How Does a Magnetized Tape Affect a Hydrogen Atom?

[TeslaCoil137]

Homework Statement

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?

Homework Equations

F= -∇(μ⋅B), ∇⋅H =-∇⋅M, Δ(∂Φ/∂r) =ΔM

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.

 

[TSny]

Welcome to PF!

Homework Equations

F= -∇(μ⋅B)

Should there be a negative sign on the right side of this equation?

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.

ok

b) In polar coordinates the general solution is Φ(r,θ) =∑ (A_l r^l + B_l /r^[l+1]) P_l (cosθ).

Are polar coordinates appropriate for this problem?

 

[TeslaCoil137]

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?

 

[TSny]

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?

Yes. (You can use simple exponential functions in place of the hyperbolic functions.)

 

[TeslaCoil137]

Ok, so with parts a) and b) out of the way how would I use the boundary conditions?

 

[TSny]

Can you write the general solution for $\Phi$ before applying boundary conditions?

 

[TeslaCoil137]

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.

 

[TSny]

If k is real, what happens to $\Phi$ as x goes to ±∞?

 

[TeslaCoil137]

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.

 

[TSny]

$\Phi$ 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 → ±∞.

 

[TeslaCoil137]

Right, just like the electric potential of an infinite plane doesn't vanish. I'm not sure what you're driving me towards then.

 

[TSny]

Right, just like the electric potential of an infinite plane doesn't vanish. I'm not sure what you're driving me towards then.

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 → ±∞"

 

[TeslaCoil137]

Ah, ok. I'll post a better reason tomorrow.

 

[TSny]

OK. You are on the right track. Just stick with the idea that $\Phi$ cannot blow up exponentially as x →±∞.

 

[TeslaCoil137]

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.

 

[TSny]

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.

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 {e^kx, e^-kx} may be replaced by the set {sinbx, cosbx}. And the set {sinky, cosky} may be replaced by the set {e^by, e^-by}.

The same result would have come from replacing the separation constant k² by -b² when solving Laplace's equation.

 

[TeslaCoil137]

Ok, so now that I have the general solution for Φ how do I implement the boundary conditions?

 

[TSny]

Think about the behavior of the general solution for $\Phi$ at the "boundaries" of x or y.

 

[TeslaCoil137]

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.

 

[TeslaCoil137]

And the normal derivative of Φ has a discontinuity equal to the change in magnetization, giving a further condition.

 

[TSny]

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.

The boundaries for x are x = +∞ and x = -∞. We already used these boundaries to determine that k is imaginary.

What are the two boundaries for y? Can you learn anything about your general expression for $\Phi$ by considering either of these two boundaries? It might be a good idea to write out your expression for $\Phi$ at this point.

 

[TSny]

And the normal derivative of Φ has a discontinuity equal to the change in magnetization, giving a further condition.

The problem gives an expression for the B field at the surface y = 0. Is there a way to relate $\Phi$ to B?

 

[TeslaCoil137]

The vacuum region is defined by y=0 and y=∞, directly at y=0 we have the B field, which relates to Φ by B= μ(-∇⋅Φ), with μ the magnetic permeability of free space. Working out the gradient of Φ and setting it equal to B/μ at y=0 I get the set of equations
1) b(C+D) (Acos(bx)-Bsin(bx)) = - B_0 sin(2πx/λ)
2) b(C-D) (Acos(bx) +Bsin(bx)) = B_0 sin(2πx/λ),
which is clearly under-determined.

 

[TSny]

Thinking about y at infinity should help determine either C or D.

On the right side of your two equations you have a sine function in each equation. Is this right?

 

[TeslaCoil137]

Thank you for pointing out that clerical error :P The x-component has a cosine function, and from asymptotic behavior C should be zero as otherwise the potential blows up exponentially.

 

[TSny]

Yes.

 

[TeslaCoil137]

Ok, so now I need one more equation to determine the potential. The magnetic 'charge' density M is what gives rise to the scalar potential so do I apply my third relevant equation at this point and grind through the algebra?

 

[TSny]

I'm not familiar with your third equation. But I don't think you need it. Note that the constant D is now just an overall constant factor in $\Phi$ that can be "lumped in" with A and B to get constants AD and BD which you can call F and G, say. So, you now have the potential in terms of just two constants which you can determine by making sure you get the right B_x and B_y at y = 0.

 

[TeslaCoil137]

Ah, thank you very much.