Quasistatic Approximation

[AdkinsJr]

I'd rather not post the exact problem since it's homework, I don't think my instructor in E&M would want me posting full problems but I will just ask relevant conceptual question...

Let's say we I have a long cylinder with time-defendant surface current density $$\vec K(t)=K_of(t)$$. So if I want to find the B-field, can I still use ampere's law? My idea was to find current from the definition of surface current density,

$$\vec K(t) =\frac{dI}{dl_{\perp}}=K_of(t)$$

EDIT: the direction of K is in the "phi" direction in cylindrical coordinate, the field is therefore oriented along the positive z-axis...

So now I have $$I(t)=\int K_of(t) dl_{\perp}=K_of(t)\int dl_{\perp}$$

I'm kind of skeptical about find the E-field everywhere by applying ampere's law to find the B-field, then finding the flux, then using faraday's law to find the field. There is a problem in my text in which a "quasistatic approximation" is used to find the B-field given a time-varying current, where basically they just treated it as you would in magnetostatics. So do I need to assess if f(t) is slow enough for the approximation or can enclosed current be a function of time?

$$\int \vec B \cdot \vec dl = \mu_oI_{enc}(t)$$

It kind of makes sense since I could just claim it is a time-dependant b-field when I apply ampere's law.

To summerize, basically I'm trying to use K to find I, then use I to find B, then use B to find flux, then use flux to find E... can this be done?

 

[tman12321]

Yes that is correct. Find I with K. Use Ampere's law to get B, then Faraday's law to get E. If I'm understanding the problem statement correctly, you have a solenoid in the limit that the number of turns per unit length goes to infinity. Is that right?

 

[rude man]

I don't think you can find the B field using Ampere's law.

The B field along the z axis is probably deteminable by Biot-Savart fairly readily though I've never done it for an infinite cylinder..

Finding the total flux inside the circulating current is prohibitively difficult unless you're willing to delve into elliptical integrals. So faraday is out as I see it (perhaps I will be refuted).

In the quasi-static case there is always j = σE where j is current density and E is the E field inside the cylinder. So since j = 0 inside the cylinder one can say that so is the E field.