Toroid with Air Gap magnetostatics problem

[Tinaaaaaa]

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.

 

[Charles Link]

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 $\oint H \cdot dl=I_{conductors}=NI$.

 

[Tinaaaaaa]

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!

 

[Charles Link]

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.