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FInd a simple expression for the vector potential which will yield this field

  1. Apr 8, 2008 #1
    em

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

    In a certain region, the magnetic field as a linear function of time is given by

    [tex]
    B = Bo \frac{t}{/tau} z hat
    [/tex]
    Bo and tau constants.

    A)FInd a simple expression for the vector potential which will yield this field.
    B)Assuming the scalar potential is a constant, find E from the above result for the vector potential. Check that your answer is consistent with the diff form of faradays law

    2. Relevant equations
    [tex]
    \nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
    [/tex]

    [tex]
    \mathbf{A} = \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{rsingle-quote},t_r)}{\cal{R}} d\tausingle-quote
    [/tex]

    [tex]
    t_r \equiv t - \frac{\cal{R}}{c}
    [/tex]


    3. The attempt at a solution

    [tex]
    B= \nabla \times A
    [/tex]

    We take the curl of B on the LHS and on the RHS we have
    [tex]
    \nabla \times (\nabla \times A )
    [/tex]

    Eventually we will get to this [tex]
    \nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
    [/tex]


    This is assuming were in the Lorentz Gauge, we don't know if we are. Or do we assume we are? We have to satisfy E = -delV -dA/dt but we don't know either of those quantities.
     
  2. jcsd
  3. Apr 8, 2008 #2
    say if

    kt y-hat = K x xhat

    What mathematical operation would I apply to show that
    K = kt z-hat

    I don't thin kthat is even relevent
     
    Last edited: Apr 8, 2008
  4. Apr 8, 2008 #3

    pam

    User Avatar

    A) Take curl(BXr)/2 and see what you get.
     
  5. Apr 8, 2008 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The simples way to go about this is to simply write explicitly [tex]
    B= \nabla \times A
    [/tex]
    in cartesian coordinates.

    You will get simple equations for A_x, A_y and A_z. It should be easy to pick functions A-x, A_y, A_z that satisfy those equations. You don't need anything else.
     
  6. Apr 8, 2008 #5
    Doing this would only yield

    dA_y/dx - dA_x/dy = Bo(t/tau) since B has only z-components.

    Solving this differential equation would still give unknown A_y A_x.

    There should be an explicit way of doing this, no?
     
  7. Apr 8, 2008 #6
    Remember that the vector potential is not physically different even if you add a gradient of a function. So you have a certain freedom available to pick the solution.
     
  8. Apr 8, 2008 #7

    nrqed

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    Science Advisor
    Homework Helper
    Gold Member

    Of course A is defined up to gauge transformations only so many different A's give the same physics. This is why the question asks you to find a simple expression for A. Just pick the simplest A that yields the correct B

    (don't forget to make sure that the other components of the equation also work)
     
  9. Apr 8, 2008 #8
    OKay problem solved. You can close this thread.
     
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