Imagine two concentric metal spherical shells (Fig. 7.44).The inner one (radius a) carries a charge Q(t), and the outer one (radius b) an opposite charge −Q(t). The space between them is filled with Ohmic material of conductivity σ, so a radial current flows:j=σE=σ 1/4Πε0 Q/r^2 r;I=-Q=∫j.

Question

Imagine two concentric metal spherical shells (Fig. 7.44).The inner one (radius a) carries a charge Q(t), and the outer one (radius b) an opposite charge −Q(t). The space between them is filled with Ohmic material of conductivity σ, so a radial current flows: [latex]J=\sigma E=\sigma \frac{1}{4\pi\epsilon _{0}}\frac{Q}{r^{2}}\hat{\pmb{r}};I=-Q=\int{J.da}=\frac{\sigma Q}{\epsilon _{0}}. [/latex] .This configuration is spherically symmetrical, so the magnetic field has to be zero(the only direction it could possibly point is radial, and [latex]\pmb{∇ · B} = 0 ⇒\oint{\pmb{B} · da} =B(4\pi r^{2}) = 0, so \pmb{B = 0}).[/latex]. What? I thought currents produce magnetic fields! Isn’t that what Biot-Savart and Ampère taught us? How can there be a J with no accompanying B?

Accepted Answer

This is not a static configuration: Q, E, and J are all functions of time; Ampère and Biot-Savart do not apply. The displacement current

[latex]J_{d}=\epsilon _{0}\frac{\partial E}{\partial t}=\frac{1}{4\pi}\frac{\dot{Q} }{r^{2}}\hat{\pmb{r}}=-\sigma \frac{Q}{4\pi \epsilon _{0}r^{2}}\hat{\pmb{r}}[/latex]

exactly cancels the conduction current (in Eq. 7.37), and the magnetic field (determined by [latex]\pmb{∇ · B }= 0, \pmb{∇ × B = 0})[/latex] is indeed zero.

[latex]∇ × B = μ_0J + μ_{0}\epsilon_{0}\frac{\partial E}{\partial t} .[/latex] (7.37)

Question 7.14: Imagine two concentric metal spherical shells (Fig. 7.44).Th...

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