Finding the Auxillary Field of a Long Copper Rod

  • Thread starter Thread starter thatguy14
  • Start date Start date
  • Tags Tags
    Copper Field Rod
Click For Summary

Homework Help Overview

The discussion revolves around finding the auxiliary magnetic field H for a long copper rod with a uniformly distributed current I. The original poster is attempting to understand the steps involved in calculating H both inside and outside the rod, as some details were omitted in their textbook example.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Amperian loops for calculating the enclosed current and question how to determine the free current inside and outside the rod. There is also a focus on the definition and application of current density in this context.

Discussion Status

Some participants have provided insights into the relationship between the Amperian loop and the total current, as well as the concept of current density. However, there is still uncertainty regarding the specific definitions and equations used to derive the free current.

Contextual Notes

The original poster notes a lack of clarity in the textbook regarding certain steps, which has led to confusion about the definitions and equations necessary for the problem. The discussion is framed within the constraints of a homework assignment, emphasizing the need for understanding rather than simply obtaining a solution.

thatguy14
Messages
45
Reaction score
0

Homework Statement


I am looking at an example problem in the text and they skipped some steps. I think I am missing somthing obvious but none-the-less I don't know what is going on.

We have a long copper rod o radius R which carries a uniformly distributed (free) current I. Find H, the auxillary field inside and outside the rod.


Homework Equations



\oint H dot dl = Free current enclosed

The Attempt at a Solution



so the LHS would be (magnitude only) H*2pis for some amperian loop. Now I don't know how to get the Free current enclosed inside and outside the rod.

Thanks for your help!
 
Physics news on Phys.org
The Amperian loops are always closed curves (that's because the line integral at LHS is closed), so they are always the circumference of a surface. In our case this is the surface of a circle with radius \displaystyle{s}. The current at RHS is the total current that goes through this surface. When you are outside of the cylinder it's easy to find this current. When you are inside, you just have to take into account that the current is uniformly distributed.
 
I mostly understand that. The answer the book gives for I free is = (I*pi*s^2)/(pi*R^2) which makes sense, but I am looking for what definition they used exactly to get that. What equation?
 
You can get this by using the current density vector \displaystyle{\vec{J}}. This is the current that flows through a unit area of the wire crossection (or for uniform distributions the ratio current/area). For this case the current flows only in a specific direction, so it's not necessary to use vectors. We can just say that:
\displaystyle{J=\frac{I}{\pi R^2}}
because the current is uniformly distributed. The current we need is given by a surface integral:
\displaystyle{I_{f,enc}=\int _{S}JdS}
where \displaystyle{S} is the circle's surface. Actually the integral is an overkill, because \displaystyle{J} is constant here. If you have seen any applications of Gauss's law, it's like using the charge density in electrostatics.
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
1
Views
2K
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K