# Calculating the B field in a small gap in a toroid

alanf

## Homework Statement

Take a steel core (K_m = 2500) electromagnet, bend it into a loop with a small air gap, and determine the B field in the gap. The cross-sectional area of the toroid is 4cm^2, and the air gap is 2.5mm. The current through the coil's 120 turns is 15 amps. The radius of the toroid is 7cm. Determine the B field in the air gap.

A full description of the problem, with a diagram, is here (MIT). It's problem 7.5.

## Homework Equations

Inside a complete toroid, B=mu_0*current*number of turns/2*pi*radius of the toroid.

## The Attempt at a Solution

I would think the B field in a gap which is much smaller than the cross-sectional area of the toroid would be nearly the same as inside the toroid itself, which I calculated at 12.86 T.

But the given answer is far lower: 0.85 T. The answer can be seen here (MIT).

I don't understand the explanation. To me, it seems internally contradictory. It says that "the magnetic field strength in the gap will be approximately equal to the magnetic field strength in the steel," and then calculates the integral of B dot dl for the toroid and the gap in a way I haven't seen before. It seems to assume that you can put an Amperean surface in the air gap itself - which is not penetrated by any current - and get an answer to integral B dot dl that is non-zero.

It also says that B in the gap is nearly the same as B in the steel core, which by my calculations can't be true. B in the steel must be the 12.86 T that I calculated, right?

Can anyone help clear this up for me?

Homework Helper
The 12.9 T is correct when there is no gap. Even the tiniest gap strongly reduces that (##2\pi R/\kappa_M##, about 0.2 mm, is already a factor ##1/2##). You can see the air gap as a big 'equivalent of resistance' in the loop. It's called (magnetic) reluctance. Check out a similar treatise here

berkeman
Homework Helper
Gold Member
[Edit: I see that BvU beat me to the punch :)]

I would think the B field in a gap which is much smaller than the cross-sectional area of the toroid would be nearly the same as inside the toroid itself, which I calculated at 12.86 T.

But the given answer is far lower: 0.85 T. The answer can be seen here (MIT).
You are right that the B field inside the gap will be approximately equal to B field inside the steel. However, the field will be much smaller than 12.86 T. The value of 12.86 T would be the B-field inside the steel for a complete toroid (no gap). With the gap, the field is only about 0.85 T. It is rather amazing that by removing a thin (2.5 mm thick) disk of steel to produce the gap, the B-field is reduced to less than 7% of that of the toroid without a gap!

I don't understand the explanation. To me, it seems internally contradictory. It says that "the magnetic field strength in the gap will be approximately equal to the magnetic field strength in the steel," and then calculates the integral of B dot dl for the toroid and the gap in a way I haven't seen before. It seems to assume that you can put an Amperean surface in the air gap itself - which is not penetrated by any current - and get an answer to integral B dot dl that is non-zero.

Ampere’s law is ## \oint \vec{B}\cdot\vec {dl} = \mu_0 I_{enc}##. When you have magnetized materials, the current on the right will include both the current in the windings and the effective microscopic currents due to the magnetization. As shown in standard texts, the contribution from the current due to the magnetization can be brought over to the left side and effectively tucked away inside the permeability ##\kappa_m## so that Ampere’s law becomes ## \oint \frac{1}{\kappa_m} \vec{B}\cdot\vec {dl} = \mu_0 I_{f,enc}##. Here, ## I_{f,enc}## is the “free current” due only to the current in the windings.

alanf
Thanks to both of you for your replies. I'll look through the text posted by BvU, but in the meantime, I am still very puzzled by one thing. If the solenoid was simply straight, and not curved into a toroid, B at the center would be 12.86 T, right? If so, I'm really surprised that curving it somehow reduces B in the solenoid itself, and then gradually brings B back up to 12.86 T as the gap is closed. I feel like I'm fundamentally misunderstanding the physics here. I keep looking for some way to put a current or a changing E field through the gap, but of course there is none.

Homework Helper
Gold Member
I am still very puzzled by one thing. If the solenoid was simply straight, and not curved into a toroid, B at the center would be 12.86 T, right? If so, I'm really surprised that curving it somehow reduces B in the solenoid itself, and then gradually brings B back up to 12.86 T as the gap is closed. I feel like I'm fundamentally misunderstanding the physics here. I keep looking for some way to put a current or a changing E field through the gap, but of course there is none.

I believe it is true that bending a solenoid with a steel core to form a toroid with a gap will cause B inside the steel to first decrease and then to increase as the gap is closed. I think you can make this plausible in the following way.

Apply Ampere’s law in the form ##\oint \frac{\vec{B}}{\mu} \cdot d\vec{l} = I_{enc}## to a long straight solenoid with a steel core where the path of integration passes through the entire length of the solenoid and then returns outside the solenoid to form a closed path. Then we get $$\frac{B_s L_s}{\mu} + \frac{B_a L_a}{\mu_0} = NI$$ where N is the number of turns of winding of the solenoid, ##I## is the current in winding, ##L_s## is the length of the steel core, ##B_s## is the average field along the part of the path inside the steel, ##\mu## is the permeability of the steel, ##L_a## is the length of the path in the air, and ##B_a## is the average field along the path in the air.

