If an ideal cylindrical inductor/electromagnet is charged and then discharged, will the energy lost by charging the inductor/electromagnet be equal to the energy gained by discharging it or will some of it be lost to radiation? Based on what I know about magnetism, there should be no loss.

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tiltat9h · Accepted Answer

An ideal inductor is lossless, so by definition it does not radiate. But an ideal inductor exists only in circuit theory, not in reality.A real inductor does radiate. We know this because it can cause interference with other circuits nearby. Shielded inductors are available, designed to minimize this effect.However, in terms of the power lost from the inductor itself, the radiation losses are typically much smaller than the resistive losses due to the imperfect conductivity of the wire used to form the inductor.Furthermore, radiation increases when the associated wavelength of the signals in the circuit become short enough to be comparable to the physical dimensions of the circuit. But inductors have other limitations (inter-winding capacitance) that limit their usability at high frequencies (with short corresponding wavelengths).Therefore you almost never have to account for radiation losses from inductors in circuit design.Also, it is possible to make a loop antenna which is essentially an inductor chosen to maximize its radiation in some frequency band. Typically these are used for receiving rather than transmitting, but reciprocity tells us that the transmitting and receiving behaviors are equal.

If an ideal cylindrical inductor/electromagnet is charged and then discharged, will the energy lost by charging the inductor/electromagnet be equal to the energy gained by discharging it or will some of it be lost to radiation? Based on what I know about magnetism, there should be no loss.

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