# "Number of turns of wire..."

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I've seen a number of formulae in electromagnetism refer to the "number of turns in a wire;" not some function that converges at infinity, but rather linear proportionality. (To things like magnetic field, magnetic inductance, etc...)

I get that, in practice, once a current carrying wire generates a magnetic field, this creates a back emf and decreases current to some equilibrium level. But in that initial instant; when the current begins to flow; does this mean you could have a theoretically infinite magnetic field/inductance/etc... with enough turns of wire?

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I've seen a number of formulae in electromagnetism refer to the "number of turns in a wire;" not some function that converges at infinity, but rather linear proportionality. (To things like magnetic field, magnetic inductance, etc...)

This formula is obtained from experiments.

You can repeat it easily by using piece of iron rod,

and surround it by Copper wire (with cable cover),

and observing how powerfulness of electromagnet will change with number of turns of wire around it.

I get that, in practice, once a current carrying wire generates a magnetic field, this creates a back emf and decreases current to some equilibrium level. But in that initial instant; when the current begins to flow; does this mean you could have a theoretically infinite magnetic field/inductance/etc... with enough turns of wire?

Wire is made of conducting material, typically Copper as it has very small resistance, therefor the largest current flowing through it I=U/R.

Wire has length d, and cross-section area A, volume A*d = V, multiply volume by density p to get mass m=p*V.

Divide mass m by molar mass of metal (63.55 g/mol for Copper) to get number of moles of material.

Multiply by 6.022141*10^23 to get number of atoms.

All of these are finite numbers, not infinite.

Edited by Sensei
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... But in that initial instant; when the current begins to flow; does this mean you could have a theoretically infinite magnetic field/inductance/etc... with enough turns of wire?

You have it the wrong way round, the current begins slowly and builds up to a steady state.

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Each turn is in series with every other turn. That is each turn is one of an array of series inductors.

There are two cases of series inductors.

Aiding where the formula for combination is L1 + L2 + 2M where L1 and L2 are the inductances of the turn and M is their mutual inductance.

and opposing where the formula for combination is L1 + L2 - 2M.

Since each turn of a coil is wound in the same direction and has the same inductance L the turns are all aiding so we have the situation

${L_{total}} = \mathop {\lim }\limits_{n \to \infty } \left( {nL + 2\left( {n - 1} \right)M} \right) = \left[ {\mathop {\lim }\limits_{n \to \infty } n\left( {L + 2M} \right)} \right] - \left[ {2M} \right]$

Which is indeed infinite.

However you should note that the series forming the sum is linear, it is not asymptotic, like many curves in electricity.

Edited by studiot
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You have it the wrong way round, the current begins slowly and builds up to a steady state.

My bad. I guess I'll have to brush up on my electromagnetism to get the full context for this discussion, then.

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• 3 weeks later...

If you had an infinite inductance you would have no current flow in the inductor, it would present an infinite impedance.

Edit

An infinite inductance would also require a long piece of wire which unless it was a superconductor would have a resistance, which over an infinite distance would have an infinite resistance also.

Edited by Handy andy
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The interwinding capitance would conduct more electricity on a infinite inductance than the coil itself

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