Does the turn density of a faraday generator matter when charging batteries/capacitors?

Thread Starter

Ray Bryant

Joined Jan 31, 2019
8
So basically, if I had a coil with say a 1000 turns per inch and I dropped a magnet through it, I could momentarily power an LED or charge a capacitor. However, if I stretched this coil out and had 10 turns per inch, the LED would no longer light since the (dt) is too high. For something like a capacitor, would a long coil work just as well as a short coil if I dropped a magnet through it, with the only difference being charging time?
 

Delta prime

Joined Nov 15, 2019
373
Hi there :) Michael Faraday was an English scientist who contributed to the study of electromagnetism and electrochemistry. His main discoveries include the principles underlying electromagnetic induction, diamagnetism and electrolysis.
Faraday received little formal education!!
With that said.
So shall you.
A rapidly changing magnetic field induces electric currents to flow in a closed circuit.
A magnet is dropped vertically through a coil linked to a centre-zero galvanometer.
what you did...with out the galvanometer.
What i did because i made one.
Anyway- in your question You noted ( dt) This is your (Delta time).
A graph of coil EMF
(Electro Motive Force)
against time shows that
1593071528086.jpg
When the first pole(S) falls through the coil EMF increases to a level then decreases.
When the middle of the magnet falls throught the coil, the EMF is at a minimum. No lines of force are being cut by the coil.
Maximum EMF is obtained when the second pole(N) falls through the coil. This is when the rate of cutting lines of force is highest, because the magnet is falling faster. As a result of the velocity being greater the period of high EMF is shorter.
As a result of the field direction being reversed when the poles drop through the coil, the induced current direction is also reversed.
So the EMF is also reversed (EMF is directly proportional to current).
Read What Mr.nsaspook sugests
& come back with more questions.
 

Thread Starter

Ray Bryant

Joined Jan 31, 2019
8
\[ v(t) = L(di/dt) \]

I found this formula, and after playing around with the tool I see how a denser number of turns leads to higher voltage.
But I'm running into a similar problem now, since I need to find the di/dt for a magnet falling through a ring or coil. I have found formulas for the flux and emf of a magnet falling through a ring.

\[ v(t) =-(d/dt)\phi(t) \]

So my question would be, say using the above formula I find that my magnet falling through one ring produces .01 volts. For an LED this may not be enough, but assuming this is a very long coil and it takes a magnet more than five time constants to fall through, can I simply substitute (E) in the formula: \( v_c = E(1-e^{-t/\tau}) \) with my emf value for one turn and get the final charge in the capacitor?
 

Delta prime

Joined Nov 15, 2019
373
So my question would be, say using the above formula I find that my magnet falling through one ring produces .01 volts
You have the equations chug & plug.meaning use a calculator.That's the beautiful thing about physics theoretical quantities can be explored confirmed or denied.& Remember Have Fun
 

Thread Starter

Ray Bryant

Joined Jan 31, 2019
8
I just want to make sure my process is valid and efficient, since I've attempted to adapt certain calculations before only to get extreme values that don't make any sense because I misunderstood certain concepts.

So is substitutin emf for E a valid method or is this wrong?
 
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