While studying intensely for your physics final you decide to take a break and l
ID: 3893770 • Letter: W
Question
While studying intensely for your physics final you decide to take a break and listen to your stereo. As you unwind, your thoughts drift to newspaper stories about the dangers of household magnetic fields on the body. You examine your stereo wires and find that most of them are coaxial cable, a thin conducting wire at the center surrounded by an insulator, which is in turn surrounded by a conducting shell. The inner wire and the conducting shell are both part of the circuit with the same current (I) passing through both, but in opposite directions. As a way to practice for your physics final you decide to calculate the magnetic field in the insulator, and outside the coaxial cable as a function of the current and the distance from the center of the cable. As an additional challenge to yourself, you calculate what the magnetic field would be (as a function of the current and the distance from the center of the cable) inside the outer conducting shell of the coaxial cable. For this you assume that the inner radius of the conducting shell is R1 and the outer radius is R2.
Explanation / Answer
The field at radius r inside the conductor depends on the current inside the circle bounded by r.
(H - magnetic field strength)
H 2 ? r = i(? r^2 / ? R1^2)
H = i ( r / (2 ? R1^2) )
B = ?o H =?o i ( r / (2 ? R1^2) ) The field increases linearly with the radius up to R1.
The field at r between the conductors R1 < r < R2 is
H 2 ? r = i
B = ?o H =?o i / (2 ? r)
The field decreases as 1/r as r increases beyond R1 until r = R2.
by ampere circuital law, The field outside the outer conductor is zero since the net current is zero
for more detail solution
It's been done already. See Example: Problem 31.15 in link.
http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter31/chapter31.html
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