Home > categories > Electrical Equipment & Supplies > Inductors > Does voltage have any affect on the E.M. field that it produces when it puts a current through an inductor?
Question:

Does voltage have any affect on the E.M. field that it produces when it puts a current through an inductor?

For example, if you put 10 amps through an inductor using 10 volts, is the field any stronger than a field produced by 10 amps going through the same inductor at 1 volt? I've been told that the only factors that effect the strength of an electromagnet are the physical specs, i.e. turns, core, guage, etc., and one electrical factor, current. I know that can't be true, because my house's water pump is able to substitute current for voltage in order to create the same output.More specifically, it runs on either 5A x 230V, or 10A x 115v. Which means that voltage must play a factor in the strength of the magnetic fields created by it's armatures. I theorized that the current going through an inductor is what determines the size, or far-reachingness of the field, while the voltage is what backs up the magnetic field, or gives it it's rigidity. So what's the answer??

Answer:

The variable that you haven't mentioned is the resistance of the inductor. At steady-state (DC), the voltage needed will be the current multiplied by the resistance. The equations for magnetic field rely only on current for simple structures, and voltage enters the picture through the resistance for DC circuits. In an AC circuit (which is what your water pump is), the current is usually established by the inductance and the amplitude and frequency of the voltage source. The resistance usually represents an undesired loss, but it will affect the current amplitude. In your water pump, you are trying to deliver mechanical power to a load (the water pressure), and this is represented as a resistance in the motor equivalent circuit. But instead of being a loss, the resistance models the delivery of power to a mechanical load. An unloaded motor draws relatively little current compared to full load, and the current that it does draw is out-of-phase with the applied voltage, which results in little real power actually consumed. Under no-load conditions, most of the power consumed is caused by the line current flowing in the winding resistance. As the mechanical load on the motor is increased, the line current increases, and becomes more in-phase, resulting in significant power consumption, although most of the power is not consumed in the motor, but actually delivered to the load. Your water pump has two sets of windings, and these are either wired in series, or parallel to determine the voltage and current needed. The magnetic field is established by the number of turns of a coil, multiplied by the current in the coil. The power rating of the motor is the constant, so at 230V, the current drawn would be half of that drawn at 115V. So at 230V, the number of turns in the coil is doubled, and the current is halved, resulting in exactly the same peak magnetic field amplitude as the 115V case.

Share to: