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Overunity Machines Forum



MH's ideal coil and voltage question

Started by tinman, May 08, 2016, 04:42:41 AM

Previous topic - Next topic

0 Members and 17 Guests are viewing this topic.

Can a voltage exist across an ideal inductor that has a steady DC current flowing through it

yes it can
5 (25%)
no it cannot
11 (55%)
I have no idea
4 (20%)

Total Members Voted: 20

Magneticitist

ok, now think about what would happen if you dropped a suitably sized magnet
down a pipe of pure bismuth.

now super cool the bismuth.

lmfao

verpies

Quote from: webby1 on May 12, 2016, 08:11:12 AM
Can you have a fixed finite inductance from an inductor that has resistance=0 and capacitance=0
Yes

picowatt

Quote from: tinman on May 12, 2016, 11:55:42 PM
If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?

Think about that,and how it relates to the ideal coil,which is also made of many superconductive loops.

As well, think about the large number of inductors in use everyday all over the world made just that way...

PW

tinman

Quote from: Pirate88179 on May 13, 2016, 12:52:10 AM
Brad:

No!  No...

Please do not introduce an ideal pipe into the mix, ha ha.  My head is about to explode from all of the other ideal items we have already.

Bill

PS  No, the magnet would not fall as Lenz would be increased proportionally to the increased conductivity of the copper pipe.

Lol-sorry Bill :D

What about this-as we are talking about the magnet falling through the copper pipe.

Now lets say the leading field(the one on the bottom of the PM as we place it in the pipe) of the PM is the north field,and the trailing field is the south field. The magnet will induce a field in that copper pipe that apposes that of the magnet,ans so apposes the motion of the magnet. So just below the falling magnet,the copper pipe will be producing a north field--that apposes the motion of the magnet. But dose that mean that a north field is also produced just above the falling magnet--so just above the south field of the magnet,that wants to also appose the motion of the magnet.
Dose this mean that the copper pipe now has produced a mono magnetic field? :o--north below the falling magnet-pushing against the falling magnet,and north above the falling magnet-pulling on the falling magnet.

This is the way it would work with a shorted coil.
A north field approaches the shorted coil,and the shorted coil produces a magnetic field that apposes the north field of the magnet. Then if we turn the magnet around,and we pull the south field away from the shorted coil,then that shorted coil will produce a north field that wants to pull the magnet back toward it.


Sorry in advance Bill :D


Brad

tinman

Quote from: Magneticitist on May 13, 2016, 01:17:23 AM
almost spit my mountain dew out at that one

At least your not drinking the cool aid ;)


Brad