savar - which they use
with certainty
204
00:23:57.080 --> 00:24:05.040
already know from the basics or from school
- we know b equals rotation A. And A have
205
00:24:05.040 --> 00:24:12.040
we just wrote it down in this form. Then of
course we can also write down B,
206
00:24:12.040 --> 00:24:19.040
as essentially the same integral but now with
the rotation operator. I have
207
00:24:19.040 --> 00:24:29.040
once again written rotation index r - so that
it is not forgotten that the operator
208
00:24:29.040 --> 00:24:36.040
here now on the r and not on the r' . And now
we can look at the integrants
209
00:24:36.040 --> 00:24:42.040
just look at it again of course. So we have
to form rotation with respect to r from j from
r'
210
00:24:42.040 --> 00:24:53.040
by r minus r'. Formally, we can simply formulate
this according to the product rule, i.e. 1
divided by r
211
00:24:53.040 --> 00:24:59.040
minus r' and then rotation applied to j of
r'. You can see how helpful it is
212
00:24:59.040 --> 00:25:05.000
that the rotation operator is with respect
to r, because j is with respect to r', ie this
term is dropped.
213
00:25:05.000 --> 00:25:17.000
And the second term is then just minus j cross
gradient 1 through r minus r' - here again
214
00:25:17.000 --> 00:25:24.000
written. And I can calculate gradient 1 through
r minus r' , that's then minus r
215
00:25:24.000 --> 00:25:30.000
minus r' divided by r minus r' amount to the
power of three. The two negative signs cancel
each other out,
216
00:25:30.000 --> 00:25:40.000
so this term remains and you will of course
insert it into the formula here above,
217
00:25:40.000 --> 00:25:49.000
so in this way we find the general Biot-Savar
law B equals Mü over 4Pi
218
00:25:49.000 --> 00:26:00.060
times the volume integral j cross r minus r'
divided by r minus r' to the power of three.
And like us
219
00:26:00.060 --> 00:26:06.060
Of course we can also have done the whole thing
for a current thread
220
00:26:06.060 --> 00:26:13.060
simplify again. Then, just like before, the
volume integral becomes only
221
00:26:13.060 --> 00:26:22.060
a path integral along the stream filament and
the integral cross-sectional area times j comes
out here
222
00:26:22.060 --> 00:26:35.060
as a stream forward. The minus sign that appeared
here is simply because here
223
00:26:35.060 --> 00:26:42.060
the two product terms in the cross product
have now been swapped here to have a form
224
00:26:42.060 --> 00:26:48.060
which is perhaps more common how it is spelled.
You can also do it the other way around
225
00:26:48.060 --> 00:26:55.060
then of course the minus is gone. And just
so that it doesn't fall into oblivion, the
term
226
00:26:55.060 --> 00:27:02.020
which we had at the bottom of the last slide
, is simply repeated here. Good,
227
00:27:02.020 --> 00:27:10.020
that would already be enough for us to actually
- if we have current courses - both that
228
00:27:10.020 --> 00:27:14.020
Calculating the E-field and also the A-field
means that you can get quite far with that.
229
00:27:14.020 --> 00:27:23.020
But of course only if we have homogeneous media
and the next step is to say:
230
00:27:23.020 --> 00:27:28.020
Okay, what if we have two homogeneous media
that are contiguous. So that leads us
231
00:27:28.020 --> 00:27:34.020
to the continuity condition at interfaces,
which of course we also consider in the general
case
232
00:27:34.020 --> 00:27:41.020
have already looked at. In the video of restrained
interfaces we already have that in general
233
00:27:41.020 --> 00:27:49.020
shown. And that's why we really only need to
repeat it here. and
234
00:27:49.020 --> 00:27:53.020
want to remind you again that we had the convention
that the normal vector
235
00:27:53.020 --> 00:28:00.080
the one that appears here, for example, is
always the one that points out from medium
1. That means from medium 1
236
00:28:00.080 --> 00:28:07.080
points into the medium 2 - this is important
for the sign position. And yes, the two conditions
237
00:28:07.080 --> 00:28:15.080
then we have is that the normal component is
steady in the transition between media and,
238
00:28:15.080 --> 00:28:20.080
that the tangent components, more precisely
all tangent tail components - because I have
239
00:28:20.080 --> 00:28:27.080
of course again several tangential components
or any number of tangential vectors,
240
00:28:27.080 --> 00:28:34.080
which are perpendicular to this normal vector.
