Magnetig flux of a straight conductor

  • Thread starter Thread starter Kruum
  • Start date Start date
  • Tags Tags
    Conductor Flux
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Kruum
Messages
217
Reaction score
0

Homework Statement



http://www.aijaa.com/img/b/00815/3678868.jpg

Homework Equations



[tex]B= \frac { \mu _0}{2 \pi r}I[/tex]
[tex]d \phi = \vec{B} \cdot d \vec {A}[/tex]

The Attempt at a Solution



I know the magnetic field of the conductor is [tex]B= \frac { \mu _0}{2 \pi r}I[/tex]. And in order to solve the magnetic flux I need to integrate [tex]d \phi = \vec{B} \cdot d \vec {A}[/tex]. But the question is, since the area is not square or rectangular, can I simply integrate over the whole area at once or do I need to cut the integral into pieces?
 
Last edited by a moderator:
Physics news on Phys.org
Kruum said:

Homework Statement



Homework Equations



[tex]B= \frac { \mu _0}{2 \pi r}I[/tex]
[tex]d \phi = \vec{B} \cdot d \vec {A}[/tex]

The Attempt at a Solution



I know the magnetic field of the conductor is [tex]B= \frac { \mu _0}{2 \pi r}I[/tex]. And in order to solve the magnetic flux I need to integrate [tex]d \phi = \vec{B} \cdot d \vec {A}[/tex]. But the question is, since the area is not square or rectangular, can I simply integrate over the whole area at once or do I need to cut the integral into pieces?

You will have different ranges along the wire that you will need to consider in the integral. Symmetry should make it easier to account for.

Let's consider this Introductory still?
 
LowlyPion said:
You will have different ranges along the wire that you will need to consider in the integral. Symmetry should make it easier to account for.

So what your saying is, that I should divide the cross into three pieces and integrate them separately? So the first integral would be from r to r+b over the area of ab. The second from r+b to r+b+a over the area of (a+2b)a. And the third one from r+b+a to r+2b+a over the area of ab.
 
Kruum said:
So what your saying is, that I should divide the cross into three pieces and integrate them separately? So the first integral would be from r to r+b over the area of ab. The second from r+b to r+b+a over the area of (a+2b)a. And the third one from r+b+a to r+2b+a over the area of ab.

3 rectangles sounds good to me.

Slice them horizontally or vertically, which ever way your Ginsu slices the easiest.