Re: Coil EMF volt vs current

On Fri, 04 May 2007 01:19:52 GMT, Rich Grise <rich@xxxxxxxxxxx> wrote:

On Thu, 03 May 2007 16:05:56 -0700, John Larkin wrote:
On Fri, 04 May 2007 08:21:17 +1000, Mike Lennis
On Thu, 03 May 2007 08:30:07 -0700, John Larkin
On Thu, 03 May 2007 19:40:37 +1000, Mike Lennis

Given the same solenoid, and assuming a fixed available wattage, what
proportion of 1) voltage vs. 2) current produces the highest possible
field strength?

How is this determined?

The coil has a resistance, and that resistance fixes the relationship
between the current and the voltage. So you can't vary them
independently, so your question is sort of meaningless.

As the voltage goes up, the current goes up, and there is a single,
unique voltage:current that dissipates the available watts.

Yes, very sorry. The fixed coil resistance was an error on my part.

What I am driving at is this. Given the same signal source, consumed
wattage can be increased by 1) lowering the coil resistance OR 2)
applying higher voltage.

Which of the two is preferrable for maximum field strength?

For example, in the extreme case, one could have microvolts (low
resistance coil) and tons of current, or heaps of voltage (high
resistance coil) and negligible current. Yet the wattage is the same.

What is the trade-off point for these two variables, ie. practical
solution?

Ampere-turns make magnetic field. For a given available wire winding
volume, the external surface area determines the cooling, so the watts
you can dump into a given coil geometry, for a given temperature rise,
is fixed.

Unless, of course, you've got some superconductor wire. ;-)

Cheers!
Rich

Y'all miss the point.

A magnetic field is potential energy. Once the field is created it
exists, that is why a permanent magnet works. It takes no energy to
maintain the field. Hence, "consuming wattage" in this case means
nothing.

Assuming we are talking about DC fields here, all that matters from
the solenoid point of view is ampere-turns. For a fixed voltage, given
a wire of uniform resistance, the MMF is independent of the number of
turns, but the consumed power is inversely proportional to the number
of turns. Hence, (for a fixed field voltage)a solenoid with an
infinite number of turns will develop the same field as a solenoid
with one turn, but with infinitesimally low power consumption. It
will, however, also have infinite inductance, so the field will take
an infinite amount of time to build up.





.

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