So for all tangent vectors we have
241
00:28:34.080 --> 00:28:41.080
these discontinuity conditions as written here.
And the discontinuity is given by
242
00:28:41.080 --> 00:28:51.080
also again a tangential component of the surface
current density and which
243
00:28:51.080 --> 00:29:00.040
Tangential component that is that follows from
this relationship t_2 equal to n cross t ie
n_t and
244
00:29:00.040 --> 00:29:11.040
t_2 form an orthogonal tripod, form a legal
system. And about that you would
245
00:29:11.040 --> 00:29:15.040
then calculate the continuity. And this is
of course also important for boundary value
problems,
246
00:29:15.040 --> 00:29:25.040
because the continuity conditions ultimately
give us the boundary terms in boundary value
247
00:29:25.040 --> 00:29:28.040
Tasks - either as Diriclet boundary conditions
or as Neumann ones
248
00:29:28.040 --> 00:29:37.040
boundary conditions. Of course, we can also
write this down in the form of refraction laws.
249
00:29:37.040 --> 00:29:43.040
This works in particular for J_F equal to 0,
where I then
250
00:29:43.040 --> 00:29:52.040
statements about the normal and tangential
components of B can be derived. I can if this
is 0
251
00:29:52.040 --> 00:29:59.040
using B equal to Mü times h, also derive a
relation for the tangential component of B.
252
00:29:59.040 --> 00:30:06.000
And with the angular relationships that are
here and the definitions Mü_1 and Mü_2, I
then get
253
00:30:06.000 --> 00:30:14.000
just such a law of refraction. We already had
that very briefly for the electric field
254
00:30:14.000 --> 00:30:21.000
written in the section stationary electric
flow field and therefore
255
00:30:21.000 --> 00:30:29.000
the whole thing is now also added here for
the magnetostatics. Next thing we have to do
is now
256
00:30:29.000 --> 00:30:35.000
provide a couple of terms. We 're not going
to do that every time we go into detail
257
00:30:35.000 --> 00:30:41.000
go, but we can make derivations again like
we did electrostatics
258
00:30:41.000 --> 00:30:46.000
to have. That's why we can actually abbreviate
it very massively here. And we'll start with
that
259
00:30:46.000 --> 00:30:55.000
Energy in the magnetostatic field and there
arises analogous to electrostatics for the
magnetic
260
00:30:55.000 --> 00:31:04.060
Energy density 1/2 times H times B and then
the total magnetic energy would be 1/2
261
00:31:04.060 --> 00:31:18.060
Volume integral H times B d_V. If we want to
express that with the current density. And
that's straight
262
00:31:18.060 --> 00:31:24.060
in magnetostatics, where we often actually
have very concentrated flow patterns,
263
00:31:24.060 --> 00:31:30.060
a good idea. Then we can do that. We do that
first for linear, homogeneous
264
00:31:30.060 --> 00:31:40.060
and isotropic media, by writing down the energy
again for this case. Then pull
265
00:31:40.060 --> 00:31:49.060
we here a factor 1 through Mü forward and
for that just here instead of B times H, just
B for
266
00:31:49.060 --> 00:31:58.060
square under the integral. And further - to
deal with this B square now - consider
267
00:31:58.060 --> 00:32:06.020
let's look at the divergence A cross B - this
is formally simply B times rotation A minus
A times rotation B.
268
00:32:07.020 --> 00:32:14.020
And rotation is of course the B itself and
here you can see why we did it, because
269
00:32:14.020 --> 00:32:22.020
this is where the B square appears. And the
other term, the rotation B that comes up here,
270
00:32:22.020 --> 00:32:35.020
that's just Mü times j Maxwell's equation.
So we just have divergence A clubs standing
271
00:32:35.020 --> 00:32:45.020
B - that's the B square, that's what we need
for this - minus Mü A times j. About this
relationship
272
00:32:45.020 --> 00:32:52.020
which we have now found here, we can replace
the B square under the integral and do so
273
00:32:52.020 --> 00:33:00.080
actually get the current density. Which might
be a little unsettling at first
274
00:33:00.080 --> 00:33:08.080
could would be that we then have another summand,
but the fact that
275
00:33:08.080 --> 00:33:16.080
we then have a volume integral over a divergence
, that gives us a bit of hope. and
276
00:33:16.080 --> 00:33:21.080
we'll see that now when we actually use it.
