Self-Induced Emf and Angular Frequency

[waley]

The context is an ac circuit with only one element - the inductor. My textbook says that the self-induced emf increases with increasing angular frequency, but I'm having trouble seeing this mathematically. If self-induced emf = ε = -L(dI/dt) and L = X/ω, then emf and ω are inversely related. What gives??

 

[sophiecentaur]

The context is an ac circuit with only one element - the inductor. My textbook says that the self-induced emf increases with increasing angular frequency, but I'm having trouble seeing this mathematically. If self-induced emf = ε = -L(dI/dt) and L = X/ω, then emf and ω are inversely related. What gives??

The emf is proportional to rate of change of current. For a given RMS current, the maximum rate of change of current is proportional to the angular frequency (and frequency of course).
d(Sin(ωt))/dt = ωCos(ωt)
if you want to use f then there are some 2π's to be put in but the same answer.

 

[ZapperZ]

The context is an ac circuit with only one element - the inductor. My textbook says that the self-induced emf increases with increasing angular frequency, but I'm having trouble seeing this mathematically. If self-induced emf = ε = -L(dI/dt) and L = X/ω, then emf and ω are inversely related. What gives??

Your problem here is that you lost track of what is the independent variable and what is the dependent variable.

If you've solved for the inductance of a solenoid, you'll see that it depends only on the physical property of the solenoid, i.e. the length, area, etc., not on the amount of current or the frequency. So L doesn't vary with angular frequency. It can't. This is similar to R not varying in R = V/I, even if I changes.

What you are forgetting is that the reactance X also varies with frequency. The quicker the current changes with time, the more back EMF the inductor will produce, and so, the greater the reactance. So it isn't just angular frequency that is changing, but also X.

Zz.