So, now we get to the point about what Melodie just mentioned.

Adding a constant to a vector potential.

So, you may remember that

scalar potential Phi was not completely specified by its definition,

as you can always find

another potential Phi prime that is equally good by adding constant C. So,

if I take a gradient of this function,

they will produce the same electric field.

So, likewise, the new potential Phi prime

gives the same electric field since the gradient C is zero.

Similarly, we can have different vector potentials,

A which give the same magnetic fields.

So, we can add to A any field which is

the gradient of some scalar field without changing the physics.

This is well-described in this box with equations,

but I'm going to ask my teaching assistant Melodie if she can explain to you in details.

Okay. So, basically what we're looking at here is we know the difference between A prime,

and A is simply the gradient of some scalar field.

If you take the curl of the gradient of a scalar field,

we know that is going to be zero because you can't have

circulation of a gradient as a gradient is always increasing or decreasing.

Good. Very good. So, let's take a look.

So, if the curl of a vector is zero,

it must be the gradient of some scalar field.

That was another theorem that we learned in the previous lecture.

That is to say, the difference between those two vector field,

A prime and A will be Del Psi is arbitrary scalar function or a scalar field.

So, if we rearrange this equations,

we can say A prime is equal to A plus Del Psi.

That means that if A is a satisfactory vector potential for a problem,

then for any Psi at all,

A prime will be on equally satisfactory vector potential leading to the same field B.

Now, so, we have some freedom of choice.

Right? So, by suitable choice of Psi,

we can make the divergence of A prime anything we wish.

So, we will do some mathematical trick.

We're going to put divergence operator in front of both of these terms,

and as we can see the divergence of A prime will be equal to divergence of

A plus divergence of gradient is Laplacian.

So it'll be Laplacian of Psi.

So, for magnetostatics, we will make a simple choice.

Later when we take up electrodynamics,

we will change our choice for our convenience.

But for now, we will only use this one where the divergence a is zero,

and you will see in the later slides why this handy and useful.

Okay? So, let's take a look at the vector potential of a uniform magnetic field.

This will be the simplest case,

where you have only one direction of

your vector and they are the same no matter where you are.

In that case, we can imagine the x and y field is zero,

and only we have z, z-field.

If we equate them with the Curl of the vector potential,

you can see only the last equation will have B Naught and the other will be zero.

Now, by inspection, we see that possible solution include,

but are not limited to these three examples.

So, let's take a look at the first example.

A_y is equal to xB Naught,

A_x is equal to zero,

A_z is equal to zero.

So, if you put them into these equations,

you will see they all satisfy the equations that are here.

Now, let's take a look at the last examples here where we have minus half yB Naught,

and A_y we have half xB Naught,

and A_z is equal to zero.

If you put them into this equation,

again see that they satisfy this equation.

But this is a typical example for a rotating field around the current carrying wire,

and we will see that example in the following slide.

So, we will see rotating vector potential for a uniform B-field,

and the third solution has some interesting properties.

Since the x component is proportional to minus y,

and the y components is proportional to plus x,

A must be at right angles to the vector from the z-axis as you can see.

So, why are they at right angles?

As you can see at any arbitrary point,

the slope of your r vector is y over x.

While the slope of your A vector is minus x over y.

If you multiply two slopes and becomes minus one,

they are at right angles.

Right? That's the case here.

So, you can understand that.

Now, if you take a look at these three components,

you can neatly arranged them into a vector form,

which is written here.

A is equal to one over two B Cross r prime.

So, I'm going to ask Melodie if

the vector potential A is equal to one over half B Cross r prime,

which direction will A pointing to?

So, A will be pointing perpendicular to both the magnetic field and the electric field.

Exactly. Because this is a cross product,

this will make this vector perpendicular or orthogonal to both of them.

So, take a look at this picture.

So, you have r and A is perpendicular to that,

and B is coming out of this plane.

So, it's also perpendicular to that as well.

The vector potential A has the magnitude

Br prime over two as you can inspect from this figure,

and it rotates about the z-axis in counter clockwise,

and that you can understand with right-hand rule.

So, if you have B-field to my direction,

you will have a counterclockwise rotation of A-field.

Now, if the B-field is the actual field inside a solenoid,

then the vector potential circulates in

the same sense as do the currents of the solenoid.

You will remember, inside the solenoid,

you have uniform B-field.

So, you can use that analogy to

understand the vector potential as well as magnetic field, right?

So continuing the discussion about the vector potential for a uniform field,

the circulation of A on any closed loop gamma,

can be related to the surface integral of the curl of A by Stoke's theorem.

At this point, I'm going to ask Melodie what Stoke's theorem was about?

Okay, so we learned about that in the prior lecture with the potato chip analogy.

So basically, what we learned is that the line integral of

the vector field on the outside is going to be equal to

the surface integral of the curl of the vector field.

Perfect. So that is now expressed in mathematical terms here,

what Melodie has just explained to you.

If you take a look at del cross A here,

what does it reminds you of?

I think we had that formula earlier,

and it should be equivalent to the magnetic field.

Exactly. So we can replace the cross product here,

or the curl of the vector field A,

vector potential A, by b.

Once we do that,

you see it becomes a flux of a magnetic field

through the area of interest which is bounded by gamma.

So, now we have this equation.

So, if we take a circular loop of radius R prime

in a plane perpendicular to a uniform field B,

the flux is just pi R prime squared B because you have a symmetry, right.

If we choose our origin on an axis of symmetry,

so that we can take A as circumferential and a function of only R prime,

the circulation will be,

this one will be two pi r prime which is periphery of your circle times

a is equal to the area of the circle which is pi r prime square times B.

Therefore, A is equal to b r prime over two.

So again you can see Br prime over two will be the magnitude of the vector potential.

So, now we will move on to a little bit more deeper subjects.

The vector potential of known's currents.

Now, we have related vector potential with a magnetic field and we

know magnetic field is related to current from the force equation.

Now we are going to relate the current directly with the vector potential.

So, let's see how that works.

Mathematically identical equations for vector and

scalar potential can be found when we rewrite

our equations in terms of vector potential A and we

can correlate that with the equations for electrostatic potential phi.

Now, this is the fourth Maxwell force equations and we just

replaced by del cross A and you remember this formula, right.

So, del cross del cross A,

is equal to gradient of del.A minus laplacian of A and you'll remember also that we

chose the boundary condition or the specific number

for our vector potential to have no divergence del.A is equals 0.

Now you see why that's convenient because then you can make one term disappear.

So now you'll only have what Laplacian.

Laplacian of A will be equal to minus j over epsilon naught c squared.

If you compare that with the Poisson's equation,

laplacian of phi is equal to minus rho dot epsilon naught you have similarity.

Similarity. You just replace A by phi by A and rho by j over c square, right.

You know this has three components.

So, you have AX AY AZ.

That's different from phi.

Phi is a scalar potential so you don't have one but you have three siblings here, right.

So, then you can just do the math and see that you

can have three independent equations that look exactly the same.

Just the differences you have different subscript x and y and z, okay.