Mutual inductance

  • Thread starter ollitsacj
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  • #1
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I'm having problems trying solve this problem.
Consider the two conductor system given below


|----------------------------------------------------------------|
| <------a-----> |
| I2-> ______________ |
| | | |
| | | | | |<-c->| L |
| | | |
| I1->__________|_____| |
| <-------b-------> |
|----------------------------------------------------------------|

There are 2 segments of 2 separate conductors. The direction of the currents(obviously) are along the conductors. Using the concept of Neumann integral formulate the mutual inductance between these conductors.

If anybody can help me with this problem I'll appreciate it. Thank you.
 
Last edited:

Answers and Replies

  • #2
Tom Mattson
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Sorry, I can't make heads or tails of that diagram. Can you attach a scanned copy or something?
 
  • #4
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Well you probably got to use
[tex]
M = I^{-1} \int_A \vec{B} \cdot \vec{n} \cdot da
[/tex]
where the surface A is probably the rectangle of width c and height L, what else can it be.
We could choose I = I1 or I = I2, it must not make a difference since the inductance is mutual. So, let's choose I = I2.
Next, B (as caused by I1) can be expressed as
[tex]B(y) = \frac{DI_1}{y}[/tex]
where y is the vertical distance from the lower wire. I wrote D because I'm really too lazy to type electromagnetic constants.
There's no problem with the scalar product since B is always normal to the surface.
We get
[tex]
M = c\frac{I_1}{I_2} \int_0^L \frac{D}{y} dy
[/tex]
I admit it can't be right since it's not symmetrical in the currents. Also, the integral is not finite. Does this help anyway?
 
Last edited:

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