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My question is very similar to this question

enter image description here 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 $$B = \int \mu_0 J(r) dr$$

but how was this obtained ?

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.

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  • $\begingroup$ 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...? $\endgroup$
    – kricheli
    Jan 9, 2023 at 20:10
  • $\begingroup$ @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. $\endgroup$ Jan 9, 2023 at 21:31

2 Answers 2

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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 $dB = \mu_0 dK = \mu_0 J dr$

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The equation $\frac{\partial E}{\partial t}=0$ is not possible, but \begin{equation} \epsilon_0\frac{\partial E}{\partial t}\simeq 0 \end{equation} 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.

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  • $\begingroup$ 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 $B =\int \mu_0 J dr$ $\endgroup$ Jan 10, 2023 at 1:24
  • $\begingroup$ You can't blame him, nothing is mentioned about $\omega$ being large in the question, and I'd have guessed the same as @HEMMI. :) $\endgroup$
    – kricheli
    Jan 10, 2023 at 17:24

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