Recent content by Troels

It's important to note that integral is not a universial solution to the poisson equation for A. Most importantly, the current distrubiton has to be localized, among a few other restrictions YOu are almost always off to safer grounds by starting directly from the poisson equation and deduce...

Don't get me wrong. The equation - like amperes law - is always true, but - likewise like amperes law - it is only useful in finding the magnetude of A along the integration curve - you have to dream up the direction from a symmetry argument. To address your other question, no. In general you...

The simplest way to do it, for a general B, is via a PDE problem: \nabla^2 \vec A =-\mu_0\vec J whilst remembering that \nabla\times \vec B =\mu_0\vec J This, togther with suitable boundary conditions, (eg. specifing B on the boundary) should give you the result you are looking for. Please...

I've stumbled upon a problem whilst doing my master thesis The problem is to construct the anisotropic conductivity tensor for a material that exhibits Anisotropic magnetoresistance. The problem has left me quite baffled, and coming to think of it, I've never seen a proper treatment of...

It's right - so just plug-in the magnetization you got from \vec M = \chi \vec H

There seems to be missing some information here. How exactly are those multi-poles arranged? equal charges? equal distances? Regular polygons? "a quadupole" or "an octopole" are very wide terms indeed

Meant Harsh of course. Well I'd say id depends very much indeed on how you define "easy" Indeed, you can't solve a system of quadratic equations using a eg. gauss elimination, but for the linear approach, you still need to construct at least 4 additional "point/slope"-equations, so when it...

Though technically correct, nonlinear is a bit rash :) Quadratric equations are easy to solve.

As I stated, it is the current density which is the total current divided by the cross-sectional area of the wire. THat is: J_0=\frac{10\,\mathrm{A}}{\pi(0.35 \,\mathrm{m})^2}

Okay - that eases thing up a bit :) Use Ampéres law for the H-field: \oint_{\partial \mathcal{S}}\vec H \cdot d\vec \ell = \int_{\mathcal{S}}\vec J_\mathrm{free}\cdot d\vec a For points outside the wire, this of course reduces to the familiar form: \oint_{\partial \mathcal{S}}\vec H...

Yes. You now have a magnetization inside the rod that prohibits use of the biot-savart law there. Also, I'm not sure what you mean by "35 cm" - are you referring to the length? If so, you should be very careful in applying ampéres law as you do not have a perfect cylindrical symmetry...

You know the wave function, and thus everything there is to ever know about the system, including the momentum. Look for an operator \hat p that yield the momentum when it operates on the wavefunction. Then use: \langle p\rangle=\int\psi_0^{*}(x)\hat p \psi_0(x)\, dx

It sounds okay - though I seem to recall a mapping that takes the exterior of the circle to the upper half plane in one step. Also when using the Schwarz-Christoffel formula, keep in mind that one of the vertices of the polygon has to map to the point infinity (I once paid dearly in loads of...

Use the equation for the circle: (x-x_0)^2+(y-y_0)^2=r^2 Inserting the three points give you three equations with three unknows: The x and y coordinates of the cetnerpoint and the radius.

Perhaps use the dipole field?