We can now use the magnetic
277
00:33:21.080 --> 00:33:32.080
Write energy as 1/2 volume integral A times
j dV and then there is a second term
278
00:33:32.080 --> 00:33:42.080
added. And that's just a volume integral over
divergence A cross B. Well, and a volume integral
279
00:33:42.080 --> 00:33:48.080
over the divergence, which is of course nothing
more than the surface integral over the
280
00:33:48.080 --> 00:33:58.080
size itself. So surface integral A Kreux B
dA. And if we now finally expanded
281
00:33:58.080 --> 00:34:07.040
look at the power distribution - that will
of course also be the norm - then we can look
at each other
282
00:34:07.040 --> 00:34:13.040
look at the sizes here in the term. And the
B that appears here is after Biot-Savar
283
00:34:13.040 --> 00:34:23.040
proportional to 1 through R squared - if we
consider R towards infinity. And the A itself
284
00:34:23.040 --> 00:34:30.040
is proportional to 1 over r, so what's under
the integral here is proportional to 1 over
r
285
00:34:30.040 --> 00:34:40.040
to the power of minus 3. And yes, if I let
the volume go to infinity, then it grows
286
00:34:40.040 --> 00:34:45.040
the surface of course only quadratic ie we
have a quadratically increasing surface, but
287
00:34:45.040 --> 00:34:52.040
an integrant that decreases with 1 through
r^3. That is, this surface integral will be
0,
288
00:34:52.040 --> 00:34:58.040
if I just choose the volume big enough. And
I can do that with a finally extended one
289
00:34:58.040 --> 00:35:05.000
Always do the power distribution, of course.
Say back here this integral falls away, so
290
00:35:05.000 --> 00:35:14.000
I can then simply write the energy as 1/2 volume
integral times A times j.
291
00:35:14.000 --> 00:35:23.000
And then we read that the energy density, the
magnetic energy density, then just ½ times
A times j
292
00:35:23.000 --> 00:35:32.000
is and that is of course very nice. Let's look
further into the terminology,
293
00:35:32.000 --> 00:35:37.000
Another term that you are of course also familiar
with is the term inductance. and
294
00:35:37.000 --> 00:35:44.000
Of course, we can now also deduce it here from
the field image. For this we consider
295
00:35:44.000 --> 00:35:52.000
two conductor loops, two paths, one loop 1,
one loop 2. We assume that we are in the
296
00:35:52.000 --> 00:36:04.060
loop 1 have a current I_1. And with that we
can of course use the field, the magnetic induction,
297
00:36:04.060 --> 00:36:12.060
calculated by Biot-Savart - that's easy
298
00:36:12.060 --> 00:36:18.060
this expression as it stands here. Current
pulled forward and then even circulation integral
299
00:36:18.060 --> 00:36:26.060
via the path C_1. The points r'_1 are the points
on this on the first loop,
300
00:36:26.060 --> 00:36:33.060
and r is of course the observation point where
we look at it. what we can do now
301
00:36:33.060 --> 00:36:39.060
is we can look at which magnetic flux is now
reaching through the second loop
302
00:36:39.060 --> 00:36:51.060
through caused by the current in the first
loop. We call this Phi_m,2. And this is
303
00:36:51.060 --> 00:37:00.020
then of course simply the flow integral - i.e.
