# Why would the peak strength of the magnetic field of an AC-electromagnet be higher than the peak field strength of a DC-electromagnet?

Elsewhere on Stack Exchange several people have said that the average, mean or RMS value of the magnetic field of an AC-electromagnet would be about the same, or slightly lower than, a DC one. But, the peak value of the AC field would be higher than the DC one. (Assuming same voltage, etc.) Why?

How could a changing current, in and of itself, produce, even very momentarily, a higher/stronger magnetic field?

• The answer to this question really depends on what you mean by "assuming same voltage". Typically there are two ways of deciding what "the voltage" of an AC supply is. If by "the voltage" you mean "the peak voltage", then an AC supply will never exceed a DC supply at "the same voltage". If, instead, you use the more common definition, which sets "the voltage" to be "the root-mean-square voltage", then, since the peak voltage of AC current is higher than the root-mean-square voltage, the peak voltage of an AC supply will exceed the voltage of a DC supply set at the rms voltage. – probably_someone Jun 14 at 15:37 An AC-electromagnet will oscillate, probably through something quite similar to a sine wave. A DC magnet has a constant strength. Often the effective strength of an AC magnet can be approximated by its root-mean-square strength, the square root of the average of the square of its instantaneous strength. For a sine wave this is $$\frac{1}{\sqrt{2}}$$ times the amplitude. This is shown in the graph. As you can see, for a portion of the curve the sine wave is above it's RMS value.
• "For a sine wave this is $\sqrt{2}$ times the amplitude" - Typo? – Hal Hollis Jun 14 at 15:28
• @KurtHikes, I can make an AC signal with 1 microamp rms current or 100 amp rms current. Or with 1 uA DC current or 100 A DC current. The 1 uA AC current has much smaller peak current than the 100 A DC current. The 1 uA AC current only has a larger peak current than the 1 uA DC current because of how we choose to express the magnitude of the AC signal. If I had a 1 uA peak-peak AC current (i.e. $i(t)=0.5\ {\rm \mu A}\sin(\omega t)$), then it would have a lower peak current than the 1 uA DC current. – The Photon Jun 14 at 15:49