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Magnetic Mirror for Neutral Atoms

  1. Aug 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider an infinite sheet of magnetized tape in the x-z plane with a nonuniform periodic magnetization M = cos(2πx/λ), where λ/2 is the distance between the north and south poles of the magnetization along the x-axis. The region outside the tape is a vacuum with no currents or time varying fields anywhere in space.
    Assume that the top surface of the tape is at y=0 and the B-field at the top is B= B_0[(y-hat)*sin(2πx/λ) -(x-hat)*cos(2πx/λ)].
    a) Show that the H field in vacuum is given by a scalar potential H =-∇Φ.
    b) Find the most general Φ by solving Laplace's equation.
    c) State the boundary conditions on B and H, at y=0 and y=∞, necessary to find B in vacuum.
    d) Find B in the vacuum by the results of b) and c).
    e) A hydrogen atom of magnetic moment μ(y-hat) approaches the tape from above along x=3λ/4. Find the force on the atom as function of distance from the tape. Does the tape attract or repel this atom?

    2. Relevant equations
    F= -∇(μ⋅B), ∇⋅H =-∇⋅M, Δ(∂Φ/∂r) =ΔM

    3. The attempt at a solution
    a) The region outside the tape has no currents so ∇xH = ∇⋅H =0 ⇒ H= -∇Φ since ∇x(∇) =0. Therefore Φ satisfies Laplace's equation in 2-D.
    b) In polar coordinates the general solution is Φ(r,θ) =∑ (A_l r^l + B_l /r^[l+1]) P_l (cosθ).
    c) The potential must be continuous at the boundary and finite everywhere. The condition on H is the third relevant equation.
    d) Here I get stuck and am not sure how to use the boundary conditions.
    e) After finding B from d) we can use the first relevant equation to find the force on the atom.
     
  2. jcsd
  3. Aug 15, 2016 #2

    TSny

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    Welcome to PF!
    Should there be a negative sign on the right side of this equation?

    ok
    Are polar coordinates appropriate for this problem?
     
    Last edited: Aug 15, 2016
  4. Aug 15, 2016 #3
    Taking your questions in turn, no there shouldn't be a negative sign and the lack of azimuthal symmetry implies that polars are not appropriate. So I should just use the general solution in terms of circular and hyperbolic trig functions, yes?
     
  5. Aug 15, 2016 #4

    TSny

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    Yes. (You can use simple exponential functions in place of the hyperbolic functions.)
     
  6. Aug 15, 2016 #5
    Ok, so with parts a) and b) out of the way how would I use the boundary conditions?
     
  7. Aug 15, 2016 #6

    TSny

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    Can you write the general solution for ##\Phi## before applying boundary conditions?
     
  8. Aug 15, 2016 #7
    Sure, Φ = (A e^kx +Be^[-kx])*(C*sin(ky) +D*cos(ky)), for k^2>0 and a similar expression for k^2<0 where k^2 is a separation constant.
     
  9. Aug 15, 2016 #8

    TSny

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    If k is real, what happens to ##\Phi## as x goes to ±∞?
     
  10. Aug 15, 2016 #9
    Right Φ would blow up exponentially. So a boundary condition we impose is that Φ should vanish as x→∞, which in this instance requires k to be imaginary.
     
  11. Aug 15, 2016 #10

    TSny

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    ##\Phi## does not need to vanish as x → ±∞. In fact, since the magnetic surface is infinitely large, you would not expect ##\Phi## to vanish as x or z → ±∞.
     
  12. Aug 15, 2016 #11
    Right, just like the electric potential of an infinite plane doesn't vanish. I'm not sure what you're driving me towards then.
     
  13. Aug 15, 2016 #12

    TSny

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    I was just trying to point out that your statement in post #9 that ##\Phi## should vanish as x → ∞ is incorrect. Therefore you can not use that as a reason for k to be imaginary.

    EDIT: Note that the statement "##\Phi## cannot blow up exponentially as x →±∞" is not equivalent to the statement "##\Phi## should vanish as x → ±∞"
     
    Last edited: Aug 15, 2016
  14. Aug 15, 2016 #13
    Ah, ok. I'll post a better reason tomorrow.
     
  15. Aug 15, 2016 #14

    TSny

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    OK. You are on the right track. Just stick with the idea that ##\Phi## cannot blow up exponentially as x →±∞.
     
  16. Aug 16, 2016 #15
    If k is imaginary then e^(kx) will be an element of U(1) and rather than blowing up it will circulate faster and faster, allowing for a finite potential at infinity and no exponential increase.
     
  17. Aug 16, 2016 #16

    TSny

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    OK. If k is imaginary then try to see that you get circular functions of x and exponential functions of y. That is, if k = ib, then the set {ekx, e-kx} may be replaced by the set {sinbx, cosbx}. And the set {sinky, cosky} may be replaced by the set {eby, e-by}.

    The same result would have come from replacing the separation constant k2 by -b2 when solving Laplace's equation.
     
  18. Aug 16, 2016 #17
    Ok, so now that I have the general solution for Φ how do I implement the boundary conditions?
     
  19. Aug 16, 2016 #18

    TSny

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    Think about the behavior of the general solution for ##\Phi## at the "boundaries" of x or y.
     
  20. Aug 16, 2016 #19
    Φ needs to be continuous across the coordinate planes, y=0 and x=0, so we impose conditions on the coefficients in the general solution.
     
  21. Aug 16, 2016 #20
    And the normal derivative of Φ has a discontinuity equal to the change in magnetization, giving a further condition.
     
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