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edit: Hi I'm trying to find the magnetic field generated by a time dependent oscillating current in the quasistatic case ($|z|,r <<c\omega$) where r is the perpendicular distance from the z-axis.

The current is flowing through a long, thing wire that is laying on the z-axis. If $z=0$, we can write the current as

$$ I(t)=I_0\sin \omega t $$ and now I am trying to find the magnetic field $B(r,t)$ at $z=0$ with this current. How can we calculate $\vec B$?

Possibly, can we calculate $\vec J$ to calculate $\vec A$ to calculate the magnetic field?

THanks a lot for your help on the last problem too.

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    $\begingroup$ Is the current $I$ flowing through some type of wire of known cross section? $\endgroup$
    – BMS
    May 21, 2014 at 5:48
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    $\begingroup$ @BMS Yes I will update that. Sorry I didn't want to be too specific to deter people. Sorry. Thanks $\endgroup$
    – Jason
    May 21, 2014 at 6:06
  • $\begingroup$ Do you focus on the B-field inside or outside the wire? I think you only really need $\vec{J}$ if you focus on the B-field inside the wire else Biot-Savart for a current thread should be sufficient for most applications. But, that also depends on the phase velocity $\omega$. $\endgroup$
    – Tobias
    May 21, 2014 at 15:34
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    $\begingroup$ Thanks for mentioning it. Yes, the above application of Biot-Savart only works for the quasi-static case. Else, you would also have to consider the current wave propagation along the wire. In this case you would need to set up $I(t,z)$. Furthermore, you would need to consider the radiation of electromagnetic waves from the wire. $\endgroup$
    – Tobias
    May 21, 2014 at 16:27
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    $\begingroup$ @Tobias Okay that makes sense. Thanks. I know for a wire the magnetic field is $$ \vec B(r,t)=\frac{\mu_0 I}{2\pi r}\hat \phi, $$ and since we have $I=I(t)$ in this case, you just plugged that into this formula. Thats how you got the magnetic field right? Thanks a lot. $\endgroup$
    – Jason
    May 21, 2014 at 16:33

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The direction of $\vec{J}$ is the direction of the current. The magnitude of $\vec{J}$ is the current per unit surface area perpendicular to the current.

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  • $\begingroup$ Thank you. If you can add the magnetic field i will check it off as answer . I can only up vote when I have +15 reputation. Upvote my question and I can :) $\endgroup$
    – Jason
    May 21, 2014 at 6:20
  • $\begingroup$ That gives the definition of $\vec J$ but it does not answer the question. You need Maxwell's equations to calculate the $\vec B$-field. If you have $\varepsilon\omega \ll \mu$ you can assume the quasi-stationary approximation $\varepsilon = 0$. I think, in the end you get something like the modified Bessel functions for the $B$-field distribution in the wire. (The descent of the field in the middle of the wire is named as skin effect in the literature.) This is much easier to compute for a conducting half-space than for a round wire. And often the half-space solution is a sufficient approx... $\endgroup$
    – Tobias
    May 21, 2014 at 8:17
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    $\begingroup$ @Tobias Does it matter that this is a long, thin wire? Not a round wire? I am familiar with Bessel functions however do you have any possible more advice as to how you got them for the magnetic field distribution? Thanks for taking your time to help $\endgroup$
    – Jason
    May 21, 2014 at 15:08
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    $\begingroup$ @Jason Look at the skin depth $\delta=\sqrt{\frac2{\omega\sigma\mu}}=\sqrt{\frac{2\rho}{\omega\mu}}$ from Skin-effect. If this skin width is much larger than the radius of your wire you can use the approximation $J\approx\frac{I}{A}$ in the full cross-section. Sorry, the above inequality should read $|\varepsilon\omega| \ll |\sigma|$ and not $\varepsilon\omega\ll\mu$. $\endgroup$
    – Tobias
    May 21, 2014 at 15:26

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