# Why does the inner conductor of a coax contribute to magnetic field outside the hollow conductor?

While studying the coaxial cable, i noticed that the magnetic field of the inner conductor can pass through the hollow conductor (can be calculated in the region 3). However, the boundary condition of the magnetic field at the surface (between dielectric and perfect conductor) of a perfect conductor is known as the tangential component of H (Ht = Js surface current density) and the normal component is Hn = 0. Inside the perfect conductor, we have Ht = 0 and Hn = 0.

So why do we superpose the magnetic field of the inner conductor and the hollow conductor when calculating the magnetic field in the region 3 ?

In this case we suppose that the current flowing in the inner conductor is i1 and in the hollow conductor is i2 (not like the coax,here i2 is not equal to -i1).

• I'm not really following what you are saying. I'm sure you have a valid question so maybe try asking more directly. I don't understand why you would want to make the shield current NOT -i1. – Andy aka Oct 25 '16 at 9:59
• the question is : why do we add the magnetic field of the inner conductor (of radius RA) to calculate the total magnetic field in the region 3 ? – aymene chafik Oct 25 '16 at 10:05
• Because that's how we find that when we superimpose the mag fields from exactly opposite currents (inner and outer), the net mag field in area 3 is zero. That's the beauty of coax. – Andy aka Oct 25 '16 at 10:09
• I'm way out of practice here, but isn't this 'simply' a question of why each conductor can be considered in isolation before superposing the two results to give a result which is also consistent with all of the conditions? Not sure there is an intuitive answer. – Sean Houlihane Oct 25 '16 at 10:11
• You question has nothing to o with coax cable, it's just a homework. You have to model the H filed caused by inner and outer conductor, then subtract each other. You have to use Biot-Savart law. – Marko Buršič Oct 25 '16 at 10:48

So why do we superpose the magnetic field of the inner conductor and the hollow conductor when calculating the magnetic field in the region 3 ?

There isn't really a good answer to this other than "that is the way it is". Electromagnetism is a linear theory. That means that if you have two wires with current flowing in them, the resulting magnetic field is the sum of the fields created by the two wires individually. The fact that one of those wires is inside the other doesn't matter.

In fact, the same is true in region "1". However, the current flowing in the outer conductor generates zero net field in region 1, so the total field is equal to the field caused by the inner conductor current.

The condition that there is no magnetic field inside a conductor is only true at high frequency. This is the skin effect. At high enough frequency, the skin effect will force a current flow on the inner surface of the outer conductor equal to -i1. If i2 != -i1, then the remainder of i2 will flow on the outer surface.

Why does the inner conductor of a coax contribute to magnetic field outside the hollow conductor?

This question is possibly best answered by considering why a conductive shield may not block an alternating magnetic field. It's all down to skin effect, a phenomenom that causes higher frequency alternating currents not to flow down the centre of a wire.

For a "thin" conductive shield (i.e. a lot thinner than the skin depth) an alternating magnetic field will not be significantly blocked. As the thickness of the shield gets bigger, the magnetic field creates larger eddy currents and these tend to zero as the shield gets significantly thicker than the skin depth. So the magnetic field at the "far side" of the shield progressively reduces.

For coax this also happens but at low frequencies, the lack of magnetic field beyond the diameter of the coax is zero because the "inner" and shield currents are equal and antiphase.