Magnetostatics: Magnetic Vector Potential

[KEØM]

Homework Statement

Give an expression for the magnetic field and show that a magnetic vector exists such as \vec{A}(P) = A(r)\hat{z} and \vec{B}(P) = \vec{\nabla} \times \vec{A}
For the infinite wire shown in figure 1.

Here is a link to the figure and problem statement. The problem is the second problem on the first page.

https://docs.google.com/fileview?id...UtOGNkYi00ZGQyLTkxOTktNWVjYzM2MGViNDg3&hl=en"

Homework Equations

\vec{\nabla} \times \vec{A} = \left(\frac{1}{r}\frac{\partial A_{z}}{\partial \theta} - \frac{\partial A_{\theta}}{\partial z}\right)\hat{r} + \left(\frac{\partial A_{r}}{\partial z} - \frac{\partial A_{z}}{\partial r}\right)\hat{\theta} + \frac{1}{r}\left(\frac{\partial(rA_{\theta})}{\partial r} - \frac{\partial A_{r}}{\partial \theta}\right)\hat{z} curl in polar coordinates

\oint \vec{B} \cdot d\vec{l} = \mu_{0}I_{enc} Ampere's Law

The Attempt at a Solution

First we must solve for the magnetic field at point P. Using Ampere's Law we get,

\vec{B}(P) = \frac{\mu_{0}I}{2\pi r}\hat{\theta} where the positive theta direction is into the page.

Now in order to find A we can set the curl equations equal to the magnetic field equation by curl(A) = B.

Doing so gives,

\frac{1}{r}\frac{\partial A_{z}}{\partial \theta} - \frac{\partial A_{\theta}}{\partial z} = 0

\frac{\partial A_{r}}{\partial z} - \frac{\partial A_{z}}{\partial r} = \frac{\mu_{0}I}{2\pi r}

\frac{\partial(rA_{\theta})}{\partial r} - \frac{\partial A_{r}}{\partial \theta} = 0

The answer to this problem (given to us by our instructor) is,

\vec{A} = \left(-\frac{\mu_{0}I}{2\pi}ln(r) + K \right)\hat{z}

and the only way I can get there is by saying that because A points most often in the direction of the current (which is the z direction in this case) then,

-\frac{dA_{z}}{dr} = \frac{\mu_{0}I}{2\pi r}.

Then solving this differential equation gives the desired result.

Can someone please point me in the right direction as to how solve this problem in a more rigorous manner?

Do I need to use the equation \vec{A} = \frac{\mu_{0}}{4\pi}\int \frac{\vec{J}(\vec{r'})}{r}d\tau '?Many Thanks in advance,

KEØM

 

[nickjer]

You already know that:

A_r = 0

and

A_{\theta} = 0

and

A_z = A(r)

since it was given to you in the problem. This simplifies your 3 diff eq's quite a bit.

 

[KEØM]

Thanks for your reply nickjer. So because A is a function of only r then all other derivatives go away and because it is only in the z direction then all other components are equal to zero. This then allows me to let -\frac{dA_{z}}{dr} = \frac{\mu_{0}I}{2\pi r}. Is there anything else I must say for that condition to be satisfied?

Thanks again,
KEØM

 

[kreil]

nope that does it:

\vec{B}(P) = \vec{\nabla} \times \vec{A}

\vec{B}(P) = \frac{\mu_{0}I}{2\pi r}\hat{\theta}

\vec{\nabla} \times \vec{A}=-\frac{dA_{z}}{dr}

thus -\frac{dA_{z}}{dr}=\frac{\mu_{0}I}{2\pi r}\hat{\theta}

 

[KEØM]

Thanks! I really appreciate the help kreil and nickjer.

KEØM

 

[KEØM]

I now have a question concerning the third problem on that page attached.

Problem Statement:

We consider now two wires of axis (O,z) and separated by the distance (2a). The currents in the two wires are +I and -I (see figure). Show that the expression of the magnetic potential vector \vec{A}(P) is given by

\vec{A}(P) = \frac{\mu_{0}I}{2\pi}ln\left(\frac{r_{1}}{r_{2}}\right)\hat{z}

(the observation point is located at P(r, \theta, z=0)). Note: The constant of integration is integration obtained by taking \vec{A}(O) = 0.

Relevant Equations:

Same as previous problem.

Attempt at a solution:

For this problem can I say the same as before? He did say that our answer should be of the form \vec{A}(P) = A(r)\hat{z}.

Thanks in advance.

 

[nickjer]

You solved for the vector potential from a single wire. So for two wires you will just add the vector potentials from each. But be careful with the direction of each of the vectors.

 

[KEØM]

I know from just looking at the answer that the -I current must be in the positive z-direction and the +I current must be in the negative z-direction but I don't know why.

 

[nickjer]

In the first problem you solved \vec{A} for a current going in the +z direction. So use that to solve this next problem.

 

[KEØM]

Thanks for your reply nickjer.

I think I got it. So because +I flows in the positive direction it must be negative and the opposite must be true for the wire with -I.

One more question:

For problem 4 on that document it says,

Problem Statement
The wires are now very close to each other. Determine \vec{A}(r,\theta) such as r >> a and by keeping the first and second order terms in (a/r).

Relevant Equations
Remember that (1 + \epsilon)^{n} = 1 + n\epsilon

and ln(1 + \epsilon) = \epsilon, \epsilon << 1

Attempt at a solution

Now I know he wants us to approximate the answer by expanding\vec{A}(P) = \frac{\mu_{0}I}{2\pi}ln\left(\frac{r_{1}}{r_{2}}\right)

with a Taylor Expansion in \epsilon = \frac{a}{r} << 1 but I don't see how to do this when the distance between the wires 2a never comes up in any of our answers.

I really appreciate all the help nickjer.

 

[nickjer]

Now you will need to write out r_1 and r_2 in terms of r and a. I suggest using the law of cosines.

 

[KEØM]

OK. Thanks again for all of your help nickjer.