Ampere's Law and Magnetic Fields

datran
Messages
13
Reaction score
0

Homework Statement


I have a coaxial cable with current density Jo in the center, with radius a, going in -z_hat direction. This generates a magnetic field. The outside of the cable, radius c, also carries a current density Jout going in the +z_hat direction. This generates its own magnetic field.

Find the value of the outside current density to make the magnetic field 0 for r > c

Homework Equations


I used Ampere's equation.


The Attempt at a Solution



I do not use Ampere's equation explicitly (starting from dot product and such), but conclude that H = 0 if the net current = 0. and then find out Jout in terms of Jo.

Is this correct thinking? I mean, if I drew a loop that was outside of the coaxial cable, the Inet would be 0, but there is still an H-field being contributed by both the current densities.

How else would I go solving this problem then?

Thanks!
 
Physics news on Phys.org
We need to know the cross-sectional area of the outide conductor.

You are right in assuming the net current = 0 for H to be 0. If the inside conductor cross-sectional area = a then you know the inside current is Joπa2 but we also need to know the outside area & it can't be πc2 obviously.
 
Ok thank you! This was more of a conceptual question than anything. So if the two current densities are in opposite direction but equal to each other in magnitude, this will cancel out the magnetic field entirely?

I'm lost in the understanding because if we take an amperian loop outside of the coaxial cable and assume that the currents were different (meaning they each generate their own magnetic field, which does not cancel each other out), the enclosed current will be 0, but that does not necessarily guarantee the magnetic field is 0 right?
 
datran said:
Ok thank you! This was more of a conceptual question than anything. So if the two current densities are in opposite direction but equal to each other in magnitude, this will cancel out the magnetic field entirely?

I'm lost in the understanding because if we take an amperian loop outside of the coaxial cable and assume that the currents were different (meaning they each generate their own magnetic field, which does not cancel each other out), the enclosed current will be 0, but that does not necessarily guarantee the magnetic field is 0 right?

It's not the current densities that cancel each other to attain H = 0, it's the currents. That's why you have to know the cross-sectional area of the outer conductor. Then (area of inner conductor) * Jo = (area of outer conductor) * Jout to give ∫Hds = I = net current = 0 so by symmetry H = 0 everywhere along any closed loop outside the outer conductor.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top