Electromagnetic force.

[ziggyfish]

Question. If i pass 1A at 100V through a coil of wire. Will it produce the same magnetic feld as passing 100A at 1V considering that both circuits have the same power consumption of 100W?

The formulas for electromagnetism dont seem to reference volts. So i assume its only the current that matters.

[Zero999]

That's a bit of a vague question.

If you have two solenoids, both with same dimensions and their coil formers full of turns, but differing numbers of turns and wire cross-sectional areas, depending on the design voltage/current. If the 100V 1A one has 100 times as many turns as the 1V 100A one, then they should both produce the same magnetic field.

If the coils were different dimensions and sizes, then the magnetic fields would differ considerably.

[Brumby]

The only thing that can be said for certain about the question as posed is that both will dissipate 100W.

Specifics about the coils are necessary for a complete answer - but to address your final point, yes - it is only the current that matters.  The voltage applied is a simple expression of Ohm's Law.

[ggchab]

You should watch Walter Lewin's lectures about electromagnetism on youtube (8.02 course). This is wonderful ! I am currently watching them. But this takes some time !

[Connoiseur]

This might be interesting

[IanMacdonald]

No. Magnetic field strength depends on current times number of turns, times a factor which depends on the nature of the magnetic materials involved and their geometry.

Voltage plays no part in creating the magnetic field, but a given amount of voltage is required to drive the current through the coil, and this has a bearing on the efficiency, in terms of field strength per watt of heat dissipated.

Using many turns of thin wire means a strong field with a relatively small current, but with a high resistance, so a high supply voltage is needed. Using thicker wire will mean less turns will fit on the core, so more current needed for the same field strength. However, proportionately less voltage is required because of the lower resistance. So, the wire gauge mainly determines the operating voltage, the power requirement being much the same regardless.

HTH.

[jmelson]

The magnetic force is Amps times number of turns.  So, a coil of fine wire with 100 turns at 1 amp will give the same force as a one-turn coil of heavy wire with 100 Amps flowing, all other things being equal.

Jon

[ziggyfish]

Just to clarify, I am not talking about changing the number of turns in the inductor, I am asking about given the same inductor, does passing 100A@1V through the inductor, produce the same magnetic field as passing 1A@100V.

For example, I may have
(Circuit A) L = 1, R = 100, V = 100, A = 1A 
and
(Circuit B) L = 1, R = 0.01, V = 1,    A = 100A

I've attached the two circuits, to illustrate this further.

[IRFP460]

In that circuit one coil gets 1A, the other one 100A, you just have a huge voltage drop across the resistor in the 1A variant.
The 100A Coil will have a much stronger magnetic field.

[Ammar]

The formulas for electromagnetism dont seem to reference volts.

Who taught you that? Gauss' law and Faraday's law both relate to voltage.

[BrianHG]

Just to clarify, I am not talking about changing the number of turns in the inductor, I am asking about given the same inductor, does passing 100A@1V through the inductor, produce the same magnetic field as passing 1A@100V.

For example, I may have
(Circuit A) L = 1, R = 100, V = 100, A = 1A 
and
(Circuit B) L = 1, R = 0.01, V = 1,    A = 100A

I've attached the two circuits, to illustrate this further.

What's the voltage across the coil in these 2 schematics.  If the coil is 0 ohms (IE an incredibly good conductor), it's going to be near 0v in both cases.  So, you are just changing the current in your coil's circuit here from 1 amp to 100 amp, but the true voltage across the coil in both cases is still near 0v regardless of what your supply voltage is.

It's the voltage across your resistor which is vastly different in both schematics, a 100v drop across the 100 ohm resistor and a 1v drop across your 10mOhm resistor, but, this is not part of your inductor.

[IanMacdonald]

If your coil is a superconductor then yes, you need the 1v, 100A circuit. Once you have the field established you can then connect the ends of the coil together and disconnect the supply. 'Free energy' might not be possible but in that case a 'free magnet' with no power input, is.

[BrianHG]

If your coil is a superconductor then yes, you need the 1v, 100A circuit. Once you have the field established you can then connect the ends of the coil together and disconnect the supply. 'Free energy' might not be possible but in that case a 'free magnet' with no power input, is.

Of course his inductor is a superconductor or at least really close to 0 ohm.  Look at the voltages, and resistors, and current.  If the inductors were anything other than 0 Ohms, then the currents would be less than 1 amp and less than 100 amp.

[Brumby]

Just to clarify, I am not talking about changing the number of turns in the inductor, I am asking about given the same inductor, does passing 100A@1V through the inductor, produce the same magnetic field as passing 1A@100V.

For example, I may have
(Circuit A) L = 1, R = 100, V = 100, A = 1A 
and
(Circuit B) L = 1, R = 0.01, V = 1,    A = 100A

I've attached the two circuits, to illustrate this further.

Your diagrams help clarify the question significantly - and they show a detail that you did not mention in your original post - the resistor.

This omission caused a lot of the reaction which may have confused you.  This reaction is one that has occurred before when people ask questions about Volts and Amps - but do not understand the relationship expressed in Ohm's Law.

While your diagrams assume "ideal" components, there is nothing inherently wrong with that when trying to understand - but it also means you need to be consistent.

With the coil being ideal, it would have zero resistance - and with a DC power supply (of any voltage), there will be zero voltage drop across the coil.  This only leaves the current flowing through the coil and this is what is important for magnetic fields.

To build these circuits to test in the real world, you would have to deal with the non-ideal nature of real world components - but if you account for these, you can still get useful results.