Deriving inductor voltage equation

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JustStudying
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Hi guys,
So I'm starting to learn about magnetic fields and electrics fields, one of the equations we use is v= -L*di/dt ... I've been trying to figure out how it's derived, in my book it says:

it starts off like this:

1. v=-dθ/dt (faraday's law) <- why is there no N(no. of turns) in this equation?

2. L = θ/i therefore θ = Li

3. putting them together gives v=-L*di/dt

so my question is, why did the course book not use v= -N*dθ/dt ?

thanks again guys
 
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JustStudying said:
Hi guys,
So I'm starting to learn about magnetic fields and electrics fields, one of the equations we use is v= -L*di/dt ... I've been trying to figure out how it's derived, in my book it says:

it starts off like this:

1. v=-dθ/dt (faraday's law) <- why is there no N(no. of turns) in this equation?
What is ##\theta##?
http://en.wikipedia.org/wiki/Electromagnetic_induction
 
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oh dear, I am terribly sorry I meant Φ
 
Simon Bridge said:
OK - what does ##\phi## stand for?

magnetic flux
 
in my equation it would mean a changing flux would cause an opposing voltage to be induced, and N means no. of coils..would it mean...(in this case) N = 1?
 
##\phi## is usually the total flux isn't it?

OK - consider a line of identical unconnected single loops with the same (changing) flux going through all of them ... if the current induced in the first one is I, what is the current induced in the second one? The third one?
 
they would all be I, wouldn't they? Because they all experience the same changing flux
 
Well done.

You can guarantee the same current through each loop by joining them up.
Do it right and the flux through the middle will always point the same way - so the flux lines from one can join up with the next one.
Notice how the flux does not change, neither does the current?
Now bring all the loops close together...
 
then the all the flux from each individual coil add together as they are in same direction, and you would get more current flow (with them all together, vs having 1 loop on its own)..right?
 
hmm, so then the N pole of 1 loop will join up with the S pole of the next loop, but their total flux is the same as the total flux in 1 loop?
 
That's what I'm trying to get you to picture.

The situation with inducing a current by changing the flux is a bit different from producing the flux from a current - I think this is where you've got mixed up.

If you think of it as this:
The ##\phi## is being applied by an outside agency to generate a current, not produced by the current.
So there is no reason for the flux to depend on the number of turns.
The current cannot depend on the number of turns either since it is the same current through all the loops since each loop is joined in series with the next.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c2
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1
 
but according to faraday's law, if we have a changing flux, and N amount of coils, we'd end up with more voltage induced if we have N turns - so more voltage induced = more current ..so wouldn't N have an effect on current
 
Yes - that's where I figured you'd be getting confused.

But notice the difference:
If you supply a constant current, you get a constant flux that depends on the number of turns (as well as current).
So you'd expect that applying a constant flux through the coils, you'd get a constant current out - that also depends on the number of turns?

But when you try it, you actually get zero current.
Therefore, you cannot just reverse the situation with Faradays law... something else needs to happen.

Did you read those links?

I suppose you could imagine the situation where a spring is wrapped around a screw-thread in such a way that when you push the screw, the spring (not the screw) turns. Push the screw so it has a constant speed - the speed through each coil of the spring is the same (the amount of screw through one coil does not add to the amount of screw through the next one), and the whole spring rotates at the same speed ... but the more coils on the spring, the harder you have to push to get the same speed.

I think I've gone as far as I can trying for a qualitative understanding - to get further, you'll have to do the math.
 
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