# Vector differential displacement - Magnetic vector potential

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1. Apr 7, 2017

### italy55

1. The problem statement, all variables and given/known data

http://imgur.com/a/k7fwG
Find the vector magnetic potential at point P1.

2. Relevant equations
Vector magnetic potential given by:
$$d \bar{A} = \frac{\mu I d\bar{l'}}{4 \pi | \bar{r} - \bar{r'} | }$$
3. The attempt at a solution
I split up the problem in 3 parts,

first solve for potential along the x-axis:
$$A_x= \frac{\mu I }{4 \pi } \int \frac{\mu I d\bar{l'}}{4 \pi | \bar{r} - \bar{r'} | } = \frac{\mu I }{4 \pi } ln(sqrt(2) +1)$$
$$\bar{r} =(a,a,0) , \bar{r'} = (x',0,0) , \bar{dl'} = \bar{e_x} dx'$$

second, I guess this one is the same except the sign:
$$A_y= \frac{\mu I }{4 \pi } \int \frac{\mu I d\bar{l'}}{4 \pi | \bar{r} - \bar{r'} | } = - \frac{\mu I }{4 \pi } ln(sqrt(2) +1)$$
in the first one, I put $$\bar{r} =(a,a,0) , \bar{r'} = (0,y',0) , \bar{dl'} = - \bar{e_y} dy' (wrong??)$$

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My questions is about the vector differential displacement $$\bar{dl'}$$, how does the vector differential displacement change between the first, second and the third one?

I'm confused in this part because my formula sheet says that $$\bar{dl'} = \bar{e_x} dx + \bar{e_y} dy + \bar{e_z} dz$$

How should I think when using the vector differential displacement in the 3 different cases?

2. Apr 7, 2017

### kuruman

You have to do three integrals, one for each side of the triangle. This means that you should write three differential displacements. For example, the one along the hypotenuse is $d \vec{l}_{hyp.}= - \frac{1}{\sqrt{2} }\hat{e}_x dx + \frac{1}{\sqrt{2} }\hat{e}_y dy.$ Stick it in the integral and integrate. The unit vectors help keep track of the components. The other two differentials should be easier.

3. Apr 8, 2017

### italy55

yes thanks, I'm aware of that I should split it up in to 3 integrals. Just a bit confused about the "differential displacement vector".

So the differential displacement depends on which direction I integrate(direction the current is going?)? For example "$$\bar{dl'} = \bar{e_x} dx + \bar{e_y} dy + \bar{e_z} dz$$" is not always like "a constant" in cartesian coordinates, instead it depends on the direction of integration, correct?

and a question about the direction of the current "I", does this decide only in which direction I should integrate or can the current become negative somehow(for examplel -I )?

4. Apr 8, 2017

### kuruman

The two questions you ask are related. Perhaps you will be able to answer them yourself if you consider that the more general expression for the vector potential is
$$\vec{A}(\vec{r}) =\frac {\mu_0}{4 \pi} \int { \frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|} dV'}$$
where $\vec{J}(\vec{r}')$ is the current density. In the special case of a current loop $\vec{J}(\vec{r}')~dV' \rightarrow I d\vec{l}'$.

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