# Finding induced magnetic field due to induced current

My question is very similar to this question

So I calculated the induced electric field which in turn induces a current density in the cylinder.

But I am confused why when one divides a cylinder into shells that can be viewed as solenoids that $$dB = \mu_0 J(r) dr$$ thus $$B = \int \mu_0 J(r) dr$$

we know from Maxwell's equations that

$$\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \partial E / \partial t$$

My questions are :

why do we say that $$\partial E / \partial t = 0$$ ? After all it is the induced electric field which induces the current we are using here, why do I only use $$J$$ in the calculation?

and shouldn't it be that $$\int B . dl = \int \mu_0 J . ds$$ from maxwell's equations? how did we get to $$B = \int \mu_0 J(r) dr$$?

Edit for clarity:

I have calculated the charge density $$J$$ induced by the induced $$E$$ field induced by the time varying magnetic field, now I wish to calculate the self-induced magnetic field through the induced $$J$$. That is the $$B$$ I am confused about here.

• It is not entirely clear to me what you want to achieve. Compare the title to the information given in the picture etc. Do you work with a given current to find a magnetic field? The other way round? For your formula to obtain $B$ from $J$ you maybe might want to have a look at Biot-Savart and express it in cylindrical coordinates...? Jan 9, 2023 at 20:10
• @kricheli I edited the post for clarity. This equation is not the Biot-savart law in cylindrical coordinates, or at least I do not see it. Please do explain further. Jan 9, 2023 at 21:31

After staring at it for a while I believe I have figured it out. The idea is that for a solenoid $$B = \mu_0 K$$ Thus here when we have a cylinder that we are slicing up into solenoids $$dB = \mu_0 dK = \mu_0 J dr$$
The equation $$\frac{\partial E}{\partial t}=0$$ is not possible, but $$$$\epsilon_0\frac{\partial E}{\partial t}\simeq 0$$$$ is valid at low frequencies. It is based on the circumstance that the displacement current is negligible at low frequencies. It would be good to calculate $$\frac{\epsilon\omega}{\sigma}\ll 1.0$$ numerically.
• I believe you are misunderstanding the question. The question is for large w find the self induced magnetic field. I want an explanation as to how $B =\int \mu_0 J dr$ Jan 10, 2023 at 1:24
• You can't blame him, nothing is mentioned about $\omega$ being large in the question, and I'd have guessed the same as @HEMMI. :) Jan 10, 2023 at 17:24