integral B_1 times d_2. d_2 would be now
304
00:37:00.020 --> 00:37:04.020
just the area vector on this loop. With the
appropriate circulation sense they have
305
00:37:04.020 --> 00:37:09.020
the direction. And let's just say first of
all - that's not spectacular - the B_1
306
00:37:09.020 --> 00:37:19.020
is of course proportional to I_1 and therefore
the Phi_m,2 is also proportional
307
00:37:19.020 --> 00:37:27.020
to 1 ie we can write it in the form like Phi_m,2
is equal to a value,
308
00:37:27.020 --> 00:37:35.020
which of course depends on loop two. So he
gets an index 2 and he hangs
309
00:37:35.020 --> 00:37:42.020
also from stream 1. So it also gets a value
of 1 - so it's just a size first of all
310
00:37:42.020 --> 00:37:53.020
with two indices M_21 times I_1. And the M_21
that's what we call the mutual inductance and
unity
311
00:37:53.020 --> 00:38:05.080
Mutual inductance is then volt-seconds per
ampere - which is also called Henry in the
SI system. and
312
00:38:08.080 --> 00:38:18.080
when calculating the flow, of course r'_2 is
on C_2. And this integral, we can
313
00:38:18.080 --> 00:38:23.080
take a closer look at us now - that 's well
written here again - by we
314
00:38:23.080 --> 00:38:33.080
now simply write B as a rotation of B_1 from
rotation A_1. Then we have an area integral
315
00:38:33.080 --> 00:38:38.080
about the rotation - this is of course nothing
other than the integral of the revolution over
the size
316
00:38:38.080 --> 00:38:46.080
myself. And we know the size - that was just
on the formula, on the last slide here
317
00:38:46.080 --> 00:38:53.080
below still on it. This means that we can also
simply use A_1 . That delivers
318
00:38:53.080 --> 00:39:02.040
Pre-factor Mü_0 by four Pi times I_1 and then
the integral over the first loop, so that
319
00:39:02.040 --> 00:39:11.040
we finally get such a double integral for the
flow through the second loop. and
320
00:39:11.040 --> 00:39:18.040
From this integral we can now actually read
what the mutual inductance is like. we
321
00:39:18.040 --> 00:39:23.040
just need to compare the two terms with each
other and then come up with the fact that the
322
00:39:23.040 --> 00:39:31.040
Mutual inductance is just Mü_0 divided by
four Pi times this double path integral. And
the
323
00:39:31.040 --> 00:39:39.040
we call the Neumann formula. With the Neumann
formula you can now calculate the mutual inductance
324
00:39:39.040 --> 00:39:51.040
actually calculate C_1 for any paths, C_1 and
C_2 for any loops. And this is
325
00:39:51.040 --> 00:39:57.040
not always particularly useful for calculations
- it works, but it's not always the case
326
00:39:57.040 --> 00:40:05.000
easiest way - but you can see two important,
fundamental properties in this Neumann formula
327
00:40:05.000 --> 00:40:14.000
the mutual inductance. Namely for the first
time, if we swap C_1 and C_2, ie if we now
328
00:40:14.000 --> 00:40:22.000
would assume that the current would not flow
through C_1 and cause a flow in C_2;
329
00:40:22.000 --> 00:40:26.000
but if we were to flow a current through C_2,
would look at each other,
330
00:40:27.000 --> 00:40:34.000
which flow does this cause in C_1, then we
would end up with the same expression. and
331
00:40:34.000 --> 00:40:37.000
this expression is also perfectly symmetrical
1 and 2. We can use the
332
00:40:37.000 --> 00:40:44.000
Swap roles of C_1 and C_2 and get the same
mutual inductance. That means this one
333
00:40:44.000 --> 00:40:53.000
Mutual inductance is not at all dependent on
the question: what is the cause and where
334
00:40:53.000 --> 00:41:02.060
do I look at the effect? Ultimately, I can
also omit the indices. and
335
00:41:02.060 --> 00:41:07.060
on the other hand, it is a purely geometric
quantity. It's really just the geometry here
336
00:41:07.060 --> 00:41:13.060
inside and nothing else. Holding on to it is
definitely worthwhile and you can see that
immediately
337
00:41:13.060 --> 00:41:23.060
on the Neumann formula. There is a problem:
if I were to assume that there weren't two
338
00:41:23.060 --> 00:41:31.060
different ways, but if I were to say C_2 is
C_1 then it's clear that
339
00:41:31.060 --> 00:41:37.060
sooner or later I will of course have problems
here with singularities, because then at some
point r'_1
340
00:41:37.060 --> 00:41:44.060
equals r'_2. That means I ca n't take the Neumann
formula well to actually use this case
341
00:41:44.060 --> 00:41:52.060
Considering C_1 equal to C_2, then it is better
to actually go back to the definition of the
342
00:41:52.060 --> 00:42:00.020
against inductance al Phi_m,1 would then be
M_11 times I_1. And in the case when we
343
00:42:00.020 --> 00:42:07.020
don't have two loops at all, just one loop
and let's see what's going to happen, so to
speak
344
00:42:07.020 --> 00:42:15.020
an flux induced by its own current, then we
speak of the self-inductance L or
345
00:42:15.020 --> 00:42:26.020
also simply from the inductance L - with the
definition Phi_m equals L times I.
346
00:42:26.020 --> 00:42:35.020
Yes, and of course we can calculate that directly
for a conductor loop. The river
347
00:42:35.020 --> 00:42:48.020
would be simply integral B d_A. And that's
because B is rotation and then I'm again
348
00:42:48.020 --> 00:43:00.080
Stokes can apply round-trip integral A ds.
