Q18P - Fundamentals Of Physics

Question

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 $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{c}\mathit{m}$?

Step-by-Step Solution

Answer

The magnitude of the magnetic field within the capacitor at  $t = 250\mu s$ is, $B=8.40\times {10}^{-13}\text{T}$ .

Given data

Series resistance,

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

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

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

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

Determining the concept

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$}\right.\right) e^{-t/T}$

$C=\frac{\epsilon A }{d}$
 

 

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.

Determining the magnitude of the magnetic field within the capacitorfor the given circuit

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.