In a circuit with an alternating emf source, resistor and inductor, the current lags the induced emf on the inductor because the inductor initially opposes a change in current. Now, consider a circuit with a resistor and an inductor, and an emf is now induced by a sinusoidally changing magnetic field. Will the induced current still lag the induced emf?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Ideally, no: it will be in phase. In reality, yes, because of leakage inductance. This phenomenon may be understood as as mismatch between the self field geometry and the applied field geometry.
-
$\begingroup$ Is the reason the induced emf and current appear simultaneously because the inductor still opposes a change in current, but 0 current through the inductor also implies 0 induced emf across the inductor as well, as opposed to a circuit with an emf source external to the inductor? $\endgroup$– PiksikiCommented Aug 10, 2022 at 14:09
-
$\begingroup$ @Piksiki Ideally, if there's no leakage flux, the self inductance of the inductor does not oppose the change in current due to an externally applied field. But in reality, some fraction of the self inductance represents leakage flux, and that fraction opposes the change in current. $\endgroup$ Commented Aug 10, 2022 at 14:21
-
$\begingroup$ I see what you mean, but I don't understand how the leakage flux can be ignored for even an ideal inductor. $\endgroup$– PiksikiCommented Aug 10, 2022 at 14:30
-
$\begingroup$ wouldn't there also be flux within the inductor? $\endgroup$– PiksikiCommented Aug 10, 2022 at 14:53
-
$\begingroup$ @Piksiki In practice, there are a couple of things that minimize leakage inductance. 1. Use a high permeability core, confining the flux. 2. In a transformer, use a "bifilar" winding where the primary and secondary wires are as close as possible to each other. $\endgroup$ Commented Aug 10, 2022 at 16:12