- #1

FranzDiCoccio

- 342

- 41

The magnetic field at a given distance [itex]r[/itex] from the common axis of the plates is calculated via Ampere's law:

[tex]

\oint_\gamma {\mathbf B} \cdot d{\mathbf s} = \mu_0\epsilon_0 \dot \Phi_S({\mathbf E})

[/tex]

If [itex]\gamma[/itex] is a circle of radius [itex]r<R[/itex], one gets

[tex]

B_t = \mu_0 \frac{r}{R^2} I

[/tex]

and otherwise

[tex]

B_t = \mu_0 \frac{1}{r} I

[/tex]

where [itex]B_t[/itex] 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 [itex]B=B_t[/itex]?

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)?