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
\begin{equation}
E = \frac{1}{2} \mu \omega n r I_0 sin(\omega t) \hat{\phi}
\end{equation}
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
\begin{equation}
B = \int_{r’}^{a} \mu J Dr
\end{equation}
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
\begin{equation}
B = \int_{0}^{a} \mu J Dr
\end{equation}
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.
Best Answer
The only part of $\mathbf{J}(r)$ that contributes to $\mathbf{B}(r')$ is current densities for which $r > r'$.
Consider the Amperian loop in this answer. Since the side cd can be chosen arbitrarily far away, the field outside of the radially symmetric current density must be zero. So you need only consider the currents that are further away from the axis than the observation point.