Flux Density in Mutual Induction Solenoid with Core

  • #1
7
0

Main Question or Discussion Point

For a standard solenoid, I've found that
B=μnI
where
μ = permeability of the core (4π×10^-7 for free space)
n = number of coils
I = current

Firstly, is the permeability of soft iron 0.08, as I found?

Primarily, however, I'm wanting to know if this still applies for a mutual induction apparatus like in
https://www.scientrific.com.au/PDFs/em2220-001.pdf
 

Answers and Replies

  • #2
140
14
I think B=μnI/l where l is the length of the solenoid.
 
  • #3
1,306
19
Keep in mind that nothing interesting will happen with DC currents. Yes, a magnetic field will be generated but you will not get any mutual induction unless your flux is changing (i.e. varying current through one of the coils.
 
  • #4
7
0
Keep in mind that nothing interesting will happen with DC currents. Yes, a magnetic field will be generated but you will not get any mutual induction unless your flux is changing (i.e. varying current through one of the coils.
It's for an alternating current.
Presumably the maximum flux density will occur at the current peak?
 
  • #5
140
14
Yes, because
f7c6a5d2ff27b8dae2e9aa5657aa5909a8bc6d8e (sorry, click on the link, I couldn't copy-paste the formula)
and here B is a linear function of the current (I) as I wrote in my first post.
hth
 
  • #6
7
0
Yes, because
f7c6a5d2ff27b8dae2e9aa5657aa5909a8bc6d8e (sorry, click on the link, I couldn't copy-paste the formula)
and here B is a linear function of the current (I) as I wrote in my first post.
hth
Thanks, that's what I though :)
Is the formula for the flux a solenoid correct though?
 
  • #7
140
14
You mean this: B=μnI/l ?
 
  • #8
140
14
It is correct assuming the solenoids length is much longer than the radius. B can be considered almost homogeneous inside the solenoid.
But in any case B is linearly proportional to I.
hth
 

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