Induced magnetic field due to current density in solenoid

Suppose there is a cylindrical solenoid with $$n$$ turns, magnetic permitivity $$\mu$$, electrical conductivity $$\sigma$$, and current $$I(t) = I_0 cos\omega t$$. Find the magnetic field produced by the induced volume current density $$J=\sigma E$$.

I have already found the electric field to be $$$$E = \frac{1}{2} \mu \omega n r I_0 sin(\omega t) \hat{\phi}$$$$

The solution states that you can treat $$Jdr$$ as a cylindrical shell which behaves as a solenoid. Then they state that the magnetic field due to this volume current density is $$$$B = \int_{r’}^{a} \mu J Dr$$$$

What I don’t understand is if the volume current density acts like a solenoid then why is the magnetic field from the current density outside of the “loop” from current density? From my reasoning I would expect the magnetic field to be $$$$B = \int_{0}^{a} \mu J Dr$$$$ Can anyone tell me why this is incorrect because I know it is but can’t figure out why the correct solution is what it is.

• I don't think the geometry of the system you are describing is entirely clear.
– Buzz
Commented May 10, 2022 at 2:02
• It’s a cylindrical solenoid with n turns around it and a current going through it. That’s the only geometry aspect of it. Commented May 10, 2022 at 4:53
• This question could use some clarification. Is the solenoid filled with a conductive material like the equations suggest? It also seems like some assumptions/simplifications were made in finding the electric field. Have you considered that the magnetic field due to the induced current density also contributes to the total electric field? Commented May 10, 2022 at 9:54
• I have a similar question but I am a bit confused, how did you get that $B = \int \mu J dr$ ? Commented Jan 9, 2023 at 18:56
• @realanswers Hi sorry for the late response! If I remember correctly it is because if you treat the cylindrical shell $Jdr$ as a solenoid, then the magnetic field due to the “solenoid” is $dB = \mu J dr$ thus the total magnetic field is $B = \int \mu J dr$ I hope that helps. Commented Jan 16, 2023 at 16:41

The only part of $$\mathbf{J}(r)$$ that contributes to $$\mathbf{B}(r')$$ is current densities for which $$r > r'$$.