Finding Magnetic Field of a Toroid with Two Circuits

[happyparticle]

Homework Statement

find the magnetic field $B$ produce by $C1$ everywhere inside the coil.

Relevant Equations

$C_1$ where $N=N_1$ and $I=I_1$

$C_2$ where $N=N_2$ and $I=I_2$

$\oint B \cdot dl = \mu_0 NI$

I have a toroid with square cross section and 2 different circuit:

$C_1$ where $N=N_1$ and $I=I_1$

$C_2$ where $N=N_2$ and $I=I_2$

I have a question that say I have to find the magnetic field $B$ produce by $C_1$ everywhere inside the coil. I assume here I have to find the magnetic field $B$ produce by $C_1$ everywhere inside the toroid.

I'm wondering if $C_2$ can affect the magnetic field caused by $C_1$ and I'm not sure how it works for a coil only around a little part on the toroid.

For exemple, If I have only a coil around the entire toroid, I can find the magnetic field using the Ampere's law.

$\oint B \cdot dl = \mu_0 NI$

$dl = 2\pi r$

$B = \frac{\mu_0 NI}{2\pi r}$

 

[Charles Link]

The modified form of ampere's law for this configuration correctly reads $\oint H \cdot dl=\sum N_i I_i$ .
Do they give the magnetic permeability $\mu$ of the core material? (where $B=\mu H=\mu_o \mu_r H$).
The question they are asking appears to be somewhat incomplete.

 

[happyparticle]

The modified form of ampere's law for this configuration correctly reads $\oint H \cdot dl=\sum N_i I_i$ .
Do they give the magnetic permeability $\mu$ of the core material? (where $B=\mu H=\mu_o \mu_r H$).
The question they are asking appears to be somewhat incomplete.

I have this statement in the question. The magnetic field inside the toroid is only affected by the distance from the center of the toroid. I think for this question I don't have to deal with the material but only with the shape of the toroid and the coils. In this case $B$ = $H$ ?

Thus,
$\oint B \cdot dl= N_1 I_1 + N_2 I_2$

$B = \frac{N_1 I_1 + N_2 I_2}{2\pi r}$

However, is that the magnetic field produce by $C_1$?

 

[Charles Link]

If it is air or vacuum, (perhaps it is), then $B=\mu_o H$, but for that case, you don't get a nearly uniform flux that goes around the entire toroid for a given $r$. It does appear they may not have given complete information. I'm also puzzled about $C_2$. In general, $C_2$ will also contribute to the $B$ field, but it is really a poorly stated problem to ask for the $B$ due to $C_1$. They would do better to ask for the $B$ given $C_1$ and $C_2$, and it is really important to have a ferromagnetic material for the core, unless the windings themselves cover the whole toroid.

They also should specify the direction that the windings are wrapped relative to each other.

It is a good introductory problem that leads into the operation of a transformer in the ac case, but they would do well to give more complete information.

 

[happyparticle]

I guess the material is ferromagnetic since we didn't work with anything else until now.

Even if the coil is only over a small area of the toroid I have to use the Ampere's law as if it was around the entire toroid, right?

 

[rude man]

As Chas. Link says, the B field depends on the permeability of the core.

You can use Ampere's law as you stated it for H, however.

 

[Charles Link]

Even if the coil is only over a small area of the toroid I have to use the Ampere's law as if it was around the entire toroid, right?

Yes. That's how it works for ferromagnetic materials=the lines of flux follow the material and the $H$ and/or $B$ are very nearly the same all the way around. That would not happen that way if it was air. For the mathematics, the formula reads $\oint H \cdot dl =\sum N_i I_i$. First the $H$ is computed, and then you compute $B$. This formula is a modified form of ampere's law, and it is based upon some very exact physics. There is always the possibility that the $H$ in the integral will be non-uniform, but for materials like an iron toroid, the $H$ is found to be nearly uniform around the toroid, even if just a couple of windings are made in a very small section of the toroid. The $H$ is simply a function of $NI$, and it doesn't matter whether the $N$ windings are done in a small section or spread out around the entire toroid.

See https://www.physicsforums.com/threa...we-apply-the-biot-savart.927681/#post-5980421 Much of this discussion is well beyond the simple problem presented above, but for future reference, you may find this "link"of interest.

