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Vector differential displacement - Magnetic vector potential

  1. Apr 7, 2017 #1
    1. The problem statement, all variables and given/known data

    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??)


    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. jcsd
  3. Apr 7, 2017 #2


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    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.
  4. Apr 8, 2017 #3
    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 )?
  5. Apr 8, 2017 #4


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    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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