Toroidal coil B field calculation

[Salvador12]

Hey folks!

Can you please help me out with a formula. I want to calculate the strength of a magnetic field inside a toroid of "X" diameter. The additional details are that the toroid is air core (empty inside) and basically just a single turn - the walls of the toroid are conducting. Given "Y" amount of current flows in the walls what would be the field strength inside?
Well I hope you got the idea. Not sure if that changes anything but the current would be RF not DC.

Thanks.

 

[wwfeldman]

how would you calculate the magnetic field in a single loop carrying DC current?
how would you calculate the magnetic field inside a toroid with many turns with DC current?

do you have a toroid, (many turns on a donut shaped form), or
do you have a hollow conducting (metal) donut?

 

[Easy peasy]

B = Uo H, = 1.256E-6 x ampere-Turns / (mean internal diameter x pi)

So, peak B is proportional to peak I

for a toroid of 100mm ave diameter, and 1 Turn, 100A peak say, the Bpk would be 399.8 uT

note this is independent of the diameter of the toroid / torus.

Note the above assumes a nearly perfect 1T "wrap" around the toroid, and even current distribution in the 1T.

 

[FvM]

The DC solution can be used for AC as long as the loop dimensions are small compared to wavelength respectively no current phase difference occurs along the loop.

It may be also helpful to review the similar "planar coil" thread linked below.

 

[Salvador12]

@wwfeldman it's a conducting donut with an empty middle (air core), only difference is that the sides have vertical slots from the center towards the periphery but a conducting hollow donut would still be a good approximation

 

[FvM]

Are you asking about the field inside the torus or in its axis?

 

[Salvador12]

@FvM yes of course I am asking about the field inside the torus.
The field on the torus center for a symmetrical torus should be almost non existent I think, with almost all the field confined in the inner toroidal volume enclosed by the conducting torus walls , or in a core as for ordinary toroidal transformers for example.

So By using the formula B=vacuum permeability x turns x current/2pi x r the result is then expressed in gauss or Tesla?
Because @Easy peasy I use the formula and I don't get such a small B field result , I am a bit confused here.

 

[FvM]

The discussed formula describes the field in the torus axis, not inside the torus.

As you say, the magnetic field inside a hollow conductor is approximately zero. The residual field is caused by second order effects and can be hardly estimated by simple formulas. A full 3D EM simulation is necessary.

For your specific RF problem, it's still unclear if a constant current (phase and magnitude) along the torus can be presumed, because the relation relation of wavelength and torus dimensions is unknown to us.

 

[Salvador12]

No I think there has been a misunderstanding. First of all wavelength at this moment is not important because I want to have a general approximate understanding of the approximate field strength so DC I guess would be ok, second of all and most importantly , the current is not flowing along the length of the torus as if the torus itself was a single turn but around the torus "minor" circumference, so the current is flowing toroidally not poloidally.
See the current direction in the link please.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
The only difference that instead of a wire the torus itself is "the wire" with it's conducting walls and the core is air.

So I want the B field inside the toroid in the area enclosed by the torus but not with the current running in poloidal direction but in toroidal direction , then I think the formula given in the link is the correct one , where B = vacuum permeability x turns x current/2pi x radius ?

 

[wwfeldman]

since you have a hollow metal doughnut, i think you need to treat it as one loop
then, as in the hyperphysics article, B = mu N I /2 pi R
where N i2 one
as for the units, I in amperes, R in meters, mu is 4 pi x 10^-7 Tm/A (tesla meter per amp)
then B is in tesla

i suggest there is only one turn because there is no wire wrapped around a form, as in the usual torroid
and because a look at the topology:
starting from a hollow metal doughnut, cut across the doughnut and straighten it, to get a cylinder
now cut along the length of the cylinder and one has a sheet
current enters the sheet along one long side and exits along the other long side

the current will spread out from the wire bringing the current in (like cars entering a multi-lane highway from a single lane)
after a very short distance, the current density/length will be constant
likewise, at exit, all of the current will "squeeze" down into the outgoing wire
hence, we can treat your torroid as a one turn device

mathematically, if we assume N turns, the current per turn will be (incoming current)/N
so in the equation, N is N and I is (incoming current/N)
N cancels out

 

[FvM]

@Salvador12 + @wwfeldman: Agreed.

 

[Salvador12]

Indeed it seems like the resultant field from a single loop is very small even under high current

I do wonder about one thing, in the calculation for mu do I have to use the vacuum free space permeability number or can I just use 1 as that is the permeability or air and the core contains air not vacuum?

 

[wwfeldman]

the vacuum free space permeability is 4 pi * 10^-1 Tm/A
the permeability of air is not 1 (unit??)
the relative permeability of air is 1 - that is, relative to the vacuum permeability

what is the current, radius, etc of your doughnut, and what B did you calculate?

 

[Salvador12]

I used 1000 amps , 1 turn, distance from center to point inside the torus also known as "r" in meters I used 1 meter.
and used the vacuum permeability of free space from wikipedia - 1.2566370614…×10−6 N/A2

I got a result of 0.00019999999 Tesla or 0.2mT or 200 microTesla