 

[happyparticle]

Great explanation, thanks!

One more thing that I'm not 100% sure.

The $I$ from the Ampere's law, is it the current inside the Ampere's loop?

 

[Charles Link]

Yes, $I$ is the current in the coil, and it is multiplied by the number of turns because in ampere's law, it is the current crossing the plane inside the ring that counts. It is a modified form of ampere's law, where the magnetization $M$, and the magnetic surface currents are taken out of the formula, and what remains instead of $B$ is the $H$ parameter. Instead of total current, which would include magnetic surface currents, what remains is the current in conductors crossing the plane inside the ring, which is the current in the coil multiplied by the number of turns.

For a derivation, see https://www.physicsforums.com/insig...tostatics-and-solving-with-the-curl-operator/ the second paragraph.

See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ which might give a little insight on where $B=\mu_o H+M$ comes from.

 

[Charles Link]

See https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/ for another fairly simple introductory problem. Upon working a couple of these, you should start to see how the formula $\oint H \cdot dl=\sum N_i I_i$ works. Notice again how the windings cover only a small region of the loop.

 

[happyparticle]

With all that, I have new questions, but I can't find any example of a toroid with 2 coils.

I'm wondering if I should consider the entire toroid to get the self inductance.$\phi_{11} = L_1I_1$

$B_1 = \frac{\mu_0 N_1 I_1}{2\pi r} = \mu_0n_1 I_1$

$\phi_{11} = B_1 \int \int ds = B_1a^2$ for a square cross section $l=a$

$\phi_{11} = \mu_0 n_1I_1a^2 (n_1l_1)$ for a coil

$L_1 = \mu_0n_1^2a^2 l$ = $L_1 = \mu_0n_1^2a^2 (2\pi r)$

Is $2\pi r = l$ is correct?

 

[Charles Link]

To get the self-inductance, you need to multiply $B A$ by the number of turns $N$ to get the flux links in the circuit. Note: $\mathcal{E}=-\frac{d \Phi}{dt}=-L \frac{dI}{dt}$. Here $\Phi =NBA$ is the flux "links."

$2 \pi r=l$ is correct.

For a toroid with two coils, the flux from each will either add or subtract, depending on whether they are wound the same or not.

There is one additional "link" that you may find of interest, and that is for the magnetomotive force: https://en.wikipedia.org/wiki/Magnetomotive_force#History

The way this is customarily presented makes it look a little like magic, but the modified ampere's law is a much more solid approach, where $\oint H \cdot dl=\sum N_i I_i$ is derived from first principles.

 

[Charles Link]

Note that it is important to get used to the concept that the magnetic material with permeability $\mu$ will enhance the applied "field" $H$, so that $B=\mu H$. In most cases the toroid will contain a ferromagnetic material such as iron.

Mathematically this enhancement can be explained by a magnetic surface current density per unit length given by $\vec{K}_m=\vec{M} \times \hat{n} / \mu_o$ on the surface of the magnetic material. The $B$ field can be computed from the current in the coils along with the magnetic surface currents using Biot-Savart, but the simpler computation is that $B=\mu_o H+M =\mu H$. This last formula comes from the magnetic pole model of magnetism, that gets the exact same result for $B$ as the computation using magnetic surface currents.

 

[happyparticle]

Thanks for all.
There's a lot of stuff that I didn't learn yet I had never used $H$ as field.

 

[Charles Link]

Most of the info I supplied is extra, but the original problem looks as if they may be slowly leading into the same configuration that is used in transformers. If you can pick up even a small portion of the material, you should be ahead of the game if and when they introduce transformers.

Edit: The $H$ field is used to represent the applied field from current in conductors, but in more detail, it also consists of a contribution from the poles in the pole model. It gets a little complicated, but if you learn it a little at a time, I think you will find it all starts to make sense.

 

[Charles Link]

Please see the Edit in the previous post. The mathematics of systems with magnetic materials is a little complicated, but it is a neat thing that the pole model and the pole model formula $B=\mu_o H+M=\mu H=\mu_o \mu_r H$ allows things to be simplified considerably. The problem in the OP is a good way to get started with these studies.