Today we will solve the problem of finding magnetic vector potential of a rotating, uniformly charged spherical shell. We won’t discuss the general idea behind the vector potential (how it follows from Helmholtz theorem, and gauge freedom etc) and how its defined. That will be part of a conceptual lecture and will be available when the same would be created. The offline version is available, but the web version will call for a special priority to be assigned.
The problem is quite well defined. We just need to follow the straightforward method of implementing the basic definitions and carrying out the required steps. But we need to be mindful of the framework in which we need to accomplish these steps.
The framework I am talking about here is the coordinate system we need to set in order to solve the problem. Notice that the problem has been stated in the spherical coordinate system (which has been discussed couple of times in this website). But we need not worry about all the aspects of this coordinate system, we will only pick on those which are immediately applicable to our problem.
While this choice of the system where the polar axis (z-axis, wrt which the polar angle θ is measured in a r, θ, φ spherical coordinate system) coincides with the angular velocity vector ω is very natural, it isn’t the most convenient for carrying out the ensuing integral for the vector potential A.
Yesterday we saw an interesting application of the Ampere’s Law (– in magnetostatics and sometimes called Ampere’s circuital law also) for the infinite uniform surface current. Today we will see yet another display of the elegance and efficacy of this law in the following problem. This problem is inherited from Griffith’s text on Electrodynamics (3rd edition)
I have tried to be a bit more explanatory than the basic solution available (in instruction manual, if you have a copy). Thats the whole idea of this labor I have taken up. I also strongly suggest anyone who want to sharpen his saber to try the problem on his/her own effort before looking into the solution. That way one can prepare oneself for the pitfalls of one’s own understanding before taking up help and damaging the opportunity of developing of a better sense of solving such problems.
A steady current I flows down a cylindrical wire of radius a. What would be the magnetic field outside the wire and inside of it? We need to find the same in two different scenarios given.
Here are the two different scenarios.
A. Its a surface current density on the outside surface and its uniform across the surface.
B. Its a volume current density and its distributed in the volume of the wire, but this time its not uniform. In-fact the volume current density J is directly proportional to s; the distance from the axis of the wire where we are referring the value of J.
The following problem is an interesting application of Ampere’s law apart from usual applications found in honors syllabus (eg infinite straight conductor, Solenoid and Torroid). This is to be found the excellent book by Griffith on Electrodynamics.
Find the magnetic field of an infinite uniform surface current K (vect) = K i-cap, flowing over the XY-plane.
Lets first visualize the problem. This will help us solve the problem. We chose a Cartesian coordinate system as shown. Our infinite surface current is a sheet that is concurrent with the XY-plane. We also show the Ampere loop which is a rectangle of length l parallel to the y-axis. This loop is half above the XY-plane and half below.
Everyone was thinking in terms of physics (i.e. force). Einstein knew one part of that is maths, pure maths. Force is physics because of mass. But acceleration is geometry, its the shape of your trajectory and its maths.
Imagine a pipe which is horizontally fitted across a wall. Now water jet is flowing through it, inside of it. Now that wall is on a rotating base. You rotated the wall downwards. The pipe is inclined downward, water streams across downward.
Now imagine you have a torch light fixed to one end cap. You get a horizontal output of light. Now the wall is rotated downward, obviously the light will also stream across downwards.
The mass of the water or light did not matter. Because they are guided by the pipe and wall.
The same thing happens in the universe. The space is what carries mass, and it carries all physical phenomena including light. When that space itself is inclined (warped, rotated or receded or whatever) light would follow a curved trajectory just like any object of mass would. Space is what holds them, matter and light. Space curves under gravity.