Question

In Fig. 32-36, a capacitor with circular plates of radius $\mathit{R}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$is connected to a source of emf $\mathit{\xi }\mathbf{=}{\mathit{\xi }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\omega }\mathit{t}$, where ${\mathit{\xi }}_{\mathbf{m}}\mathbf{=}\mathbf{220}\mathbf{}\mathbf{\text{V}}$and $\mathit{\omega }\mathbf{=}\mathbf{130}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$. The maximum value of the displacement current is ${\mathit{i}}_{\mathbf{d}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{60}\mathbf{}\mathit{\mu }\mathbf{\text{A}}$ . Neglect fringing of the electric field at the edges of the plates. (a) What is the maximum value of the currenti in the circuit? (b) What is the maximum value of$\mathbf{d}{\mathit{\varphi }}_{\mathbf{E}}\mathbf{/}\mathbf{dt}$ , where${\mathit{\varphi }}_{\mathbf{E}}$ is the electric flux through the region between the plates? (c) What is the separation dbetween the plates? (d) Find the maximum value of the magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$ between the plates at a distance$\mathit{r}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$from the center.

Short answer

(a) The maximum value of currenti in the circuit is $7.60\mu \text{A}.$

(b) The maximum value of role="math" localid="1663142669802" $\frac{d{\varphi }_E}{dt}$ is $859\text{kV.}\frac{\text{m}}{\text{s}}$

(c) The separation between the plates is 3.39 mm

(d) The maximum value of the magnetic field between the plates at a distance $r=11\text{cm}$ from the center is $5.16\text{pT}.$

Step-by-step solution

Step 1: Given

The radius of circulating plates,$R=18\text{cm}$

The magnitude of emf,$\xi ={\xi }_{m}sin\omega t,$ where,${\xi }_m=220\text{V}$

Angular frequency,$\omega =130\text{rad/s}$

Maximum displacement current, $i_d=7.60\mu \mathrm{A}$

Determining the concept

At any instant, the displacement current $i_d$ in the gap between the plates is the same as the conduction current i in the wires. Using the displacement current formula, find the maximum rate of change of electric flux. Use the relation of electric field and potential difference in displacement current to find the separation of the plates. Find the enclosed current at the required distance, and substituting this current value in Ampere’s law, find the maximum value of the magnetic field.

Formulae are as follows:

$i_d={\xi }_0\frac{d{f}_E}{dt}$

$V=Ed$

$\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{i}_{enc}$

where, $i_d$ is the displacement current, $\varphi$is the flux, V is the potential difference, B is the magnetic field.

(a) Determining the maximum value of current  i  in the circuit.

The maximum value of current i in the circuit:

At any instant, the displacement current $i_d$ in the gap between the plates is the same as the conduction current i in the wires.

Therefore,

$i_{max}=i_d=7.60\mu \text{A}$

Therefore, the maximum value of currenti in the circuit is $7.60\mu \text{A}.$

(b) Determining the maximum value of   dϕEdt

Maximum value of$\frac{d{\varphi }_E}{dt}$:

The displacement current and electric flux are related as,

$i_d={\xi }_0\frac{d{\varphi }_E}{dt}$

Therefore,

$\begin{array}{rcl}{\left(\frac{d{\varphi }_E}{dt}\right)}_{max}& =& \frac{i_d}{{\xi }_0}\\ {\left(\frac{d{\varphi }_E}{dt}\right)}_{max}& =& \frac{7.60\times {10}^{-6}\mathrm{A}}{8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}}\\ & =& 0.859\times {10}^{\text{6}}\text{V.}\frac{\text{m}}{\text{s}}\\ & & {\text{=8.59×10}}^5\text{V.}\frac{\text{m}}{\text{s}}\\ & & \text{=859kV.m/s}\end{array}$

Therefore, the maximum value of $\frac{d{\varphi }_E}{dt}$ is $859\text{kV}·\text{m/s}$

(c) Determining the separation d  between the plates

The separation d between the plates:

