How do you calculate ampere turns for a toroid with an air gap?

[hiiragizawa]

I'm very sorry if this questions seems easy to you guys but it's been giving me a hard time.

An iron ring has a uniform cross-sectional area of 150mm^2 and a mean radius of 200mm. The ring is continuous except for an air gap of 1mm wide.

Calculate the ampere turns at the air gap when B= 0.5T, and relative permeability is 250.

I have tried searching for another example of a 'toroid-but-with-an-air-gap' question in several books including Fawwaz T. Ulaby's Electromagnetic for Engineers but to no avail. The only formula that i could think of is H = (-N*I)/(2*pi*r) and that would mean leaving out the area of the cross section of the ring.

What does it mean when the toroid has an air gap like that? So far, there isn't anything like that covered in my syllabus. I have only covered magnetic field for an infinitely long wire and the toroidal coil.

Any help is greatly appreciated.

Thanks

 

[Astronuc]

Are you familiar the relationship between electric current and magnetic field (Ampere's Law)?

N i = H l_c where c designates the magnetic core.

Also B = $\mu$ H

Now with a gap in a magnetic conductor

N i = H_c l_c + H_g g, where g is the gap (distance between faces).

The $\phi$ = B * A, where $\phi$ is the magnetic flux.

 

[hiiragizawa]

Hello there. Thanks for replying.

The only relationship i know is as below :

$$\int$$ B.dl = $$\mu$$I

N i = H lc where c designates the magnetic core. Does small i denote the current?

Do we have to use the boundary condition for this type of case? Where Bgap=Bint?

 

[Astronuc]

Yes, i is the current.

$\Large\int_S J\cdot da = \oint H \cdot dl$


I believe this will be of use in understanding the problem - http://services.eng.uts.edu.au/~joe/subjects/eet/eet_ch4.pdf pdf file.