A parallel plate capacitor made of circular p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The displacement current $I_d$ is given by $I_d = \epsilon_0 \frac{d\phi_e}{dt} = \epsilon_0 \frac{d}{dt}(EA \cos 0^{\circ})$.Since $E = \frac{\si

Vedclass · Accepted Answer

non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Vedclass · Answer

(C) The displacement current $I_d$ is given by $I_d = \epsilon_0 \frac{d\phi_e}{dt} = \epsilon_0 \frac{d}{dt}(EA \cos 0^{\circ})$.Since $E = \frac{\sigma}{\epsilon_0}$,we have $I_d = \epsilon_0 A \frac{d}{dt}(\frac{\sigma}{\epsilon_0}) = A \frac{d\sigma}{dt}$.Given that the surface charge density $\sigma$ increases at a constant rate,$\frac{d\sigma}{dt}$ is constant,so $I_d$ is constant.Using Ampere-Maxwell law,the magnetic field $B$ at a distance $r$ from the axis (inside the plates,$r Thus,$B = \frac{\mu_0 I_d r}{2\pi R^2}$.For $r > R$ (outside the plates),$B(2\pi r) = \mu_0 I_d$,so $B = \frac{\mu_0 I_d}{2\pi r}$.The magnetic field is non-zero everywhere and reaches its maximum value at the boundary $r = R$ (the imaginary cylindrical surface connecting the peripheries of the plates).

$A$ parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is

Similar Questions

Write the $SI$ unit of ${\epsilon _0}\left( {\frac{{d{\Phi _E}}}{{dt}}} \right)$.

An insulator plate is passed between the plates of a capacitor. Then the displacement current

You are given a $2 \,\mu F$ parallel plate capacitor. How would you establish an instantaneous displacement current of $1 \, mA$ in the space between its plates?

Vedclass Test Series

Exam Paper Generator

Online Exam Module