Maker Pro
Maker Pro

Mutual Inductance, How can I convince him.

J

Jian

Jan 1, 1970
0
As you know, when you place a single loop coil made of
infinitively thin wire perpendicular to another same coil
the mutual inductance between the two is zero.

(as the Neumann integer is zero, dl x dl'= 0 every where,
and you can prove it easily with a network analyzer).

But how do I convince my buddy how insists that: the magnetic
field line produced by of one coil is passing through the other,
how can M = 0?
 
J

James Meyer

Jan 1, 1970
0
As you know, when you place a single loop coil made of
infinitively thin wire perpendicular to another same coil
the mutual inductance between the two is zero.

(as the Neumann integer is zero, dl x dl'= 0 every where,
and you can prove it easily with a network analyzer).

But how do I convince my buddy how insists that: the magnetic
field line produced by of one coil is passing through the other,
how can M = 0?

Show him that for every line of flux from the first coil going in one
direction in the second coil that there is another one going in the opposite
direction.

Jim
 
R

Reg Edwards

Jan 1, 1970
0
If your buddy can't see what's bloody obvious then it's time to change
buddies.
 
N

N. Thornton

Jan 1, 1970
0
As you know, when you place a single loop coil made of
infinitively thin wire perpendicular to another same coil
the mutual inductance between the two is zero.

(as the Neumann integer is zero, dl x dl'= 0 every where,
and you can prove it easily with a network analyzer).

But how do I convince my buddy how insists that: the magnetic
field line produced by of one coil is passing through the other,
how can M = 0?


Hi

Maybe the answer is just like the problem: mutual inductance. If you
let him induce you to believe something, its a lot easier to then
induce him to beleve something. same with coils I guess. Sometimes
this approach works.

Regards, NT
 
C

Chris Carlen

Jan 1, 1970
0
Ken said:
How about this:

With the inductors parallel, the mutual inductancs is, lets say "A". If
we turn one coil around 180 degrees, the mutual inductance is "-A". We
have gone from +A to -A therefor we must have passed through zero. If you
don't believe that the mutual inductance is zero at the 90 degree point,
you must believe that it is zero at some other angle.


I have a big problem with the idea of negative mutual inductance, or M.
M is dependent on permeance and the number of turns of the coil
involved, neither of which can be negative. At a mathematical level and
assuming one sets up the derivations using fields, currents, voltages,
contours, and surface normals that obey Faraday's law, mutual inductance
must always be positive.

If it is not, it must be because you improperly defined the reference
polarity for one of the coils.


Good day!


--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
[email protected] -- NOTE: Remove "BOGUS" from email address to reply.
 
J

John Larkin

Jan 1, 1970
0
As you know, when you place a single loop coil made of
infinitively thin wire perpendicular to another same coil
the mutual inductance between the two is zero.

(as the Neumann integer is zero, dl x dl'= 0 every where,
and you can prove it easily with a network analyzer).

But how do I convince my buddy how insists that: the magnetic
field line produced by of one coil is passing through the other,
how can M = 0?

Ask him the sign.

John
 
J

Jim Thompson

Jan 1, 1970
0
Ask him the sign.

John

Just draw a pretty picture with flux lines from the 1st coil into and
out of the 2nd coil canceling each other.

...Jim Thompson
 
T

Terry Given

Jan 1, 1970
0
As you know, when you place a single loop coil made of
infinitively thin wire perpendicular to another same coil
the mutual inductance between the two is zero.

(as the Neumann integer is zero, dl x dl'= 0 every where,
and you can prove it easily with a network analyzer).

But how do I convince my buddy how insists that: the magnetic
field line produced by of one coil is passing through the other,
how can M = 0?

I can think of several ways. Firstly, a simple experiment. I wouldnt use
infinitely thin wire, as it takes an infinitely long time to draw it that
thin - use ordinary wire instead. A signal generator and a scope will
clearly show the mutual inductance reducing to almost zero as one coil
become normal to the other. Almost zero of course as the coils are not
infinitely thin, nor will they be perfectly symmetrical, nor can you get
them exactly normal.....but the coupled signal ought to drop by at least a
factor of 100, perhaps more. Any real engineer will tell you that < 1% is
approximately zero.

Secondly, get your copy of "Fields & Waves" by Ramo et al, and hit him
repeatedly on the head with it until he promises to read and understand it.
Kraus's Electromagnetics would also work (the heavier the book, the more
effective this technique becomes)

I have had several problems like this in the past. There was the electrical
engineer who refused to accept my "approximate" inductance formulae for
large parallel closely spaced plates - regardless of the references i
provided, or the measurements I took that showed the inductance was within a
few percent of the amount I had calculated.......aaargh. I eventually solved
that problem by telling him to **** off.
 
K

Ken Smith

Jan 1, 1970
0
I have a big problem with the idea of negative mutual inductance, or M.
M is dependent on permeance and the number of turns of the coil
involved, neither of which can be negative. At a mathematical level and
assuming one sets up the derivations using fields, currents, voltages,
contours, and surface normals that obey Faraday's law, mutual inductance
must always be positive.

If it is not, it must be because you improperly defined the reference
polarity for one of the coils.

No, a minus sign of the mutual inductance just means that the windings are
bucking. There is nothing "improper" about "bucking windings" even though
it does sound slightly suggestive.

There is a type of variable inductor which uses a coil pair much like a
Helmholtz coil. Inside of these, is another coil made to be nearly a
sphere. Through this inside coil runs a shaft with a knob and a pointer
on it. By turning this knob, you can rotate the internal coil from aiding
to bucking the outter coil. The two individual coils are connected in
series. By rotating the knob, you can vary the inductance from a lot more
than the sum of the two inductors down to almost zero. You have the
option of stopping a rewriting the equations when you turn the knob to 90
degrees or you can allow mutual inductance to change signs.
 
K

Kevin Kilzer

Jan 1, 1970
0
On Thu, 28 Aug 2003 23:08:39 -0700, "Terry Given"
Secondly, get your copy of "Fields & Waves" by Ramo et al, and hit him
repeatedly on the head with it until he promises to read and understand it.
Kraus's Electromagnetics would also work (the heavier the book, the more
effective this technique becomes)

Although I never saw Professor Kraus hit someone with his book, he did
sometimes hit them with a good does of reality.

As is often the case, many students would skip class until the day
prior to an exam. On one occasion the exam was on Monday. On the
preceeding Friday, in anticipation of those people, Professor Kraus
distributed something that looked very much like test questions and
answer books, announced that this was it, and played the part to its
fullest. Those that came just for the day (at least 10 or 15 guys),
rose at the rear of the class, announced their displeasure with the
situation, stating they were going to speak to the Dean, and left the
room en masse. After about 15 minutes of labor by the remaining
dozens of us, the Professor calmly exposed his charade and thoroughly
explained every question on the test. The real test on Monday, of
course, was very similar to Friday's practice test.

As I have since learned, in real engineering, the question rarely
comes when it is most expected, and sometimes is comes in a completely
different context.

Kevin
 
Top