And that should be L times I, meaning we get
for them
349
00:43:00.080 --> 00:43:10.080
Self-inductance L then the simple expression
1 by I times the round-trip integral A ds.
So we have
350
00:43:10.080 --> 00:43:16.080
a direct relationship actually between the
round-trip integral A ds and the inductance.
That
351
00:43:16.080 --> 00:43:27.080
is only once again normalized to the current
that flows in this loop. And we can
352
00:43:27.080 --> 00:43:34.080
link this with the magnetic energy, of course.
that's very nice
353
00:43:34.080 --> 00:43:39.080
because then we get formulas that we know from
the basics again. If we go again
354
00:43:39.080 --> 00:43:53.080
write, what is the magnetic energy , i.e. 1/2
volume integral A times J dV. And now
355
00:43:53.080 --> 00:44:02.040
just rewrite the volume integral J as an integral
over the cross-sectional area. That is
356
00:44:02.040 --> 00:44:10.040
yes then the current times the path integral
- then we have this expression 1/2 times the
path integral A ds. and
357
00:44:10.040 --> 00:44:15.040
this path integral A ds - we had just seen
that - this is essentially - up to
358
00:44:15.040 --> 00:44:21.040
on the normalization - just the inductance.
And if we use that, we get the
359
00:44:21.040 --> 00:44:30.040
from the basic well-known expression, 1/2 L
times I^2 for the magnetic energy.
360
00:44:32.040 --> 00:44:41.040
A notion that we should also introduce is the
magnetic moment. We have at the very beginning
361
00:44:41.040 --> 00:44:48.040
Yes, it has already been said that there will
not be a monopoly in the multipoly development
362
00:44:48.040 --> 00:44:55.040
and therefore the dipole term is of course
a particularly important one. And to look there
now
363
00:44:55.040 --> 00:45:02.000
Let's look again at the magnetic vector potential.
And yes, very characteristic
364
00:45:02.000 --> 00:45:10.000
of course we have this term 1 through r minus
r' which we have already looked at,
365
00:45:10.000 --> 00:45:19.000
how we can develop it. It has namely the development
1 through r plus r' times r
366
00:45:19.000 --> 00:45:34.000
r^3 plus and so on. And of course we can use
this development up here - what we
367
00:45:34.000 --> 00:45:41.000
leads to this shape. And without me formally
proving it now, let 's leave it at that
368
00:45:41.000 --> 00:45:47.000
actually simply the monopoly term that stands
in front. You can show that he disappears
369
00:45:47.000 --> 00:45:55.000
but without proof, let's just omit it for now
. And then come up with a development for
370
00:45:58.000 --> 00:46:17.060
Vector potential in the form. And we can now
do this integrand here again
371
00:46:17.060 --> 00:46:26.060
to write. Again, we won't explicitly demonstrate
how to do this; but that's easy
372
00:46:26.060 --> 00:46:31.060
just written down and if we rewrite it in this
form, then we can use this term,
373
00:46:31.060 --> 00:46:47.060
which now only depends on r', to which we give
m at location r. And this way we can
374
00:46:47.060 --> 00:46:55.060
the vector potential also as a development
write about such terms and is important
375
00:46:55.060 --> 00:47:04.020
here above all the leading term, the dipole
term, the one with the magnetic moment M here
376
00:47:04.020 --> 00:47:09.020
is written down again. And just like we did
with the electric field,
377
00:47:09.020 --> 00:47:14.020
where we have then derived a dipole field,
we can actually do it with the same steps
378
00:47:14.020 --> 00:47:21.020
derive a development for the magnetic induction
. And the first term that appears
379
00:47:22.020 --> 00:47:33.020
that would simply be the dipole term here.
Well, that's what you should do here now
380
00:47:33.020 --> 00:47:43.020
actually enough for the basics. As I said,
we will now in another post
381
00:47:43.020 --> 00:47:53.020
then go into the properties of matter again.
But we will be more
382
00:47:53.020 --> 00:47:58.020
actually don't need it for magentostatics,
because we from electrostatics - which we are
very
383
00:47:58.020 --> 00:48:05.080
have done in great detail - in particular,
we can and do take over all the solution methods
384
00:48:05.080 --> 00:48:11.080
what we said about boundary value problems
actually works one-to-one
385
00:48:11.080 --> 00:48:21.080
transfer. Well, that just leaves me to thank
you for your attention. I
386
00:48:21.080 --> 00:48:25.080
refer to the website, where you can also find
further information. Until next time.