How Do You Calculate the Magnetic Field in a Cylindrical Conductor?

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Homework Help Overview

The discussion revolves around calculating the magnetic field in a hollow cylindrical conductor carrying a uniformly distributed current. The original poster seeks to derive an expression for the magnetic field in the region between the inner and outer radii of the cylinder.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the magnetic field and current distribution, with some suggesting the use of Ampère's law. Questions arise regarding the direction of the current and the necessary steps to derive the magnetic field.

Discussion Status

There is active engagement with participants providing hints and corrections to each other's reasoning. Some guidance has been offered regarding the use of specific equations and the importance of determining the current density.

Contextual Notes

Participants note the need for clarity on the current's direction and the setup of the problem, indicating that assumptions about the geometry and current distribution are under discussion.

wisper
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Problem said:
A conductor is made in the form of a hollow cylinder with inner and outer radii a and b, respectively. It carries a current I, uniformly distributed over its cross section.

Question said:
Derive an expression for the magnitude of the magnetic field in the region a < r < b.

I understand that it needs to relate to the following equation but that is it. [tex]B=\frac{\mu_{0} I}{2\Pi r}[/tex]

I am unsure how to derive an answer for this, I am drawing a blank on solving this. Can anyone give me any helpful hints?
 
Last edited:
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what is the direction of the current? post the whole question pls...
 
That is the full question, the current is running through a hollow cylinder. From how I understand the current is flowing through direction of the hieght of the cylinder.
 
find the [itex]\ver{A}[/itex] first, then B
 
[tex]\oint{\vec{B}\vec{dl}}=\mu_0\int{\int_S{j}}ds[/tex]
and from this you can simply derive

[tex]B(r)2\pi r=\mu_0 j \cdot \pi(r^2-a^2)[/tex]

where
[tex]j=\frac{I}{\pi (b^2-a^2)}[/tex].
 
Thanks to both of you. Clive you pointed out a part in which I made an error in my thinking of my previous answer. Appreciate the help.
 

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