The circuit in Fig. consists of switch S, a $\mathbf{12}\mathbf{.}\mathbf{0}\mathit{V}$ideal battery, a $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{M}\mathit{\Omega }$ resistor, and an air-filled capacitor. The capacitor has parallel circular plates of radius $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$ , separated by $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathit{m}$ . At time $\mathit{t}\mathbf{=}\mathbf{0}$ , switch S is closed to begin charging the capacitor. The electric field between the plates is uniform. At $\mathit{t}\mathbf{=}\mathbf{250}\mathit{\mu }\mathit{s}$, what is the magnitude of the magnetic field within the capacitor, at a radial distance $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$?

Series resistance,

$$\begin{gathered} R=20.0M\Omega \\ =20.0\times {10}^6\Omega \end{gathered}$$

Battery Voltage,$\epsilon =12.0\text{V}$

The radius of parallel circular plate capacitor,$5.00cm=5.00\times {10}^{-2}m$

Parallel plate separation distance,$d=3.00cm=3.00\times {10}^{-2}m$

To find the magnetic field in the capacitor, the current should be evaluated for a given time. For the given RC circuit current can be determined using$\mathit{R}$, $\mathit{C}$and, using the time constant.The capacitance depends on factors like plate area, separation distance, and permittivity of the separator. These are not normally affected by a magnetic field.

Formulae are as follows:

$B=\frac{{\mu }_0{i}_{d}r}{2\pi R^2}$

$i=\left(\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$R$}\right.\right)e^{-t/T}$

role="math" localid="1663220025271" $C=\frac{\epsilon A}{d}$

where, B is the magnetic field,i is the current and,R is the inside radius,r is the outside radius, C is the capacitance,A is the area.

Calculate the capacitance for the given dimension of the parallel plate circular capacitor and it is given as,

$C=\frac{\epsilon A}{d}$

$C=\frac{{\epsilon }_0{\mathrm{\pi r}}^2}{d}$

$C=\frac{8.85\times {10}^{-12}\times \pi \times {\left(0.05\right)}^2}{3.00\times {10}^{-2}}$

$C=2.32\times {10}^{-11}\text{F = 23.2pF}$

The capacitive time constant is given as$\tau$,

$\tau =RC$

$\tau =20.0\times {10}^6\Omega \times 2.32\times {10}^{-11}F$

$\tau =4.64\times {10}^{-4}s$

At $t=250\mu s$, the capacitive current is given by

$i=\left(\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$R$}\right.\right)e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$T$}\right.}$

$i=\frac{12}{20.0\times {10}^6}{e}^{\raisebox{1ex}{$-250.0\times {10}^{-6}$}\!\left/ \!\raisebox{-1ex}{$4.64\times {10}^{-4}$}\right.}$

$i=3.5\times {10}^{-5}A$

In capacitor,$i=i_d$, to find the magnitude of the magnetic field within the capacitor at given time $t=250\mu s$,

$B=\frac{{\mu }_0{i}_{d}r}{2\pi R^2}$,

where$i_d$ is the charge displacement current calculated at$t=250\mu s$.

Hence, the magnitude of the magnetic field B can be calculated as,

$B=\frac{4\times \pi \times {10}^{-7}\times 3.5\times {10}^{-7}\times 0.03}{2\times \pi \times {0.05}^2}$

$B=8.40\times {10}^{-13}\text{T}$

Hence, the magnitude of the magnetic field is$B=8.40\times {10}^{-13}\text{T}$.

The magnetic field for the parallel circular plate capacitor at a given time by evaluating the capacitor value, the time constant and the capacitor current can be found.