I find something not clear about coaxial cables. All I know is that coaxial cables have about four layers. The innermost is a copper wire that is intended to carry the AC current or a signal. The next adjacent layer is a dielectric or an insulator. The third one is another copper shield layer that carries the current back to the source (may be). Anyway, the question is: Does the magnetic field intensity (H) of the innermost copper wire reach the third layer (the copper shield)? In another way, can I use ampere's law that says < Integral(H).dl = Ienclosed > as follows :
The H in the inner radius (a) of the copper shield (third layer) is calculated this way ---> integral H.dl = Ienclosed
where I enclosed is the current carried by the internal wire.
H(2pi(a)) = Ienclosed
H= Ienclosed/(2pi(a))
Thnx in advance!!
Answers & Comments
Verified answer
I remember that formula from school, but not well enough to tell you exactly how it applies to coaxial cable. Generally, a coax cable should be used for signals, not so much for power transition. The idea of the shielding is to ground any voltages induced by external sources of EMI so that it doesn’t end up interfering with the signal on the center conductor. In your house, for instance, if your cable for internet/TV runs near a power cable, the shielding keeps the 120 Hz from bleeding into your signal.
Conversely, any EMI that radiates from the center conductor will also be grounded by the shielding so that it doesn’t interfere with anything else.The reason why they need the insulation between the center conductor and the shielding is to prevent the signal on the center conductor from shorting to ground.