Electromagnetic Induction Without Air Core

AI Thread Summary
The discussion centers on the concept of electromagnetic induction and the effectiveness of tightly wound coils without air cores. Participants debate the relationship between the number of turns in a coil (N), the area (A), and the magnetic flux (B) in generating induced EMF (voltage). It is emphasized that while increasing the number of turns may seem beneficial, a smaller area can lead to reduced magnetic flux, ultimately diminishing EMF output. The importance of balancing coil design to maximize both turns and area is highlighted, suggesting that calculations and experiments are necessary to determine optimal configurations. The conversation concludes with the acknowledgment that understanding these principles is crucial for achieving desired EMF levels.
billmanhillma
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Scientifically, it doesn't make sense why you can't just wind up wire without really much air in the middle. In order to get more wire closer to the magnetic flux, why wouldn't you just wind up the coil tighter? I keep seeing ferromagnetic or air cores. Why not a tightly would wire so more windings can be achieved?
 
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What are you trying to achieve? Or what do you think you can achieve?
 
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I'm trying to achieve more EMF(voltage) per coil. According to the equation:
EMF = -N((delta B * delta A)/delta T)

My goal is to play with each part of the equation with real world prototypes. I'm starting with 'N' as it pertains to my original question.
 
billmanhillma said:
I'm trying to achieve more EMF(voltage) per coil. According to the equation:
EMF = -N((delta B * delta A)/delta T)

My goal is to play with each part of the equation with real world prototypes. I'm starting with 'N' as it pertains to my original question.

EMF=-N \frac {\partial \Phi}{\partial t}=-N\frac{\partial (AB)}{\partial t}=-NA\frac{\partial B}{\partial t}, \;\hbox { where A is cross section area of the coil and B is scalar magnetic field density.}

If you wind so tight and there is no area, A=0. Where does that get you?

In English, the induced EMF across the two terminals of the coil is equal to the magnetic field density B through the area enclosed inside the coil. The bigger the area A, the larger the induced EMF! You are going the opposite direction.
 
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Your reply is puzzling to me.
Let's suppose you have 1 loop of wire where the length of that wire is 40 meters. Now, let's suppose we make that loop to be a perfect square. The area of that loop if 20 meters2. Now let's suppose the magnetic flux going through that area is .2T which passes the loop in .5 seconds. The induced EMF that I get is:
A = (40/4)*2
-1(20*.2)/.5 = -8 volts

If however I take the same 40 meters of wire and create 200 loops where there's virtually no air gap (granted the magnetic flux is an average through the coil because it thicker), it's a smaller area, it's just multiplied by a factor of N. The length of the wire is the exact same. I think the point is to get as much magnetic flux through the coil.

If I want to convert the coil into a single square loop, than the area of that loop will always be L/2 where L is the length of wire.

Am I thinking about the wrong area? Is it the area of the air core? or the cross sectional area of the coil?
 
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You mean you have a fixed length of wire and you can wind one turn with big area in the middle or many turns with smaller area in the middle? That was not clear in your original question.

For that, you calculate the EMF in each case and see. Maybe you can write a little excel program to find out whether there is a peak EMF at certain area vs turns and all. You don't need to ask here. I don't have an answer and it is not apparent one way or the other.
 
EMF = -N((delta B * delta A)/delta T)

bill

it's not the proximity to the flux that counts
but the amount of it you have fenced in with your coil.

Remember B is flux per unit area
so if you halve the diameter of your coil, indeed you get twice as many turns out of the same amount of wire
but they only encircle a quarter as much flux
so you lost ground.
 
If I want to convert the coil into a single square loop, than the area of that loop will always be L/2 where L is the length of wire.

say what?

##L## feet of wire around four fenceposts makes a square with each side ##L/4## feet
which would be area ##L^2/16## square feet not L/2.

Takes a mile of fence for forty acres
but only four miles for six hundred & forty.
 
jim hardy said:
bill

Remember B is flux per unit area
so if you halve the diameter of your coil, indeed you get twice as many turns out of the same amount of wire
but they only encircle a quarter as much flux
so you lost ground.
You got it.
That's the bottom line, Jim.
No more need be said!
(But I'm sure it will. :rolleyes:)
 
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