torid question..

toroid question..

https://i31.tinypic.com/2061tue.jpg

i tried to solve A like this:
[latex]B=\frac{\mu _0NI}{2\pi r}\\[/latex]
[latex]\phi=BA=LI\\[/latex]
[latex]A=(b-a)h\\[/latex]
[latex]\phi=\frac{\mu _0NI}{2\pi a}(b-a)h=LI[/latex]
I (current) cancels out
where is the mistake here

the solution is totaly different

Since B is not constant, you have to do an integral to get the flux.

φ = ∫BdA

dA = hdr

Also Nφ = LI

so i need to do a sum with respect to the center point
if it were a 2d system i could do a sum
but here we have a volume
the toroid is a curved 3d bar.

if i convert to polar coordinated.
for each theta angle i have a 2d slice

so it goes to triple integral
because we need another angle to scan the slice of the teroid
where is the mistake in my logic

so i need to do a sum with respect to the center point
if it were a 2d system i could do a sum
but here we have a volume
the toroid is a curved 3d bar.

if i convert to polar coordinated.
for each theta angle i have a 2d slice

so it goes to triple integral
because we need another angle to scan the slice of the teroid
where is the mistake in my logic

Click to expand...

No!! Just do what I told you to do.

You know B and I have given you dA. Put them together. Everything is a constant except r. You will have to integrate 1/r which yields ln r. That's it.

And the limits on the definite integral are a and b.

[latex]B=\frac{\mu_0NI}{2\pi r}\\[/latex]
[latex]N\phi=\int BdA=\int_{a}^{b}\frac{\mu_0NI}{2\pi r}hdr=\frac{\mu_0NI}{2\pi }h\ln (\frac{b}{a})=LI\\[/latex]
[latex]L=\frac{\mu_0N}{2\pi }h\ln (\frac{b}{a})\\[/latex]
this gives me a correct answer.

but when we find the flux from the toroid
we calculate the flux from one tiny slice,close section
a 2d square.

i thought we should slice the toroin into small square
and then sum all the fluxes which go threw each square.

so i get the idea that the total flux going trew the toroid can be represented into
the total flux going from one of its slices .