No.

Maybe. Okay. So, let me give you some hint.

Do you remember the potato chip or arbitrary shape?

You can put it into a rectangular pieces and circulation of

those rectangular pieces will add up to make

the equivalent circulation of any arbitrary loop?

Use that fact, to understand this problem.

Okay. So, we can put

our equation in vector form if we define the direction of the vector mu,

to be the normal to the plane of the loop with

a positive sense given by the right-hand rule.

So, if I'm rotating like this,

magnetic dipole will be upright.

So, here you see the rotatings in

counterclockwise we will make sure that your magnetic moment is in positive z direction.

Let's take a look at the equation here.

The vector potential a is equal to one over four pi epsilon not c squared,

times mu the vector cross product r over r cubed,

which is equal to one over four pi epsilon not c squared

mu cross-border e sub r over r square.

Using this equation and inserting it into bx by bz,

you will be able to get all of these equations by yourself.

Where, dot dot dot mean mu over four pi epsilon r c square.

All right. So, let's take a look at these amazing table.

So, we are comparing magnetic versus electric dipole moments.

The components of the B-field behave exactly like those of the E-field for dipole

oriented along the z-axis and take a look at this, B sub z,

E sub z, and you see one over r cubed minus three z

square over r to the fifth is the exactly the same,

only the difference you see mu is replaced by b,

epsilon not c square is replaced by epsilon not

and the rest you can see them by yourself.

So, let's discuss a little bit about the magnetic dipole moment.

So Melody. Do we have to pause for a magnetic dipole?

No.

We don't have that, right?

So, the word dipole is slightly misleading when

applied to magnetic field because there are no magnetic poles,

as Melody answered, corresponding to electric charges.

So, the magnetic dipole field is not produced by

two charges but by an elementary current loop.

It is curious though that starting with

completely different laws we can end up with the same kind of field as we saw from

the comparison between the components

of magnetic field and electric field from each typo. Why should that be?

I think it has a lot to do with the math we're using to

analyze it like the simplification to me in discussing the system.

Exactly. So, we are not thinking of the higher-order terms and as Melody just mentioned,

it is because the dipole fields appear only when we are far away from

all the charges or currents meaning we are doing approximation.

So, through most of the relevant space the equations for E and B are identical.

So they give the same solutions.

However, be careful to sources

whose configuration we summarized by the dipole moments are fiscally quite different.

But this is also strings.

So, you can forget about the real physics,

you can only focus on the math and then you

can go get out to the physics and make sense of them.

Okay. So, we're going to discuss the vector potential of a circuit.

Circuit is our current loop where you have no starting point and ending point.

So, we're often interested in the magnetic field produced by

circuits of a wire in which diameter virus is very small,

compared with the dimensions of the whole systems,

like printed circuit system board or

even the flexible circuit board that is using smartphones.

So for thin wire, we can write our volume element as dv is

equal to Sds where the capital S is the area

of the cross-section of the wire or ds is

a small length component of your wire along its length direction.

Since the vector ds is in the same direction as j

the current density we can write a vector of equation jdV is equal to jSds.

As js is current i, in a wire,

the vector potential becomes a of one is equal to one over four pi epsilon not c

square times integral of i dx2 over r12.

So now, we can use some measurable current into the equation to get the vector potential.

So now, that leads us to the law of Biot Savart.

So, installing electrostatics we found that electric field of

a known charge distribution could be obtained directly

by an integral that is depicted here,

e of one is equal to one over four pi epsilon not times integral of

rho of two times E sub one to dv2 over r12 squared.

This is an arbitrary loop that we just made a small loop inside there.

You see next to Melody,

there's an observation point one.

So, the distance between those two will be the r12 to

many times it takes much easier to do the integral for the potential and that's great.

That was the case for let us statics.

It is not always the case for magneto statics.

Right. We see there's a similar

integral which relates to magnetic field B to the current I.

As you can see B of one is equal to

del cross a of one which is the curl of vector potential,

is equal to del cross and this is the equation we know.

We will transform that equation,

to use the current.

So, the curl operator means taking the derivatives of

a's of one that is to say it operates only on the coordinates x1,

y1 and z1 and if you look at this equation only the denominator has subscript one,

denominator doesn't have subscript one.

So, we have for the x-component of B,

b sub X is equal to

round az over round y1 minus round ay over round z1.

If you lay this out expand it,

you will be able to prove that can be arranged in this neat form of the vector form.

So, the integral gives me directly in terms of

the known currents so you can translate this into this one.

Where you just have to know the current along

the circuit and do the integration to the get the B-field at the observation point.

So now, another homework will be,

what is the advantage of the vector potential if

we can find B directly with a vector integral?

All right. That's it and I hope to see you again in the next lecture.

Thank you and have a nice day.