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Vector Potential

  1. Mar 7, 2005 #1
    How to find the Vector Potential of an infinite solenoid with n turns per unit length,radius R and current I.
    since here current extends to infinty..
    How will it be done
    Pls Help
     
  2. jcsd
  3. Mar 7, 2005 #2

    Astronuc

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    Heman, please try to write the equation for vector potential. Then we can better point you in the right direction.

    You can use LateX - see the thread on Math applications of LateX - https://www.physicsforums.com/showthread.php?t=8997 (General physics forum).
     
  4. Mar 7, 2005 #3

    Tom Mattson

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    The infinite solenoid has a uniform magnetic field parallel to the solenoid axis. Let that axis be the z axis, for concreteness. So [itex]\mathbf {B}=B \mathbf {k}[/itex]. Now the vector potential is [itex]\mathbf {B}=\nabla \times \mathbf {A}[/itex]. The vector potential is not unique, so any vector-valued function [itex]\mathbf {A}[/itex] whose curl gives you [itex]B \mathbf{k}[/itex] will fit the bill.
     
  5. Jan 23, 2010 #4
    You may consider using the following:
    gradXA = B
    then use Stoke's theorem to write surface int(gradXA.dS) = line int (A.dl), with the closed curve chosen suitably.

    This gives line int(A.dl) = int(B.ds).

    The surface integral is straight forward while a reasonable choic eof the closed curve makes the line integral staright forward.

    Of help?
     
  6. Jan 23, 2010 #5
    The line integral of A around a closed path is equal to the flux of B through the path. For an infinitely long solenoid, B = (n)(I)/((eps0)(c^2)) at internal points. (B = 0 at all points outside of the solenoid.) Thus the flux of B inside the solenoid is (pi)(R^2)(n)(I)/((eps0)(c^2)). At a distance r>R from the solenoid's axis, (2)(pi)(r)(A)=flux of B. That is,

    A=(n)(I)(R^2)/((2)(eps0)(c^2)(r))
     
  7. Jan 23, 2010 #6

    fluidistic

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    I don't think so. In fact the thread is 5 years old. It might be of help for further people though.
     
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