Calculating Ampere Turns for Toroid with Air Gap: Exploring Magnetic Fields

In summary, the conversation discusses a question about calculating the ampere turns at an air gap in an iron ring with a magnetic core, given the values of B and relative permeability. The formula H = (-N*I)/(2*pi*r) is mentioned, as well as the relationship between electric current and magnetic field (Ampere's Law). The concept of magnetic flux is also brought up. The conversation ends with a resource for further understanding the problem.
  • #1
hiiragizawa
2
0
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
 
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  • #2
Are you familiar the relationship between electric current and magnetic field (Ampere's Law)?

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

Also B = [itex]\mu[/itex] H

Now with a gap in a magnetic conductor

N i = Hc lc + Hg g, where g is the gap (distance between faces).

The [itex]\phi[/itex] = B * A, where [itex]\phi[/itex] is the magnetic flux.
 
  • #3
Hello there. Thanks for replying.

The only relationship i know is as below :

[tex] \int[/tex] B.dl = [tex] \mu[/tex]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?
 
  • #4
Yes, i is the current.

[itex]\Large\int_S J\cdot da = \oint H \cdot dl[/itex]


I believe this will be of use in understanding the problem - http://services.eng.uts.edu.au/~joe/subjects/eet/eet_ch4.pdf [Broken] pdf file.
 
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1. What is magnetostatics?

Magnetostatics is a branch of electromagnetism that deals with the static behavior of magnetic fields. It studies the interactions between moving electric charges and magnetic fields, and the resulting forces and energy. In simple terms, it is the study of how magnets behave when they are not in motion.

2. What are the key concepts in magnetostatics?

The key concepts in magnetostatics include magnetic field, magnetic flux, magnetic moment, magnetic dipole, and magnetic force. These concepts help in understanding the behavior of magnetic materials and their interactions with electric currents and other magnetic fields.

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Magnetostatics deals with the behavior of magnetic fields and materials in the absence of any changing electric fields, while electromagnetism studies the interaction between electric and magnetic fields and their effects on charged particles. In other words, magnetostatics is a simplified version of electromagnetism that focuses only on the static behavior of magnetic fields.

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Solving problems in magnetostatics involves applying the fundamental laws and equations of electromagnetism, such as Ampere's Law and the Biot-Savart Law. It also requires a good understanding of vector calculus and knowledge of some basic concepts such as magnetic field lines and magnetic flux. Practice and familiarization with different types of problems are essential for mastering the subject.

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