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Expert-verified Found in: Page 967 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Suppose that a parallel-plate capacitor has circular plates with a radius ${\mathbit{R}}{\mathbf{=}}{\mathbf{30}}{\mathbf{}}{\mathbf{\text{mm}}}$ and, a plate separation of ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{mm}}}$. Suppose also that a sinusoidal potential difference with a maximum value of ${\mathbf{150}}{\mathbf{}}{\mathbf{\text{V}}}$ and, a frequency of ${\mathbf{60}}{\mathbf{}}{\mathbf{\text{Hz}}}$ is applied across the plates; that is,${\mathbit{V}}{\mathbf{=}}\mathbf{\left(}\mathbf{150}\mathbf{}\mathbf{\text{V}}\mathbf{\right)}{\mathbf{\text{sin}}}\mathbf{\left[}\mathbf{2}\mathbf{\pi }\mathbf{\left(}\mathbf{60}\mathbf{}\mathbf{\text{Hz}}\mathbf{\right)}\mathbf{t}\mathbf{\right]}$(a) Find ${{\mathbf{\text{B}}}}_{\mathbit{m}\mathbit{a}\mathbit{x}}\left(\mathbit{R}\right)$, the maximum value of the induced magnetic field that occurs at ${\mathbit{r}}{\mathbf{=}}{\mathbit{R}}$. (b) Plot ${{\mathbf{B}}}_{{\mathbf{max}}}\left(\mathbf{r}\right)$ for ${\mathbf{0}}{\mathbf{<}}{\mathbit{r}}{\mathbf{<}}{\mathbf{10}}{\mathbf{}}{\mathbf{\text{cm}}}$.

1. The maximum value of the induced magnetic field that occurs at $r=R$ is $B=1.9×{10}^{-12}.$
2. The plot is given in the calculation section.
See the step by step solution

## Step 1: Given

The radius of plates, $R=30\mathrm{mm}=0.03\mathrm{m}$

Plate separation, $d=0.005\mathrm{m}$

Maximum potential difference, $V=150\mathrm{V}$

Frequency, $f=60\mathrm{Hz}$

$V=\left(150\right)\mathrm{sin}\left[2\pi \left(60\mathrm{Hz}\right)t\right]$

## Step 2: Determining the concept

By using the Maxwell equation, finding the magnetic field for the maximum potential, and plotting the graph maximum B vs r Maxwell's equations represent one of the most elegant and concise ways to state the fundamentals of electricity and magnetism.

Maxwell’s law of Induction-

$.\stackrel{\to }{B}.d\stackrel{\to }{A}={\mu }_{0}{\mathcal{E}}_{0}A\frac{dE}{dt}$

The electric field is given as-

$E=\frac{V}{d}$

Where, B is the magnetic field, is the area enclosed by the Amperian loop, E is the electric field, t is the time, V is the potential difference and, d is the distance.

## Step 3: (a) Determining the maximum value of the induced magnetic field that occurs at r=R

The electric field is given as-

$E=\frac{V}{d}$

The magnetic field induced by the changing electric field is given by the relation,

localid="1663162361810" style="max-width: none; vertical-align: -15px;" $\stackrel{\to }{B}.d\stackrel{\to }{A}={\mu }_{0}{\mathcal{E}}_{0}A\frac{dE}{dt}$

Where, localid="1663162418635" $A$ is the area enclosed by the Amperian loop, which is, localid="1663162390281" $A=\pi {d}^{2}$ .

So that, for $r,

$\begin{array}{rcl}B\left(2\pi r\right)& =& {\mu }_{0}{\mathcal{E}}_{0}\pi {r}^{2}\frac{dE}{dt}\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}r}{2}\frac{dE}{dt}\end{array}$

But,

$E=\frac{V}{d}$

So,

$\begin{array}{rcl}B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}r}{2d}\frac{dV}{dt}\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}r}{2d}\frac{d}{dt}\left({V}_{max}\mathrm{sin}\omega t\right)\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}r}{2d}{V}_{max}\omega \mathrm{cos}\omega t\end{array}$

For the maximum value of potential ${V}_{max}=150\mathrm{V}$,

$B=\frac{{\mu }_{0}{\mathcal{E}}_{0}r}{2d}{V}_{max}\omega$

The $r=R=0.03\mathrm{m}$

localid="1663161555184" $\begin{array}{rcl}B& =& \frac{\left(4\pi ×{10}^{-7}\text{H/m}\right)\left(8.85×{10}^{-12}\text{F/m}\right)×0.03\text{m}}{2×0.005\text{m}}×150×2\pi ×60\text{Hz}\\ B& =& 1.9×{10}^{-12}\mathrm{T}\end{array}$

Therefore, the maximum value of the induced magnetic field that occurs at $r=R$ is $B=1.9×{10}^{-12}\mathrm{T}.$

## Step 4: (b) Determining the required plot

The maximum value of $B$, the magnetic field is induced by the changing electric field so that,

$\begin{array}{rcl}{B}_{max}& =& {\left(\frac{{\mu }_{0}{\mathcal{E}}_{0}{R}^{2}}{2r}\frac{dE}{dt}\right)}_{max}\\ {B}_{max}& =& {\left(\frac{{\mu }_{0}{\mathcal{E}}_{0}{R}^{2}}{2rd}\frac{dV}{dt}\right)}_{max}\\ {B}_{max}& =& {\left(\frac{{\mu }_{0}{\mathcal{E}}_{0}{R}^{2}}{2rd}{V}_{max}\omega \mathrm{cos}\omega t\right)}_{max}\\ {B}_{max}& =& {\left(\frac{{\mu }_{0}{\mathcal{E}}_{0}{R}^{2}}{2rd}{V}_{max}\omega \right)}_{max}\end{array}$

So, all values are constant except r.

B is dependent on the value of r, so plot the graph B vs r.

$\begin{array}{rcl}{B}_{max}& =& \frac{\left(4\pi ×{10}^{-7}\text{H/m}\right)\left(8.85×{10}^{-12}\text{F/m}\right)×{\left(0.03\text{m}\right)}^{2}}{2×0.005\text{m×r}}×150×2\pi ×60\text{Hz}\\ & =& 5.7×{10}^{-14}×\frac{1}{r}\end{array}$

Here, r varies from 0 to 0.1. Hence, plotted the graph ${B}_{max}$ vs $r$, for varying from 0 to 0.1. ### Want to see more solutions like these? 