Question

(Please dont type out the answer in normal computer type, its hard to see the math symbols used. Please make the answer clear, legible an easy to read showing all of the steps needed to get the solution, it helps me learn the material better)

A uniform current density, J J2,flows down a long cylindrical wire of radius a. The current density is uniform for 0 s p s a (a) Determine expressions for the B field (magnitude and direction) in cylindrical co-ordinates in the (8) two regions p a and p a. (b) Calculate the force per unit length (magnitude and direction) on the wire as a function of position due to the interaction of its B field with its current density (6) (c) By whatever means you can think of, obtain expressions for the vector potential in the two regions p a and D a, assuming that A 0 at p 0. (11)

Answer 1

(a)

The formalism is not so simple, and the typing system takes a lot of time. I'll try to write the main steps in order to understand it:

use Ampere's theorem, for a circular countour.

$\oint_{C}\vec{H}d\vec{r}=I_{int}=\oint _{S}\vec{J}d\vec{S}$; $J=\frac{dI}{dS_{\perp }}$

$\vec{H}=\frac{\vec{B}}{\mu }\: \left ( 1 \right )$

Chose a circular contour, concentric, of radius r, which will be a line of field. H depends only on the distance r:

$\oint_{C}\vec{H}d\vec{r}=H\oint_{C}ds=H\cdot 2\pi \rho$

[I considered $\vec{r}$= the vector of $\rho$]

Within the interior of the conductor: $H\cdot 2\pi \rho =I_{int}=I\frac{\rho ^{2}}{a^{2}}$

B will be (use eq(1)): $B=\mu H=\frac{\mu I}{2\pi a^{2}}\rho ,\: \rho \leqslant a$

In the exterior of the conductor: $H\cdot 2\pi \rho =I\Rightarrow B=\frac{\mu I}{2\pi \rho }, \: \rho \geqslant a$

In vectorial forms:

In interior: $\vec{H}=\frac{1}{2}\vec{J}\times \vec{r}$    ($J=\frac{I}{\pi a^{2}}$)

In exterior: $\vec{H}=\frac{a^{2}}{2\rho ^{2}}\left ( \vec{J}\times \vec{r} \right )$

Use eq(1) to convert to B.

(b)

I did not hear/read about such a request/calculation. I know B created by a current influences OTHER currents not the same that generated it. And the formula I know cannot be "accommodated" to this situation.

(c) The potential vector has a formula that connects it with vector B:

$\vec{B}=\bigtriangledown \times \vec{A}$