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Toroid with Air Gap magnetostatics problem

  • #1
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Homework Statement



consider a toroidal electromagnet with an iron ring threaded through the turns of wire. The ring is not complete and has a narrow parallel-sided air gap of thickness d. The iron has a constant magnetization of magnitude M in the azimuthal direction. Use Ampere's law in terms of the magnetic field vector H, along with the boundary conditions on B and H at the interface, to show that the magnitude of the field H within the iron at a distance r from the centre of the ring is given by
Hiron=(NI-Md)/2πr[/B]


Homework Equations


B=μ0(M+H)
- Ampere's Law

The Attempt at a Solution


So far I have calculated the B field in the closed toroid which would be, using Ampere's integral law: B=μ0ΝΙ/2πr.

Based on that plus the fact that Bair⊥=Biron⊥, we assume that for the air the B is the same. Thus, I equated
B=μ0(M+H)→(B-μ0M)/μ0=H→H=(NI-M)/2πr

This is clearly not the correct answer so I would appreciate it if you could please show me how to reach the correct answer.
 

Answers and Replies

  • #2
Charles Link
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You need another form of Ampere's law which is ## \oint H \cdot dl=NI ##. The ## H ## is not continuous though in this problem, while the ## B ## is assumed to be continuous. ## H ## will take on a different value in the air gap=call that ## H_1(r) ##, and ## H_2(r) ## will be the value in the iron. Expressing the continuity of ## B ## at a distance ## r ## from the center: ##B(r)=\mu_o H_1(r)=\mu_o (H_2(r)+M) ##. ## \\ ## Next, write ## \oint H(r) \cdot dl =NI ## in terms of ## H_1(r) ## and ##H_2(r) ##, and then with just ##H_2(r) ## by substituting in for ##H_1(r) ##. With these hints, you should be able to complete it, by solving for ## H_2(r) ##. ## \\## Note: The standard form of Ampere's law is ## \oint B \cdot dl=\mu_o I_{total} ## where ## I_{total}=I_{conductors}+I_{m} ##, where ## I_{m} ## is from magnetic currents and magnetic surface currents. The ## I_m ## requires extra computation, and it is far simpler to just use ## \oint H \cdot dl=I_{conductors}=NI ##.
 
Last edited:
  • #3
7
2
You need another form of Ampere's law which is ## \oint H \cdot dl=NI ##. The ## H ## is not continuous though in this problem, while the ## B ## is assumed to be continuous. ## H ## will take on a different value in the air gap=call that ## H_1(r) ##, and ## H_2(r) ## will be the value in the iron. Expressing the continuity of ## B ## at a distance ## r ## from the center: ##B(r)=\mu_o H_1(r)=\mu_o (H_2(r)+M) ##. ## \\ ## Next, write ## \oint H(r) \cdot dl =NI ## in terms of ## H_1(r) ## and ##H_2(r) ##, and then with just ##H_2(r) ## by substituting in for ##H_1(r) ##. With these hints, you should be able to complete it, by solving for ## H_2(r) ##. ## \\## Note: The standard form of Ampere's law is ## \oint B \cdot dl=\mu_o I_{total} ## where ## I_{total}=I_{conductors}+I_{m} ##, where ## I_{m} ## is from magnetic currents and magnetic surface currents. The ## I_m ## requires extra computation, and it is far simpler to just use ## \oint H \cdot dl=I_{conductors}=NI ##.
Thank you!!
 
  • #4
Charles Link
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I should point out, in the pole theory of magnetism, there is a good reason for the discontinuity in ## H ##. Magnetic pole density ## \rho_m ## is given by ## \rho_m= -\nabla \cdot M=\nabla \cdot H ##. This makes a magnetic surface charge density on the end faces at the gap given by ## \sigma_m =\vec{M} \cdot \hat{n} ## which are sources of ## H ## that result in this discontinuity. The calculation can also be done with this surface charge density ## \sigma_m ## and Gauss's law, but using Ampere's law in the modified form, and simply assuming an ## H_1(r) ## and an ## H_2(r) ## allowed us to sidestep this calculation, which would be found to be in complete agreement.
 

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