B field between the plates of a charging capacitor (Ampere's law)

[FranzDiCoccio]

A standard example consider a capacitor whose parallel plates have a circular shape, of radius R, so that the system has a cylindrical symmetry.
The magnetic field at a given distance r from the common axis of the plates is calculated via Ampere's law:
<br /> \oint_\gamma {\mathbf B} \cdot d{\mathbf s} = \mu_0\epsilon_0 \dot \Phi_S({\mathbf E})<br />

If \gamma is a circle of radius r&lt;R, one gets
<br /> B_t = \mu_0 \frac{r}{R^2} I <br />
and otherwise
<br /> B_t = \mu_0 \frac{1}{r} I <br />
where B_t is the component of the magnetic field that is tangent to the circular loop.

My question is: how do we know that there is no radial component of the magnetic field, so that we can conclude B=B_t?
I do not think that the cylindrical symmetry of the system is enough to warrant this.

Could the reason be some "matching condition" between the magnetic field inside and outside the capacitor (I mean, in the regions where the current carrying wires are)?

 

[hutchphd]

Because there is another Maxwell equation for B which is (Gauss) that the flux of B is zero through any closed surface.

 

[FranzDiCoccio]

Oh, right. Thanks!
Silly me. Serves me right for working until 2 AM. I should have gone to bed and started fresh in the morning.

So, right, if I consider a cylindrical "pillbox" with the same axis as the plates, and the same radius as the loop, the flux of B through it should be 0. And this is possible only if B has no radial component.

I'd say that the same line of reasoning would apply to the analogous problem where a solenoid is charging up.
The uniform B field inside the solenoid is increasing, and there is a nonzero E_t along a circular loop that has the same axis as the solenoid. Again, Gauss' theorem and the fact that there is no net electrical charge in the solenoid allows us to rule out an E_r.
Neat.

Thanks again!