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Uniform changing magnetic field does not accelerate a charge?

  1. Apr 21, 2014 #1
    Consider a charge in free space. There is initially no magnetic field and the only electric field is the one caused by the charge. Now introduce a B field which increases with time:

    [itex]\vec{B}=B\:t\:\hat{z}[/itex]

    Faraday's Law:

    [itex]\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}[/itex]

    states that a curling electric field would be induced. The integral form of the law would imply that for any loop I choose in the x-y plane, there would be a changing magnetic flux through it and thus a curling electric field around it. But for any loop whose right edge I place on the charge to apply some electric force to it, I can choose an identically sized loop whose left edge is also on the charge. By symmetry there should be no force on the charge in the x or y direction. But since the x-y plane and those parallel to it are the only planes through which there is changing magnetic flux, then it would have been the only plane on which the charge could have had an electric field applied to it. Since there is none there, and I see no reason to accelerate the charge in the z direction, is it safe for me to say the charge is not accelerated?

    edit: Just want to make clear that I know the magnetic field itself would do no work. My question really is if the induced electric field from the changing magnetic field is doing any work.
     
    Last edited: Apr 21, 2014
  2. jcsd
  3. Apr 21, 2014 #2
    I can't see the equations but if dB/dt is in the z direction and the magnetic fields is homogenous in the x-y plane them there will be no locally-induced electric field because you're right: the curls will cancel.

    A bigger issue is that no such field could ever exist, because there exists no possible current density that can create a uniform field. The question is inconsistent with maxwell's equations.
     
  4. Apr 21, 2014 #3
    Thank you for your reply. Your comment regarding there existing no possible current density that can create a uniform field got me thinking and made me recall an example in Griffiths (Example 5.11 on page 236 of the third edition of "Introduction to Electrodynamics"). In this example, he considers a spherical shell of radius R carrying a uniform surface charge σ that is spinning at a constant angular velocity ω. It turns out that the magnetic field inside such a sphere is uniform and is equal to:

    [itex]\vec{B}=\frac{2}{3}\mu_{0}\:\sigma\:R\:\vec{\omega}[/itex]

    (I understand you yourself cannot see this equation, mikeph, but I wrote it out in case someone can)

    Now I ask my question again, but this time with a charge inside that sphere as it increases its rate of rotation so that the field is uniform in the sphere but changing with time. Is it true that no acceleration of the charge occurs there either?
     
    Last edited: Apr 21, 2014
  5. Apr 21, 2014 #4

    jtbell

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    This is the operating principle of the betatron:

    https://teachers.web.cern.ch/teachers/archiv/HST2001/accelerators/teachers notes/betatron.htm

    IIRC Griffiths has an example (or maybe an exercise) which has a cylindrical region with a uniform axial magnetic field that varies with time, and calculates the induced E as a function of r. I don't have my copy at hand so I can't give you a page reference.
     
  6. Apr 21, 2014 #5

    Meir Achuz

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    It works for the betatron because the B field has cylindrical symmetry, and vanishes for r>R, where can be any radius, large or small. Then the E field is given by symmetry to be rX(dB/dt)/2, where r is the distance from the center of symmetry. If the B field does not vanish beyond a radius R, the E can't be gotten by symmetry, but your example of a B that is non-vanishing everywhere does suggest E=0. It is interesting, as a feature of the symmetry, that the behavior of B at huge distances can affect E close in.
     
    Last edited: Apr 21, 2014
  7. Apr 22, 2014 #6

    jtbell

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    According to the integral form of Ampère's Law:
    $$\oint {\vec E \cdot d\vec l} = - \frac{d}{dt} \int_A {\vec B \cdot d\vec a}$$
    knowing the rate at which the magnetic flux through the loop is changing (the right-hand side of the equation) tells you only the (total) line integral of ##\vec E## around the loop. By itself, it tells you nothing about ##\vec E## at any specific point on the loop. For that, you need to know the direction of ##\vec E## from symmetry arguments, and (usually) identify a path along which its magnitude must be uniform.

    It's Example 7.7 in the third edition.
     
  8. Apr 23, 2014 #7

    Jano L.

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    This is not a valid reasoning. If B is posed as uniform all over the space as specified in the first post, the equations
    $$
    \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}
    $$
    $$
    \nabla \cdot \mathbf E = 0
    $$
    have infinity of solutions, none of which is identical zero vector field. The lack of boundary condition breaks uniqueness of the solution, but does not mean that the solution is a zero vector field.
     
  9. Apr 23, 2014 #8
    Thank you for the feedback. What about the more realistic scenario I posed in post #3? A charge placed in a sphere with uniform surface charge that is undergoing angular acceleration. The magnetic field inside the sphere would be uniform and increasing with time. What happens to the charge?
     
  10. Apr 23, 2014 #9

    Jano L.

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    The sphere determines the field uniquely, because we have boundary conditions now. The electric field should be similar to that in between poles of a magnet, i.e. rotating vector field whose lines are circles in the plane of rotation. So if there is a charged particle inside the sphere, it should experience electric force perpendicular to its radius vector (beginning in the center of the sphere) and parallel to the equatorial plane.
     
  11. Apr 24, 2014 #10
    This problem reminds me of the problem of a infinite volume filled with a uniform charge density. What's the force on a point test charge?
     
  12. Apr 30, 2014 #11

    Meir Achuz

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    I thought about this a little. For the magnetic field case,
    [itex]{\bf E}=\frac{1}{2}{\bf r}\times(\partial_t{\bf B})+\nabla\psi[/itex],
    where [itex]\psi[/itex] is any scalar field. The scalar field is determined by the symmetry.
    If B vanishes beyond a sphere of radius R (which may be infinite), then the gradient is zero if the origin for r is taken to be the center of the spherical symmetry.
    For the uniform charge density, [itex]{\bf E}=\rho{\bf r}/3+\nabla\times{\bf V}[/itex],
    where V is any vector. For spherical symmetry, the curl is zero if the origin for r is taken to be the center of the spherical symmetry.
     
  13. May 1, 2014 #12
    All seems correct except the part about R being possible infinite. Where do you suppose is the center of an infinite sphere? The problem for the infinite sphere doesn't have a unique solution because any point may be taken as the center of symmetry
     
  14. May 1, 2014 #13

    clem

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    An "infinite sphere" is a sphere of radius R, with R becoming infinite.
    Check the meaning of 'infinity'.
     
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