Coil EMF volt vs current

Sound like a homework question!

Field strength is entirely determined by the current through the coil. The voltage across the coil reflects both the rate of change of current, which generates a voltage across the inductance of the coil, and the instanteous current through the coil which generates a voltage drop across the resistance of the coil.

At very high frequencies, one has to take into account the current flowing throuhg the capacitance between the turns of the coil, which by-passes both the resistance and the inductance of the coil, and has no effect on the filed strength.

This is a rather ill-posed question, which suggests that it isn't homework.

-- Bill Sloman, Nijmegen

What determines the steady-state current in the solenoid? (What are the electrical properties of the solenoid?)

"Mike Lennis

So, with coil resistance R and wattage W

W is given by I squared R

So I = sq rt ( W/R )

And V = I.R

....... Phil

This is not really correct. The current that passes through a capacitance between the windings still makes a magnetic field. It is just a lot smaller because it travels a shorter path.

The rules are straight forward. When a current flows, it makes a magnetic field around its path. The effects of multiple current flows add linearly.

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.

John

"John Larkin" Mike Lennis

....... Phil

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?

Thanks again.

Mike Lennis

Thanks again.

Mike Lennis

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.

So, imagine you replace a N=100 turn coil with a 200 turn coil. The cross-section area of the wire must drop by 2:1 to fit into the same space. Resistance per foot of wire doubles, and you have twice the wire length, so net resistance goes up by 4:1. Since power is I^2 * R, we need to drop the current in half to get back to the same power dissipated. So N*I is unchanged!

So, it doesn't matter. In real life, very fine wire will lose more winding area to the insulation, so you may start to lose pull with lots of very fine wire.

Square wire helps, both to pack more copper into the available space and because the heat flow will be better. Anything to improve cooling will allow you to dump more watts into the coil, too. High temperature insulation will allow more watts, too.

Then there's the old Austin-Healey overdrive solenoid trick: apply lots of power to pull in the solenoid, and then once it's seated, back off to some lower holding current.

John

Make the coil resistance zero by making it super conducting. Infinite voiltages are much harder to do.

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.

If the winding area is constant, increasing the number of turns by a factor of N increases the resistance by N^2. Since power is E^2/R, then for constant E, power falls as N^2, not N. Field strength goes as

That assumes you don't change the wire size, so it will also be infinitely large.

John

what

possible

True enough. I was not concerned with constant area. I presupposed a fixed wire size, too.

Yup.

I was simply trying to show that steady-state power consumption is no predictor of field strength. It is all wasted, once the field is created.

no, power consumption is about the same (unless the solenoid is larger)

It'll also have infinite resistance, beacause the wire's thinness and length.

More turns won't make solenoids more efficient, more copper can but that has limits.

Bye. Jasen

Ignoring build up:

If you hold the pounds of copper constant, changing the number of turns doesn't change the power. The resistance of the windings runs as the square of the number of turns and the required current decreases proportionally so I^2*R remains the same. It is better to think in terms of how much conducting material you can afford.

Thinking about build up:

The outter most turn can be quite a bit further from the coil center than the inner most. Adding turns to the outside of a coil increases resistance more rapidly than N because they are bigger turns. For a very thick coil, the resistance grows as N^2. The field from each of the new turns is decreasing by a 1/N curve.

Holding current constant, we can find the rate that power and the field increases with N.

is proportional to.

P = I^2 * R N^2

Field integral(1/N) ln(N)

Therefor for a very thick coil:

Field/P ln(N)/ N^2

For some large amount of wire, adding more turns stops helping. This is another good reason to go with super conducting wire.

Good points.

Big electromagnets used to be water cooled and now mostly superconductive. The NMR magnets I work around have 20T fields, are stable to a ppm field per day, and have zero power consumption.

John

Scaling arguments--fun.

This assumes that the core is a small perturbation on the inductance of one turn at the outer edge, which is true for open cores but not for closed ones. For closed-loop cores, the field is determined by the total current threading the loop. Position doesn't matter, at least not until order (iirc) 1/mu**2. If "solenoid" here has the physicsy meaning of "a coil wound like a Miniductor", this is true, but if it means "a push-pull magnetic actuator whose magnetic circuit is nearly closed at all times" then it isn't.

Cheers,

Phil Hobbs

To add more turns, once the core is full you need a bigger core. You get into the same bind with a core once you start going to larger and larger cores. All the turns get longer when you go to a bigger core.

I understand that...it's just that to make a scaling argument, you have to be clear what you're keeping constant.

Cheers,

Phil Hobbs