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torid question..

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toroid question..

http://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
 
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skyhawk

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

φ = ∫BdA

dA = hdr

Also Nφ = LI
 
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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
 

skyhawk

New Member
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
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.
 
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[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 .
 
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