# Solenoid and toroid inductance puzzle

## Main Question or Discussion Point

Im working with building a reluctance motor and ive posed a problem to myself that i cannot answer without questioning some long accepted facts about the inductance of a solenoid.

The simple solenoid and toroidal solenoid are two shapes that seem to have simple equations for calculating the inductance/B field.
e.g:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html

for my puzzle see attachment...

Panel 1:
I start with the inductance of a toroid, a completed magnetic circuit which has a inductance and B field for which I agree with the online equation as quoted.

Panel 2:
I add an airgap in the toroid. The magnetic circuit now has a break in it, so the complete circuit needs to take into account for the air and the iron that it passes through. This is done using the reluctance of the two materials, the metal and the air. the result presented as i understand it, now 1/4 of the flux path is in air.

Panel 3:
A 'simple' solenoid where over half the flux path must be through the air. since the relative permeability of the air is so much smaller this part should be the dominent term. making the B field considerably smaller than the value normally quoted!

I have not neglected to include effects of the obvious change in area that the flux is passing through when it is not 'confined' to the metal, so this is oversimplified... BUT this does not explain away the piece of physics that I am questioning here.

Have I made a foolish mistake here or has virtually every physicist/engineer in history learnt something thats fundamentally wrong? which i find hard to believe.

This has confused me for far too long, which is it?

Woody

#### Attachments

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The inductance of a gapped toroid can be calculated by equating the stored electrical energy ½LI2 and the stored magnetic energy ½∫BH dV:

$$\frac{1}{2}LI^2=\frac{1}{2}\int_V BHdV = \frac{\mu\mu_0}{2}\frac{A_i_r_o_nN^2I^2}{\ell_i_r_o_n}+\frac{\mu_0}{2}\frac{A_g_a_pN^2I^2}{\ell_g_a_p}$$

where $$H=NI/\ell$$ and B = μH in gap and μμ0H in iron, Airon and Agap are the cross sectionional areas of the flux, and volume V =2 πrA, and $$2\pi r = \ell_g_a_p + \ell_i_r_o_n$$. Rewriting, we get

$$L = \mu_0N^2[\frac{\mu A_i_r_o_n}{\ell_i_r_o_n}+\frac{A_g_a_p}{\ell_g_a_p}]$$

Agap is larger than Airon because Blongitudinal is continuous in the iron and gap, while Hparallel is continuous across the iron-gap boundary.
This leades to "refraction" and Snell's Law across the iron-gap boundary, just like light in glass.

Bob S

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$$L = \mu_0N^2[\frac{\mu A_i_r_o_n}{\ell_i_r_o_n}+\frac{A_g_a_p}{\ell_g_a_p}]$$
So.. If we imagine that lgap is very small, then the equation should become equal to the original toroid equation. Agap will approximately equal to Airon. So... the second term in that equation appears to blow up rather than diminish to zero as should be expected.

and yes... I had an equation that produced that mistake at some point, which I think is in part what lead me to the original problem described.

You are right. The correct solution is shown in the thumbnail. Basically, although

$$\int_C H d\ell = NI$$

H is not continuous around the loop C, so H≠ NI/ l. Only B is continuous. Equation (5) in the thumbnail shows the correct result.

Bob S

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So... back to my original question:

Does this imply that the inductance of a solenoid, frequently quoted in textbooks, is actually wrong? as in panel 3 in my original post

Calculating the inductance of a solenoid with iron is very difficult, but for a magnetic circuit with a small gap like the thumbnail in my previous post, the calculation is the following. Use the relation

$$\frac{1}{2}LI^2=\frac{1}{2\mu\mu_0}\int_VB^2\;dV_v_o_l=\frac{1}{2\mu\mu_0}\int_VB^2\;dV_i_r_o_n+\frac{1}{2\mu_0}\int_VB^2\;dV_g_a_p$$

for the stored energy in the iron and a small gap, and use (from the thumbnail in my previous post)

$$B=\frac{\mu\mu_0NI}{\ell_i_r_o_n+\mu\ell_g_a_p}$$

for the magnetic field in both the iron and the gap. Here, $$\ell = \ell_i_r_o_n +\ell_g_a_p$$. Turn the crank and get an expression for the inductance L.
This does not correctly handle the fringe field in the gap for a large gap, but an adjustment in the cross sectional area in the gap should improve the accuracy.

In an air-filled solenoid, half the stored magnetic energy is inside the solenoid, and half outside. For a solenoid with the interior filled with iron, nearly all the magnetic energy is outside.

Here is the on-axis field inside a short air-filled solenoid.

http://www.netdenizen.com/emagnettest/solenoids/?solenoid

For short air-filled solenoids (single-turn loops), you should look at Smythe Static and Dynamic Electricity (Third Edition) page 339. For N turns, multiply by N2.

[added] For the integration, use $$V_i_r_o_n=A_i_r_o_n\ell_i_r_o_n\;\;and\;\;V_g_a_p=A_g_a_p\ell_g_a_p$$

For small gaps, $$\;\;\;A_i_r_o_n=A_g_a_p$$

Bob S

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