How Does the Stress Tensor Change at Point P Near Current-Carrying Wires?

vwishndaetr
Messages
86
Reaction score
0
So I'm working on yet another problem, and have come to a minor stump.

2 wires run parallel with the Z axis in the xz plane, one with current I-1, other with current I-2.

I need to determine the components of Maxwell's stress tensor at a field point P, where P is a point on the yz plane (x=0), a distance y above the xz plane.

We know,

T_{ij} = \epsilon_0 \left( E_i E_j - \frac{1}{2} \delta_{ij} E^2 \right) + \frac{1}{\mu_0} \left( B_i B_j - \frac{1}{2} \delta_{ij} E^2 \right)

I calculate B due to wire 1 to be,

\overrightarrow{B_1} = -\frac{{\mu}_0 I_1y}{2\pi(d^2/4+y^2)}\hat{x} -\frac{{\mu}_0 I_1d}{4\pi(d^2/4+y^2)}\hat{y}

And B due to wire 2,

\overrightarrow{B_2} = -\frac{{\mu}_0 I_2y}{2\pi(d^2/4+y^2)}\hat{x} + \frac{{\mu}_0 I_2d}{4\pi(d^2/4+y^2)}\hat{y}

I know the the equation for the stress tensor involves Electric field also, but is there an E-field created by to wires with current? Or is there no E-field and E=0?

Also, since P lies on yz plane, and x=0, does that mean all tensor components with x in the indices is also zero?

Some help to guide me?

Thanks!
 
Last edited:
Physics news on Phys.org
Any ideas?
 
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