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Mutual Inductance?

by azaharak
Tags: inductance, mutual
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azaharak
#1
Jul14-10, 10:44 PM
P: 152
Consider two solenoids, an often familiar formula for the induced emf from 1 solonoid in the 2nd solenoid is . My question is.

Doesn't this only account for the emf induced in solenoid # 2 from solenoid #1, and not the emf induced in solenoid #2 from solenoid #2's own self inductance.

The total flux in 2 should be the sum of contributions from coils in #1 and coils in #2. This emf produced from mutual inductance seems to only account from flux contribution from coils in #1, and the total emf must include another term?



Here is the formula url if it doesnt show above
http://www.mediafire.com/?ymfmhzjgqkgenjz
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AJ Bentley
#2
Jul15-10, 01:00 AM
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P: 665
The simple answer is that the emf induced in the second coil is not producing a current.
(The coil is assumed to be open circuit so that you can measure the emf)

Since no current is flowing it doesn't produce a B field.
azaharak
#3
Jul15-10, 07:30 AM
P: 152
Quote Quote by AJ Bentley View Post
The simple answer is that the emf induced in the second coil is not producing a current.
(The coil is assumed to be open circuit so that you can measure the emf)

Since no current is flowing it doesn't produce a B field.
Makes sense...


Now what about the first coil. Am I correct in saying that there will be an inductive voltage due to the self inductance of the 1st coil, which seeks to generate a current in opposition to its original driving current. The induced current will be 90 Deg out of phase, so that the Net current in the 1st coil is the sume of two sinosoidal functions.

Since the induced current and driving current in the 1st coil are out of phase, does this mean that the Amplitude or magnitide of the current remains unaffected, it just gets shifted by some phase angle?

And can you say the same about the 2nd coil and its own self inductive current once it is connected to a load.? Thanks

Az

AJ Bentley
#4
Jul15-10, 07:54 AM
AJ Bentley's Avatar
P: 665
Mutual Inductance?

Quote Quote by azaharak View Post
The induced current will be 90 Deg out of phase
In opposition means 180 degrees so there is no 90 degrees sum to worry about.

With two coils sharing the same magnetic field, the basic relation to keep in mind is:-

[tex]
\int V_{1} dt = \Phi_{1} = L_{1} I_{1} + M I_{2}
[/tex]
[tex]
\int V_{2} dt = \Phi_{2} = L_{2} I_{2} + M I_{1}
[/tex]
azaharak
#5
Jul15-10, 08:06 AM
P: 152
Quote Quote by AJ Bentley View Post
In opposition means 180 degrees so there is no 90 degrees sum to worry about.

With two coils sharing the same magnetic field, the basic relation to keep in mind is:-

[tex]
\int V_{1} dt = \Phi_{1} = L_{1} I_{1} + M I_{2}
[/tex]
[tex]
\int V_{2} dt = \Phi_{2} = L_{2} I_{2} + M I_{1}
[/tex]
So these are the induced voltages in each coil, due to contributions from self and mutual inductances.

What I'm having difficulty grasping is this feedback effect.

If we introduce I1, and then V1 is induced due to self and mutual inductance, there ought to be a opposing current induced (180 Deg) out of phase in coil 1. But isn;t this immediately going to change the value of I1 which should change the value for V2 and so on forever.


same thing for the 2nd coil.


Thanks for your help
azaharak
#6
Jul15-10, 08:24 AM
P: 152
What I orginally thought was that the induced voltage is 90 degrees out of phase (which is still assume is true). If this induced voltage is of phase, then the opposing current that It creates must be 90 out of phase.


So the opposing current does not change the magnitude of the overall current, it simply shifts it by some phase angle. If you add two sinusoidal functions that are 90 out of phase, shouldn't you get a new sinusoidal function with the same amplitude.
AJ Bentley
#7
Jul15-10, 08:50 AM
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P: 665
Nature doesn't work that way.

For example, if you place an object on a table, it's weight presses down with a force of (say) n Newtons.
Then there must be a reaction force that pushes up with a force of n Newtons. So there must be a reaction to that force from the object of another n Newtons making 2n Newtons. So the table pushes back with another 2 Newtons...
You see where this is going? That's the logic you are using.

Those equations show how the currents and voltages balance out between the two coils regardless of the waveform or voltages/currents you feed to one coil or the other. They are simply experimental results - a statement about how the world behaves.
azaharak
#8
Jul15-10, 09:05 AM
P: 152
Quote Quote by AJ Bentley View Post
Nature doesn't work that way.

For example, if you place an object on a table, it's weight presses down with a force of (say) n Newtons.
Then there must be a reaction force that pushes up with a force of n Newtons. So there must be a reaction to that force from the object of another n Newtons making 2n Newtons. So the table pushes back with another 2 Newtons...
You see where this is going? That's the logic you are using.

Those equations show how the currents and voltages balance out between the two coils regardless of the waveform or voltages/currents you feed to one coil or the other. They are simply experimental results - a statement about how the world behaves.
...

I understand your logic, I'm just going to assume that your reasoning is correct.

Forget about the 2nd coil for a second
How or what would one state for the total value of current in the 1st coil.

Is it going to be the sum of the applied current and the induced current?
If so then what value do I use to calculate the induced voltage, the appied current or the (net current)


OR

Is my question wrong.? Am I not able to say (Applied current) because there is only 1 current which assumes its value when we encorperate the coil.

If this is so... then I1 (the total current in the 1st circuit or coil) automatically incluedes the effects of the (opposing current).

Thanks again
AJ Bentley
#9
Jul15-10, 09:56 AM
AJ Bentley's Avatar
P: 665
What you have is two coils, one of which is usually the 'driven' one. (If you try to drive the second one as well it all gets very complicated). If we say that the driven coil is L1

The current in the driven coil is I1 as stated in the formulae. You can see that it is going to depend on M, L1, L2 and I2

To make life simpler at this point you can stipulate that the second coil is open circuit so no current can flow.
So you can write:-
[tex]
\int V_{1} dt = \Phi_{1} = L_{1} I_{1}
[/tex]
The last term involving L2 is zero.

If you take the special case of a sinusoidal input you can see that the current will be 90 degrees out of phase to the driving voltage because integral cos(wt) = sin(wt).

Basically the first coil behaves as if the other wasn't there.

If you do allow current to flow in the second coil, it is simply 'bled off' from the current that would have flowed in the driven coil. (You have to bring in ohms law for the second coil to work out exactly how much - you simply have three equations instead of just two - which allows you to work out all of the unknowns.)

But you can see from the symmetry of the equations that both voltages will be in phase with each other and both currents will be in phase with each other. Voltage and current being out of phase in both coils.


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