2) In charging a circular parallel plate capacitor, it can be shown that the magnetic flux density between the plates is of the form-\vec{B}(r, t)=\frac{\mu_{0} \Delta \phi}{2 \pi R} \frac{r}{r_{0}^{2}} \exp \left(-\frac{t}{R C}\right) \hat{\varphi}-where R is a resistance in series with the capacitor, C is the capacitance of the parallel plate capacitor of radius ro, r is the radial distance from the centre of the capacitor in cylindrical co-ordinates, t is time and A is the potential difference across the capacitor.-a) Using this result, determine an expression for the displacement current density inside the capacitor while it charges.-b) Consider the case where the material between the electrodes has a small but non-zero conductivity. Briefly,and with explicit reference to the appropriate Maxwell equations and their dependent variables such as current and time, describe the electric and magnetic fields present inside the capacitor, including any relevant information about the field directions.-c) It ro = 3 mm, the gap between the parallel plates is 50 µm and filled with a material with a relative permittivity 58,determine the series resistance required so that the magnetic flux density drops by 37% in 1 ms.

Mechanics(Physics)

Expert answer

100% Verified

Back

+

1/4

x

Sign Up to continue

You are just one step away from getting connected
to an expert tutor ! Hurry Up !!

Post your query, Signup and get connected
to a human operator instantly.