Physics Asked on May 4, 2021
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!!
(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 ? 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.
Answered by Philip Wood on May 4, 2021
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