91Ó°ÊÓ

Series resistance,

$$\begin{gathered} R=20.0M\mathrm{Î\copyright } \\ =20.0×{10}^6\mathrm{Î\copyright } \end{gathered}$$

Battery Voltage,$\mathrm{Î\mu }=12.0\text{V}$

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

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

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:

$B=\frac{{\mathrm{μ}}_0{i}_{d}r}{2\mathrm{π}{R}^2}$

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

role="math" localid="1663220025271" $C=\frac{\mathrm{Î\mu }A}{d}$

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,

$C=\frac{\mathrm{Î\mu }A}{d}$

$C=\frac{{\mathrm{Î\mu }}_0{\mathrm{Ï€°ù}}^2}{d}$

$C=\frac{8.85×{10}^{-12}×\mathrm{π}×{\left(0.05\right)}^2}{3.00×{10}^{-2}}$

$C=2.32×{10}^{-11}\text{F = 23.2pF}$

The capacitive time constant is given as$\mathrm{Ï„}$,

$\mathrm{Ï„}=RC$

$\mathrm{Ï„}=20.0×{10}^6\mathrm{Î\copyright }×2.32×{10}^{-11}F$

$\mathrm{τ}=4.64×{10}^{-4}s$

At $t=250\mathrm{μ}s$, the capacitive current is given by

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

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

$i=3.5×{10}^{-5}A$

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

$B=\frac{{\mathrm{μ}}_0{i}_{d}r}{2\mathrm{π}{R}^2}$,

where$i_d$ is the charge displacement current calculated at$t=250\mathrm{μ}s$.

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

$B=\frac{4×\mathrm{π}×{10}^{-7}×3.5×{10}^{-7}×0.03}{2×\mathrm{π}×{0.05}^2}$

$B=8.40×{10}^{-13}\text{T}$

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

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.