Coaxial cables and displacement current

AI Thread Summary
In a coaxial cable with alternating current flowing through its inner core and surface, the discussion centers on whether the magnetic field outside the cable is zero. The user questions if both the current and displacement current effects cancel out beyond the outer radius of the cable. It is suggested that due to the symmetry and opposing currents, the net magnetic field outside the cable would indeed be zero. The clarification sought pertains specifically to the implications for the magnetic field generated by the cable in this scenario. Thus, the conclusion is that outside the cable, the magnetic field remains zero due to the cancellation of currents.
DaTario
Messages
1,092
Reaction score
46
Hi All,

Consider a situation in which a coaxial cable have electric current i(t) = Imax * cos(w*t) flowing through its cilindrical inner cylindrical core (with radius a) and the inverse current flowing on its surface cover (with radius b).
I would like to know if it is correct to say that, outside de cable (r > b), we will see not only zero current effects (due to obvious cancelation) but also zero displacement current effects.

I hope my explanation could have achieved the necessary clarity.

Best wishes,

DaTario
 
Physics news on Phys.org
By "effects"in the OP I mean effects in the magnetic field B generated by the cable. In other word my question is:
with current i varying in time, the magnetic field produced by the cable outside it is zero?
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top