# Understanding mutual inductance

1. Jun 10, 2013

### ShreyasR

In mutual inductance, energy transfer takes place from one circuit to another through magnetic flux. But while trying to understand the equations involved from the attached image, I am getting confused.
If the current in the first coil gives rise to a flux $\phi$1 and only a part of it ($\phi$12) links with the second coil, is the remaining flux ($\phi$11) only responsible for the back emf produced in the first coil? Or is the back emf produced by the main flux $\phi$1 as a whole?

Another doubt I have: When there are two magnetically linked coils (like in a transformer), when a time varying current is passing through the first coil, it creates a time varying flux which induces an emf in the second coil. But in this case, is there no self induction taking place in the first coil? I mean this is not considered in the attached image. So i am not pretty convinced about this concept.

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• ###### BEE1.jpg
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2. Jun 10, 2013

### jim hardy

Wow I do better with pictures than with text.
That book seems to me a little too brief.

Please excuse my non-professional demeanor in this post. I often resort to "word pictures" and write just the way I would talk in an informal one-on-one discussion.

The emf induced in any coil is due to the flux through it from whatever source.
E = N * d$\Phi$/dt
So in your example the first coil sees voltage arising from $\Phi$1.

Some of $\Phi$1 meanders over to coil 2 and induces voltage there.
That's what he called $\Phi$12.

If there is any additional flux coming back to coil 1 from current flowing in coil 2,
well then
$\Phi$1 isn't all the flux in that coil 1, is it?
He called that additional flux $\Phi$21.
So the total flux in coil 1 would be the sum of $\Phi$1 and $\Phi$21.
The voltage induced in coil 1 would be in proportion to that sum..

That's why load current flowing in the secondary of a transformer lowers the counter emf in the primary, allowing the source to push in more current and that restores the balance.
My old 1880's text book, written when they were figuring out how to teach this stuff, calls that phenomenon "...beautifully self regulating" . I agree.

Academics sometimes forget that we beginners have to struggle up the learning curve and repeat those thought processes from 1800's. That results in brief explanations that not everybody gets.
I don't disparage them for that - after something becomes ingrained it is difficult to remember when you didn't know it.
But I am such a slow learner I have to just remember the basics and figure the maths out from there.
The questions you ask indicate a desire to intuitively understand. Train your brain to take rigorous little baby steps in your thinking. (Pssst - You don't have to admit to it.)
And work tons of problems so the math becomes second nature.

See if this hyperphysics link helps you --
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indmut.html
I find their explanations usually to be very good. I think it was Sophie who put me onto them.

Remember E = N d$\Phi$/dt and $\Phi$ is all the flux in that coil.

old jim

3. Jun 11, 2013

### ShreyasR

The link that you sent seems to be pretty useful!!! With all those mind maps! Thanks a lot!

You said "So the total flux in coil 1 would be the sum of $\Phi$1 and $\Phi$21."

In the attached image, He says $\Phi$1 is the flux due to the current in first coil, and part of $\Phi$1 which is $\Phi$12 links with 2nd coil. Remaining part $\Phi$11 is confined to coil 1 itself...
Does he mean to say the total flux in coil 1 would be the sum of $\Phi$11 and $\Phi$21 which i guess is a bit different from what you have stated... The difference lies in $\Phi$11 and $\Phi$1.

I have to agree with you. So the flux linking with the second coil will be the sum of $\Phi$12 and $\Phi$2, and the emf in the coil 2 will be:
e2 = N2 d($\Phi$12+$\Phi$2)/dt

but equation 3.16 in the attachment says: e2 = N2 d$\Phi$12/dt

This was the thing which was driving me nuts!!! So he's wrong there isn't he?

Last edited: Jun 11, 2013
4. Jun 11, 2013

### jim hardy

I have to think he follows one train of thought down through eq 3-17, then reverses on us.
At eq 3.17 he is showing how current in coil 1 induces voltage over there in coil 2, and he has not yet considered current in coil 2.

Now - an open circuited coil makes a decent flux detector. (Well, rate of change of flux detector..).
It has to be open circuited, else its own current changes the flux you are tying to measure.
So I think, though he didn't state so, that in his explanation he has not yet allowed current in coil 2 up to this point. Coil 2 is an open circuited flux detector at this point in his development.

Observe before 3.18 he used word "Similarly". So in eq 3.18 he's starting his train of thought again, this time setting current in coil 2 and observing the flux that arrives in coil 1 resulting from i2.. Note 3.18 is first appearance of i2 in his equations.
So at that point he's using coil 1 as his flux detector, with zero current there.

He's demonstrating symmetry of mutual inductance, not the complete solution of fluxes and voltages. (edit - I said symmetry, he said reciprocity which is probably a better term)

So - You're right - voltage in coil 2 is due to all the flux that's there. That's Faraday.
So - was your author wrong, or just too brief?

I think he didn't clearly state that in his description,
he allowed only one coil to carry current in each thought train ,
and he used the other coil as an open circuit flux detector.

Does above help?

That old 1880's textbook of mine is currently reprinted in India. Its explanations are, if anything, as much too long as the one above is too brief. I've spent hours in my copy of the 1901 edition and it was great help to me.
Of course we've gone from many British to SI units since then, but you may find this a useful resource :

"Dynamo Electric Machinery" by Sylvanus P Thompson
The drawings of old machinery are delightful.
He also wrote "Calculus Made Easy". a classic still popular here.

old jim

5. Jun 11, 2013

### ShreyasR

Yes! That helped a lot! I did not start the train of thought again. I considered this to be a transformer. That was the mistake I made! But then I did not understand what he meant by "The remaining flux $\Phi$11 is confined to coil 1 itself. Isn't is $\Phi$1 as a whole which is confined to coil 1?

And about the book, I dint find a link to the e-book. And the price was about Rs.2000 (about 40 US$). Is that true? Does it cost so much in the US? 6. Jun 11, 2013 ### jim hardy $\Phi$11 seems to me unnecessary - it's just $\Phi$1 - $\Phi$12, that is how much of $\Phi$1 doesn't meander over to link coil 2. Too much arithmetic is sometimes as perplexing as too little explanation. Amazon show it$30 US
but I don't know which edition it is.

https://www.amazon.com/Dynamo-Electric-Machinery-Manual-Students-Electrotechnics/dp/B00A2PHJY8

EDIT found another list of them
https://www.amazon.com/s/ref=nb_sb_...ery: A Manual for Students of Electrotechnics

............................................................

Mine is the 1901 printing, I think seventh edition. . I rescued it from an old building that was being demolished in about 1967, it is very old and tattered . I was very(pleasantly) surprised to see it enjoying a revival..

Being so old it is out of copyright and I believe you can find it archived for student use here:

http://archive.org/details/dynamoelectricma00thomrich
http://archive.org/stream/dynamoelectricma00thomrich#page/n0/mode/2up

and

http://openlibrary.org/books/OL2046...nery_A_Manual_for_Students_of_Electrotechnics

seems to have several editions.

I wouldn't advise you to spend that much money until you've looked it over and decided whether it fits your tastes.

I used mine a lot over the years.