Consider the set-up illustrated in the attachement. The radius of the capacitor plates is(adsbygoogle = window.adsbygoogle || []).push({}); a. The field btw the plates varies according to

[tex]\vec{E}(t) = \frac{It}{\epsilon_0 \pi a^2}\hat{z}[/tex]

(z is to the right)

So since the current enclosed is 0, Ampere-Maxwell law in its integral form after evaluation of the integral of [itex]d\Phi_E/dt[/itex] reads...

[tex]\oint \vec{B}\cdot d\vec{l} = \mu_0 I \frac{s^2}{a^2}[/tex]

s being the radius of my amperian loop.

What is the argument according to which B is solenoid and constant along the path of integration? In the case of magnetostatic, it was the right-hand thumb rule (i.e. the Biot-Savart law) that allowed us to determine the orientation of B. But now what is it that permits to conclude?

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# Application of Ampère-Maxwell equ.

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