How Can a Cylindrical Cavity Create Constant Magnetic Flux Density?

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The discussion centers on the creation of a region with constant magnetic flux density using an off-center cylindrical cavity in a long cylindrical conductor carrying a uniform current density, specifically described as J=a[SIZE="1"]z j. The key focus is on determining the magnitude and direction of the magnetic field (B) within the cylindrical cavity, which is displaced from the axis of the conducting part by a distance d. Participants suggest using Ampere's Law to calculate the magnetic field and emphasize the importance of understanding the geometry of the conductor and cavity.

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It is a physical problem.. please help~

In certain experiments it is desirable to have a region of constant magnetic
flux density. This can be created in an off-center cylindrical cavity that is
cut in a very long cylindrical conductor carrying a uniform current density.
The uniform axial current density is J=az j
Find the magnitude and direction of B in the cylindrical cavity whose axis
is dispalced that of the conducting part by a distance d


I am depressed because of this problem. I have know idea... ㅜㅜ

so... I submit this forum. please help~
 
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Welcome to the Physics Forums, fufufuha. This sounds like a homework problem, and if it is, it should be posted in the PF Homework forums:


https://www.physicsforums.com/forumdisplay.php?f=158


What textbook is this problem in? It sounds like they want you to calculate the B field inside a hollow cylindrical conductor, where there is a uniform DC current flowing up the cylinder wall. It sounds like the answer should show that the B field is fairly uniform off-axis from the center axis of the cylinder. My guess is that you can use Ampere's Law for this, but my E&M books are all at work at the moment, and I'm home...

Edit -- after re-reading your post, I don't think that I understand the geometry of the conductor and the hollow cylinder portion after all. Is there a diagram associated with this problem?
 
Last edited:
I'm sorry that I'm not good at this forum.

Anyway~ Thanks berkeman
 

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