Magnetic flux through a wire cross-section using Ampere's law

[AngelofMusic]

Hi,

I'm having trouble visualizing the way to calculate magnetic flux inside a wire and I was hoping someone here could help me.

$$\Phi_B = \oint B \cdot dA$$

Inside a wire, using ampere's law, I got:

$$B = \frac{\mu_0 i}{2\pi}\frac{r}{R^2}$$

And that's where I get stuck. http://img23.photobucket.com/albums/v68/AngelOfMusic/wire.jpg" of the wire, it seems that B is perpendicular to dA everywhere.

In the book's solution, they had this perspective of the wire:

http://img23.photobucket.com/albums/v68/AngelOfMusic/wire2.jpg" And it suddenly makes sense.

I was just wondering if someone could point out to me how the two diagrams relate? I'm having trouble going from one perspective to the next.

 

[jdavel]

AngelofMusics said: "I'm having trouble visualizing the way to calculate magnetic flux inside a wire..."

You don't really calculate flux "in" something; you calculate flux through a surface. The magnetic flux through the surface of a cylindrical current carrying wire is zero. But I'm not sure why you want to know this.

 

[AngelofMusic]

I should have been more clear. I'm calculating the magnetic flux through a wire in a set up where there are two wires parallel to each other. The solution manual says:

The field a distance r from the axis of the wire is given by: $$B = \mu_0ir/R^2$$ and the flux through the strip of length L and width dr at that distance is: $$\frac{\mu_0 ir}{2\piR^2}Ldr$$. Thus, the flux through the area inside the wire is:

$$\Phi_B = \int_{0}^{R} \frac{\mu_0iL}{2\piR^2}rdr = \frac{\mu_0iL}{4\pi}$$

A question, though: Under what circumstances is the magnetic flux through a wire zero? It seems from the book's explanation that it shouldn't be zero at all.