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Would the magnetic field produced by a current carrying conductor be greater or smaller if the conductor is a bundle of small copper wires or a single, larger rod - assuming the same voltage?

Further, would the magnetic field differ (stronger, weaker, or greater diameter of field) between a small copper rod or a larger one (say a .125"Dia rod vs a .25"Dia rod) carrying the same current?

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The magnitude of the magnetic field produced around the current carrying conductor is only a function of the current passing through the conductor AND its radial distance with respect to the central axis corresponding to the conductor.

For part one of your question: Just because there is same voltage, it doesn't mean that there is same current in two scenarios. So, first check the current in each scenario and then see which one is larger and which one is smaller. Here by current we mean the enclosed current inside radius r which would be the total current due to all pieces.

However, for the second part of your question, the answer would be certainly no because you are explicitly assuming that there is same current in two scenarios.

Thanks,

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  • $\begingroup$ Thank you for the answer, and I take your point that the current will very depending on the conductor itself. Put another way, I am interested in creating the greatest magnetic field possible with the available charge from an external source. Would I be better off with bundled wire, a solid wire, or a solid wire/rod of greater diameter? My assumption was the later as the larger rod would have less resistance and, as the charge normally flows near the perimeter/surface, the magnetic field would be of greater radius. $\endgroup$ – MDH Mar 10 '16 at 13:51
  • $\begingroup$ Yes, please do a research on Large Hadron Collider (CERN) experiments and see how they come up with such gigantic magnetic fields so that they can accelerate the charged particles near to the speed of light. Then, possibly you can come up with a toy-model in a much smaller scale in the backyard. Hopefully this will give you a motivation but please be safe, $\endgroup$ – Benjamin Mar 10 '16 at 22:21
  • $\begingroup$ Thank you for the surest ion. Can anyone offer a another answer? $\endgroup$ – MDH Mar 16 '16 at 18:43

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