If connect a battery to an electromagnet, would the magnetic force be equal stronger or weaker if I used a battery with a capacity of 5000V and 0.5A (2500Watt) or 0.5V and 5000A?

Or will it be equally strong at some distance but the field of the 5000V battery reach further than the 5000A setup?


2 Answers 2


First, a battery is basically a voltage source. The current rating is how much current the battery can deliver when the load demands it. It is not the current the battery always puts out. So, just because the 5 kV battery can deliver up to 500 mA, it doesn't mean that the coil will draw that much current. That is up to the coil. In steady state, the coil would have to have a resistance of 5kV / 500mA = 10 KOhms to draw the 500mA. More, and it will simply draw less current. Less, and the battery's current capacity is exceeded and its output voltage will probably droop, it could vanish into a puff of greasy black smoke, or whatever.

The magnetic field strength generated by any electromagnet is directly proportional to the current thru it. Let's say your coil had a DC resistance of 10 KOhms and therefore drew 500 mA from the 5 kV battery. The current it would draw from the 500 mV battery would be 500mV / 10kOhms = 50 uA. That means the magnetic field will be 10000 times weaker. The fact that the battery could have delivered 5 kA is irrelevant since the coil isn't trying to draw that much current.

Earlier I said that the magnetic field of a electromagnet is proportional to its current. That is true for any one electromagnet. Different electromagnets can be constructed with different proportionality constants. For example, it's quite possible to make one electromagnet that draws 1A at 10V and another one that draws 2A at 5V that have the same magnetic field strength. The first one would use a longer but thinner wire, but both could be wrapped around the same core.


The resistance in the first case would be 10k\$\Omega\$, in the second case it would be 0.1m\$\Omega\$. If you use the same wire the 10k\$\Omega\$ magnet will have 100 000 000 times as many turns as the 0.1m\$\Omega\$. Suppose 100 000 000 turns for the former, and 1 turn for the latter.
MMF (MagnetoMotive Force) is expressed in Ampere-turns, then the high voltage magnet will have an MMF of 100 000 000 turns \$\times\$ 0.5A = 50 000 000 Ampere-turns. The low voltage version will have 1 turn \$\times\$ 5000 = 5000 Ampere-turns. The high voltage version wins.

To see a certain flux you'll be further away from the magnet for the more powerful magnet than for the same flux from the less powerful. So the more powerful magnet's field will reach further.

Olin and I interpreted the problem a bit differently. He used the same magnet for both situations, while I need two different magnets to make the figures come out. But our result is the same. That's because in Olin's case the resistance is fixed, and so is the number of turns, so the MMF ratio is equal to the voltage ratio. In my case the number of turns ratio is 100 000 000:1, but the current ratio is 1:10 000, so in Ampere-turns this is also 10 000:1.

  • \$\begingroup\$ Doh, you beat me by 48 seconds this time! \$\endgroup\$ Jul 1, 2011 at 18:25
  • 2
    \$\begingroup\$ @Olin - that's because you always give such extensive answers! But I really appreciate them, there's always something to be learned. \$\endgroup\$
    – stevenvh
    Jul 1, 2011 at 18:42

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