Magnetic scalar potential and function expansion

[jfy4]

Homework Statement

Consider two long, straight wires, parallel to the z-axis, spaced a distance [itex]d[/itex] apart and carrying currents [itex]I[/itex] in opposite directions. Describe the magnetic field [itex]\mathbf{H}[/itex] in terms of the magnetic scalar potential [itex]\Phi[/itex], with [itex]\mathbf{H}=-\nabla \Phi[/itex]. If the wires are parallel to the z-axis with positions [itex]x=\pm d/2,\; y=0[/itex] show that in the limit of small spacing, the potential is approximately that of a two dimensional dipole
[tex]
\Phi\approx -\frac{Id\sin\phi}{2\pi \rho}+\mathcal{O}(d^2/\rho^2)
[/tex]

Homework Equations

For 2D, the general solution for a polar coordinates problem is
[tex]
\Phi(\rho,\phi)=a_0 + b_0\ln\rho + \sum_{n=1}^{\infty}a_n \rho^n \sin(n\phi+\alpha_n)+\sum_{n=1}^{\infty}b_n \rho^{-n} \sin(n\phi +\beta_n)
[/tex]

The Attempt at a Solution

Well, I already have the solution for this problem doing it a different way... but I was thinking about it some more and I was wondering if it's possible to solve it using the equation above as a solution to Laplace's equation
[tex]
\nabla^2 \Phi=0
[/tex]
and writing down a solution as a series solution. But I don't know how to properly implement the BCs (if any...) to begin solving it this way. Is it possible to get a solution to Laplace's equation this way, and if so, is it close to (or faster) that simply taking the scalar potential of a wire and using superposition? Also, in the event that its practical, can someone give me a hand in setting it up?

Thanks,

 

[mark726]

I would approach this problem by first understanding the physical setup and the governing equations. In this case, we have two long, straight wires parallel to the z-axis with opposite currents, which will create a magnetic field around them. The magnetic field \mathbf{H} is related to the magnetic scalar potential \Phi by the equation \mathbf{H}=-\nabla \Phi. This means that the magnetic field is the negative gradient of the scalar potential.

To solve for the magnetic scalar potential, we can use the fact that it satisfies Laplace's equation, \nabla^2 \Phi=0, in the region outside of the wires. This means that we can write the potential as a series solution in polar coordinates, as given in the homework equations.

Next, we need to consider the boundary conditions. In this case, we have the wires located at x=\pm d/2 and y=0. This means that the potential must be continuous across these boundaries, which can be written as \Phi(x=\pm d/2,y=0)=constant. We also know that the potential must go to zero at infinity, which can be written as \Phi(\rho=\infty,\phi)=0.

Using these boundary conditions, we can determine the coefficients of the series solution for the potential. We know that the potential must have a logarithmic term, b_0\ln\rho, to satisfy the boundary condition at infinity. We can also use the symmetry of the problem to determine that the potential must be an even function of \rho and an odd function of \phi. This means that all the a_n coefficients must be zero, and only the b_n coefficients will contribute to the potential.

Using these conditions, we can solve for the potential and find that it is given by

\Phi(\rho,\phi)=-\frac{Id\sin\phi}{2\pi \rho}+\mathcal{O}(d^2/\rho^2).

This is the same result that was given in the homework statement. This shows that in the limit of small spacing, the potential is approximately that of a two-dimensional dipole, as stated in the problem.

In conclusion, while it may be possible to solve this problem using the series solution to Laplace's equation, it is not necessary in this case. Understanding the physical setup and using the boundary conditions to determine the coefficients of the series solution is a