Solving Magnetic Field Problem: Reitz-Milford-Cristy 9.15 (4th Ed.)

facenian
Messages
433
Reaction score
25

Homework Statement


This is a problem from Reitz-Milford-Cristy-Problem 9.15(fourth Ed.) .There is a long wire carrying a current "I" above a plane,inside the plane we have to cases a)μ=∞; and b)μ=0. We must find the field above the plane.

Homework Equations


H=∇∅
Δ∅=0


The Attempt at a Solution


To solve the problem we split the field sources: 1)wire + 2)Magnetic plane. Part 1) is already known and to find 2) we put H=-grad∅ and solve laplace equation Δ∅=0. To find ∅ we consider an image current in case a) it is a parallel current and in case b) it is an antiparallel one.
I understand case b), in this case we have B=0 inside the plane so the normal component of H must vanish so the normal derivate of ∅ must cancell that of "I" and the problem is uniquely solved.
However in case a) we have H=0 and the tangencial component of H must vanish, we can do this with a parallel image however in this case what we know is the tangencial derivate of ∅ and the unicity theorem for laplace equation does not apply.
Can someone shed some light on this?
 
Physics news on Phys.org
facenian said:
However in case a) we have H=0 and the tangencial component of H must vanish, we can do this with a parallel image however in this case what we know is the tangencial derivate of ∅ and the unicity theorem for laplace equation does not apply.
Can someone shed some light on this?

If the tangential derivative of ∅ vanishes on the boundary plane, what does that tell you about how ∅ varies on the boundary plane?
 
TSny said:
If the tangential derivative of ∅ vanishes on the boundary plane, what does that tell you about how ∅ varies on the boundary plane?
The problem is that the tangencial derivative of ∅ does not vanish but it is equal an opposite to the tangencial component of source 1) (the wire) so the field generated by the to sources combined add up to a null tangencial field
 
OK. I see. ∅ is the magnetic scalar potential of the image current alone.

From the known tangential component of H on the boundary plane due to the true current, you know the tangential component of H due to the image current. So, as you said, you know the value of the component of the gradient of ∅ in the direction of the tangential component of H, ∇t∅(x,y), at each point (x,y) on the boundary plane.

Taking ∅ = 0 at infinity, can you use ∇t∅(x,y) to find (in principle) the value of ∅ at any point of the plane? Does uniqueness follow?
 
TSny said:
Taking ∅ = 0 at infinity, can you use ∇t∅(x,y) to find (in principle) the value of ∅ at any point of the plane? Does uniqueness follow?

Now I see, determining the tangencial component in this case is equivalent to determine the value of ∅ on the boundary surface so uniqueness can be applied. Thank you TSny
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top