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G A
on 11 Aug 2021

Will the FDTD 3D method help you? See e.g.

https://www.mathworks.com/matlabcentral/fileexchange/54557-3d-finite-different-time-domain-first-order-mur-boundary-condition

David Goodmanson
on 12 Aug 2021

Edited: David Goodmanson
on 12 Aug 2021

Hello MC,

From the plot it is not that easy to see what is going on since the two field colors are similar. Could you post a separate plot for each field?

If you can successfully produce the B field by using a Biot-Savart integration then you can produce an E field in the same way, especially since it's easier. Here I use the common assumptions [1] the current is changing slowly, so that there is a negligible amount of change to the system in the time it takes for light to propagate across the loop. Therefore retarded time effects can be neglected, [2] the current has the same value all the way around the loop so that there is no resulting charge density that can be the source of an electric field. In that case the B field is

B = (mu0/4pi) Integral J(r',t) x (r-r')/|r-r'|^3 d^3r'

where r,r' and J are vector quantities and (r-r') extends from the source point to the observation point. If the current is confined to a small filament with constant current I, then

B = (mu0/4pi) I Integral ds' x (r-r')/|r-r'|^3

where vector ds' is a line element that is everywhere tangent to the loop. I assume you are doing an integral like this to find B.

The electric field in this circumstance is

E = -(mu0/4pi) Integral 1/|r-r'| dJ(r',t)/dt d^3r'

and

E = -(mu0/4pi) dI/dt Integral ds'/|r-r'|

For an oscillation at a single frequency, described by e^(-iwt), then d/dt can be replaced by -iw in which case

E = (mu0/4pi) iw Int 1/|r-r'| J(r') d^3r'

and

E = (mu0/4pi) iwI Int ds'/|r-r'|

The factor of i just means that B and E are 90 degrees out of phase as functions of time.

The E field lines go in circles and look similar to the current loop.

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