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So I came across this question in Griffith's Electrodynamics which described a long co-axial cable which carried a current flowing through its inner cylinder of radius $\mathbf a$ to outer words cylinder of radius $\mathbf b$. (a<b) It was given that, to find the magnetic field, Ampere's law has to be used which gives:
$B(2$ $\pi$ $s)$=$\mu_0$$\mathbf I$

But how is the enclosed current in some arbitrary circular loop of radius '$\mathit s$', $\mathbf I$ ? (Magnetic field is circumferential) but $\mathbf I$ is distributed over the entire cylindrical volume between the two co axial cylinders. Also, it is also said that the magnetic field inside the smaller cylinder of radius $\mathbf a$ will be zero. But isn't there's a current flowing inside it as well to produce some magnetic field? Please help!!

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  • $\begingroup$ A co-axial cable is generally an inner conductor (usually a solid wire), surrounded by a cylindrical outer conductor (often braided). The two are separated by an insulator. There is probably another insulator outside of the outer conductor. If the current flows down one conductor and back on the other, there is almost no electric or magnetic field outside of the cable which might cause noise in nearby circuits. $\endgroup$
    – R.W. Bird
    Dec 13, 2020 at 17:55

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(a) "I is distributed over the entire cylindrical volume between the two co axial cylinders"

I is the current. It is not a vector. Nor is it distributed over the entire volume between the co-axial cylinders, but only over the smaller cylinder, where it represents the flow-rate of charge parallel to the cylinder axis.

(b) "the magnetic field inside the smaller cylinder of radius 𝐚 will be zero."

This will be the case if the small cylinder is hollow, that is a cylindrical shell. If it is a solid cylinder with the current density uniform across its cross-section, then B inside it will be circumferential and proportional to $r$, the distance from its central axis. We can deduce this from Ampère's law.

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  • $\begingroup$ For the (a) part, current is the current density J integrated over all volume. J is a vector and I is also a vector representing volume current. This is given as per Griffith's Electrodynamics book. Isn't this true? $\endgroup$
    – Ruchi
    Dec 13, 2020 at 17:54
  • $\begingroup$ I don't have Griffiths's book, and I haven't come across the expression 'volume current'. Current is the integral (essentially J.dS) of current density, J, over a surface and is a scalar. I assumed that $I$ stood for (ordinary) current, because the equation that you quote, $B(2 \pi s) =\mu_0 I$ works with the ordinary scalar current. Perhaps someone who has a copy of Griffiths will come to the rescue. $\endgroup$
    – Philip Wood
    Dec 13, 2020 at 23:01
  • $\begingroup$ "current is the current density J integrated over all volume. J is a vector and I is also a vector representing volume current." Are you sure that this is exactly what Griffiths says? $\endgroup$
    – Philip Wood
    Dec 13, 2020 at 23:10
  • $\begingroup$ I just checked it out, current I is J integrated all over the surface, but nevertheless, in the book, this is given: When the flow of charge is distributed throughout a three-dimensional region, we describe it by the volume current density, J. $\endgroup$
    – Ruchi
    Dec 14, 2020 at 6:03
  • $\begingroup$ So Griffiths does not say exactly what you claimed he said! My answer still stands; but you must bear in mind that what I call current density, J, Griffiths calls volume current density. $\endgroup$
    – Philip Wood
    Dec 14, 2020 at 7:51

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