The first term on the left tends to be small due to the large value of ##\mu##. The second term tends to be small due to the fact that in the air the B field spreads out and becomes weak. For a long, thin solenoid the average field, ##B_a##, along the part of the path in the air becomes so small that the first term dominates. Solving for ##B_s## gives ##B_s \approx \frac{\mu NI}{L_s}##.

Now suppose the ends of the solenoid are bent around to form part of a toroid with a large air gap. ##L_a## decreases while ##B_a## increases in the gap due to less spreading (fringing) of the field in the gap. As the gap is made smaller, ##B_a## actually increases faster than ##L_a## decreases and the second term on the left of the equation starts to become significant. Since the right hand side remains constant, you can see that ##B_s## must decrease as the gap is made smaller.

When the gap gets small enough so that fringing of B is negligible in the gap, you can say ##B_s \approx B_a##. Then the equation leads to ##B_s \approx \frac{\mu N I}{L_s+ \frac{\mu}{\mu_0}L_a}##. This is essentially the equation used in the problem (note ##\mu = \mu_0 \kappa_m##).

If the gap ##L_a## is made extremely small, then ## \frac{\mu}{\mu_0}L_a## begins to become negligible compared to ##L_s##. Then you see that ##B_s \approx \frac{\mu NI}{L_s}## which is the same expression we had for the long, straight solenoid. So by the time the gap is completely closed, ##B_s## has increased back to its original value in the straight solenoid.

It is important to keep in mind that ##\mu## is not a constant for ferromagnetic materials. It depends on the value of B and even the history of the variation of B in the steel. So, while the gap is changing, ##\mu## will also be changing. The value of ##\mu## when there is a 2.5 mm gap might not be the same as the value of ##\mu## in the long straight solenoid or in the toroid without a gap.

Last edited:
cmb
I have a follow up question on this sort of problem.

In this question there is an air gap, and a coil further around the toroid of 120 turns around a 4cm^2 steel toroid.

Within that solenoid the steel will have the same magnetic flux as the air gap, say it is 0.85T x 4cm^2 = 0.34mWb for the sake of a number (disregarding fringing).

So all that is needed is for the coil to generate the 0.34mWb. My question is whether that can be achieved with the same number of turns and the same current, but on a narrower part of the steel toroid?

I mean, let's say the saturation of the steel is 2T (for the sake of a number) and I put this 120 turn coil around a part of the toroid on the opposite side to the air gap that has been smoothly narrowed to 2cm^2. So, I now apply the 15 A, it scoots around 120 turns, and I get 1.7T in my narrowed part (which assumes the ur of the steel is negligible compared with the air gap) and I still get 0.85T in the air gap.

The benefit being I have used a half of the amount of wire I needed to, and therefore only use a half of the ohmic power through the wire for the same air gap field (maybe very slightly more than half, to compensate for the extra magnetic energy in the narrowed part of the steel toroid).

I can keep doing that, narrowing the steel centre of the solenoid as small as I can before it saturates at 1800Amp.turns.

Is that correct?

cmb
Bump.

As above. Can the coil be around a part of the solenoid with a smaller cross-section, and achieve the same field for the same current, but therefore through less wire?

(So long as nothing saturates.)

Homework Helper
Gold Member
I am still very puzzled by one thing. If the solenoid was simply straight, and not curved into a toroid, B at the center would be 12.86 T, right? If so, I'm really surprised that curving it somehow reduces B in the solenoid itself, and then gradually brings B back up to 12.86 T as the gap is closed. I feel like I'm fundamentally misunderstanding the physics here. I keep looking for some way to put a current or a changing E field through the gap, but of course there is none.

Homework Helper
Gold Member
You bring up an excellent point. Start with a flexible hi-mu toroid, no gap. Calculate the H field (I prefer to talk about H instead of B) and you get HL = Ni, L = circumference, N = no. of turns, i = current.

Now, make a cross-sectional cut in the toroid and stretch the thing into a straight solenoid. Calculate the H field again: HL = Ni, L = length of rod, same L as before.
So, same H, huge gap!
The explanation is that computing the H field in a solenoid the way introductory physics texts do is a gross, I mean GROSS, misapplication of Ampere's law.
We are usually admonished to restrict numerical calculations using this law to symmetrical situations, yet here is about as asymmetrical a situation as imaginable! The basis of this swindle is the assumption that the H field outside the solenoid can be ignored. Pretty puny argument considering that H flux lines have to be closed & so go around from one end of the solenoid to the other outside the solenoid as well as inside. On top of that the wiring required to connect the solenoid to a source of emf would further distort the H field outside the solenoid.

And it doesn't matter how long the solenoid is, again contrary to the assumption we are told to make.

Bottom line, the circulation of H is far greater than the line integral inside the solenoid. So H will be far less inside the solenoid. And unfortunately the computation would be horrendous, which is why this misleading methodololgy is introduced in the first place.

In contrast, the H field in a hi-mu toroid, even with a short gap, is very well contained and the computations are consequently pretty accurate.

BvU