The relation between displacement current and electric flux is,

$i_d={\xi }_0\frac{d{\varphi }_E}{dt}$

Electric flux${\varphi }_E=EA$

$i_d={\xi }_{0}A\frac{dE}{dt}$

The relation between potential difference and electric field is given by,

$$\begin{gathered} V=Ed \\ E=\frac{V}{d} \end{gathered}$$

Therefore,

$i_d={\xi }_{0}A\frac{d}{dt}\left(\frac{V}{d}\right)=\frac{{\xi }_{0}A}{d}\left(\frac{dV}{dt}\right)$

The potential difference developed across the capacitor is the same in magnitude as the emf of the generator.

$V={\xi }_m\sin \omega t$

$\frac{dV}{dt}=\omega {\xi }_m\cos \omega t$

Thus,

$i_d=\frac{{\xi }_{0}A}{d}\left(\omega {\xi }_m\cos \omega t\right)$

This will be maximum when.$\cos \omega t=1$

$\begin{array}{rcl}{i}_{dmax}& =& \frac{{\xi }_{0}A\omega {\xi }_m}{d}\\ d& =& \frac{{\xi }_{0}A\omega {\xi }_m}{i_{dmax}}\end{array}$

A = Area of circular plates

$A=\pi R^2;R=18\mathrm{cm}=0.180\mathrm{m}$

$\begin{array}{rcl}d& =& \frac{\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)\mathrm{\pi }{\left(0.180\mathrm{m}\right)}^2\left(130\mathrm{rad}/\mathrm{s}\right)\left(220\mathrm{V}\right)}{\left(7.60\times {10}^{-6}\mathrm{A}\right)}\\ & =& 3.39\times {10}^{-3}\text{m}\\ & =& 3.39\mathrm{mm}\end{array}$

Therefore, the separation between the plates is 3.39 mm.

(d) Determining the maximum value of the magnetic field between the plates at a distance r = 11 cm  from the center

The maximum value of the magnetic field between the plates at a distance from the center:

Draw the Amperian loop at r = 11 cm from the center parallel to the plates, and the current enclosed in this region is some part ofthe displacement current.

As this displacement current is uniform between the gaps,

$\frac{i_d}{A}=\frac{i_{denc}}{A_{enc}}=\text{constant}$

r = 11 cm

Since,$A=\pi R^2,A_{enc}=\pi r^2$.

$\begin{array}{rcl}{i}_{denc}& =& i_d\left(\frac{A_{enc}}{A}\right)\\ i_{denc}& =& i_d\left(\frac{\pi r^2}{\pi R^2}\right)\\ i_{denc}& =& i_d\left(\frac{r^2}{R^2}\right)\end{array}$

Now using Ampere’s law,

$\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{s}& =& {\mu }_0{i}_{denc}\\ B\left(2\pi r\right)& =& {\mu }_0{i}_{denc}\\ B& =& \frac{{\mu }_0{i}_{denc}}{2\pi r}=\frac{{\mu }_0}{2\pi r}{i}_d\left(\frac{r^2}{R^2}\right)\\ B& =& \frac{{\mu }_0}{2\pi r}{i}_d\left(\frac{r}{R^2}\right)\end{array}$

The maximum magnetic field will be,

$\begin{array}{rcl}{B}_{max}& =& \frac{{\mu }_0{i}_{dmax}r}{2\pi R^2}\\ B_{max}& =& \frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(7.6\times {10}^{-6}\mathrm{A}\right)\left(0.110\mathrm{M}\right)}{2\mathrm{\pi }{\left(0.180\mathrm{m}\right)}^2}\\ & =& 5.16\times {10}^{-12}\mathrm{T}\\ & =& 5.16\mathrm{pT}\end{array}$

Therefore, the maximum value of the magnetic field between the plates at a distance r = 11 cm from the center is 5.16 pT

Using the displacement current formula, the separation of the plates and also the rate of change of electric flux can be found. Using Ampere’s law, the magnetic field at any distance can be found.