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How to find vector magnetic potential given magnetic field? 
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#1
Mar911, 12:05 PM

P: 3,883

If we are given B, how can we find A? I can fine the magnitude of A by:
[tex] \int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}[/tex] So given B and the surface area, you can get the magnitude of A. But how do you get the direction information? All I know is A is orthogonal to B and [itex]\vec B = \nabla X \vec A [/itex]. But still I cannot find a formula to nail down the direction. Anyone can help? 


#2
Mar911, 02:26 PM

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P: 1,256

It's a bit like long division. There is no general result. For a specific B, you try to find a vector A whose curl is that B. If you can write B in a coordinate system, then you can write partial differential equations whose solution give A.
You can also use divB=del^2 A and solve that equation for A. 


#3
Mar911, 09:59 PM

P: 969

Also, I don't see how: [tex] \int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}[/tex] gets you the magnitude of A. 


#4
Mar911, 10:00 PM

P: 969

How to find vector magnetic potential given magnetic field?



#5
Mar911, 11:23 PM

P: 3,883

[tex]\hat x \;X\; \hat y = \hat z [/tex] [tex] \Phi = \int_{s'} \vec B \cdot d\vec{s'} \;\Rightarrow \;\int_{C} \vec A \cdot d\vec{l}= \Phi[/tex] So If you are given the B and given the surface that the flux cut through. You get the A within that surface. 


#6
Mar911, 11:49 PM

P: 969

As for your second equation, you can get the line integral of A around the perimeter of the surface, but I don't think you can get A at a particular point on the perimeter, or at the center point of the surface. addendum: here's a way to visualize why you can't get A at a particular point on the perimeter. This is a triforce taken from a popular classic video game: http://www.google.com/imgres?imgurl=...ed=0CB0Q9QEwAA You know the line integral of the middle triangle. But you want the field on the left side of that middle triangle. So how do you isolate just that part? Well you know the line integral of the leftmost triangle, and the rightmost part of the leftmost triangle is what you're looking for. So by considering the leftmost triangle, you do get an extra equation involving the left side of the middle triangle, but unfortunately this equation causes you to have two more sides (the remaining sides of the leftmost triangle), so in general you don't have enough information to get A at all. I hope this is right  someone correct me if I'm wrong! 


#7
Mar1011, 06:19 AM

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#8
Mar1011, 11:40 AM

P: 125

[tex]\nabla^2 \vec A =\mu_0\vec J[/tex] whilst remembering that [tex]\nabla\times \vec B =\mu_0\vec J[/tex] This, togther with suitable boundary conditions, (eg. specifing B on the boundary) should give you the result you are looking for. Please note, that an analytical solution may not be obtainable. The "Amperes Law for A" method; [tex]\int_S \vec B\cdot d\vec a = \oint_{\partial S} \vec A\cdot d\vec l[/tex] That you suggested in your OP works only if you have some suitable symmetry in the problem (ie A and dl are parallel at all points along the integration curve). The only case I can think of off the top of my head is the magnetic potential inside an infinitely long solenoid 


#9
Mar1011, 05:25 PM

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I am still reading the materials in the book. 


#10
Mar1011, 05:27 PM

P: 3,883

[tex]\int_S \vec B\cdot d\vec a = \oint_{C} \vec A\cdot d\vec l[/tex] can only be true with certain symetry like a long wire carrying current or a long solenoid? Yes this is from a problem of a long solenoid. 


#11
Mar1011, 05:59 PM

P: 3,883

I read Griffiths page 234 in Magnetic Vector Potential in STATIC condition where [itex]\nabla X \vec B = \mu_0 \vec J [/itex].
[tex] \nabla X \vec B= \nabla X \nabla X \vec A = \nabla(\nabla \cdot \vec A )  \nabla^2 \vac A = \mu_0 \vec J [/tex] We invoke gauge where [itex] \nabla \cdot \vec A = 0 [/tex] [tex]\Rightarrow\; \nabla^2 \vec A =\mu_0 \vec J [/tex] Does this imply in all cases of static condition, A and J are in the same direction in all static case? On the other note, for time varying condition where: [tex] \nabla X \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac {\partial {\vec E}}{ \partial t}[/tex] A and J are not in the same direction. Am I correct? 


#12
Mar1111, 05:43 AM

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Del^2A need not be in the same direction as A.



#13
Mar1111, 11:17 AM

P: 3,883

[tex] \nabla^2 \vec A = \mu_0 \vec J \;\Rightarrow\; \vec A(\vec r) = \frac {\mu_0}{4\pi}\int_{v'} \frac {\vec J(\vec r)}{\vec r} dv' [/tex] And [tex] \vec A = \frac {\mu_0}{4\pi}\oint_{C'} \frac {\vec I}{\vec r} dl' [/tex] That shows A is same direction of J. Is this right? 


#14
Mar1111, 11:58 AM

P: 125

To adress your other question, no. In general you cannot assume that A and J are parallel at all points, but this is where symmetry comes in. In case of a a solenoid, you have current running in the [tex]\hat \phi[/tex]direction only and thus the PDE becomes: [tex]\nabla^2 A_r = 0[/tex] [tex]\nabla^2 A_\phi = J_{\phi}[/tex] [tex]\nabla^2 A_z = 0[/tex] A trial solution of A_{r}=A_{z}=0 satisfies the first and last equation and thus, by the uniqueness theorem, are the solutions, thus A has only a phicomponent, in this case. 


#15
Mar1111, 04:15 PM

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#16
Mar1111, 04:39 PM

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[tex] \vec A = \frac {\mu_0}{4\pi}\oint_{C'} \frac {\vec I}{\vec r} dl' [/tex] [tex] \Rightarrow\; (\hat A)\; A \;=\; (\hat I)\;[\frac {\mu_0}{4\pi}\oint_{C'} \frac {I}{\vec r} dl'] [/tex] [tex] \Rightarrow\;\hat A =\hat I [/tex] 


#17
Mar1111, 07:00 PM

P: 192

Lookup Helmholtz decomposition for vector fields, which I think is what you're tiptoeing around.
http://en.wikipedia.org/wiki/Helmholtz_decomposition It will give you the vector potential from the field, or field from the potential. 


#18
Mar1111, 07:53 PM

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The unit vector I CAN NOT be pulled out of the integral.
It can, and commonly does, vary in direction during the integration. 


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