Suppose that a parallel-plate capacitor has circular plates with a radius $\mathit{R}\mathbf{=}\mathbf{30}\mathbf{}\mathbf{\text{mm}}$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,

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

(a) Find${\mathbf{\text{B}}}_{\mathit{m}\mathit{a}\mathit{x}}\left(\mathit{R}\right)$, the maximum value of the induced magnetic field that occurs at $\mathit{r}\mathbf{=}\mathit{R}$.

(b) Plot${\mathbf{B}}_{\mathbf{max}}\left(\mathbf{r}\right)$ for$\mathbf{0}\mathbf{<}\mathit{r}\mathbf{<}\mathbf{10}\mathbf{}\mathbf{\text{cm}}$.

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, $f=60\mathrm{Hz}$

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

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

Maxwell’s law of Induction-

$.\overrightarrow{B}.d\overrightarrow{A}={\mu }_0{\mathcal{E}}_{0}A\frac{dE}{dt}$

The electric field is given as-

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

Where, B is the magnetic field,is the area enclosed by the Amperianloop,Eisthe electric field, t is the time, Vis the potential difference and, dis the distance.

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;" $\overrightarrow{B}.d\overrightarrow{A}={\mu }_0{\mathcal{E}}_{0}A\frac{dE}{dt}$

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

So that, for $r<R$,

$\begin{array}{rcl}B\left(2\pi r\right)& =& {\mu }_0{\mathcal{E}}_0\pi r^2\frac{dE}{dt}\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0}r}{2}\frac{dE}{dt}\end{array}$

But,

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

So,

$\begin{array}{rcl}B& =& \frac{{\mu }_0{\mathcal{E}}_{0}r}{2d}\frac{dV}{dt}\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0}r}{2d}\frac{d}{dt}\left(V_{max}\sin \omega t\right)\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0}r}{2d}{V}_{max}\omega \cos \omega t\end{array}$

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

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

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

$\begin{array}{rcl}{B}_{max}& =& {\left(\frac{{\mu }_0{\mathcal{E}}_0{R}^2}{2r}\frac{dE}{dt}\right)}_{max}\\ B_{max}& =& {\left(\frac{{\mu }_0{\mathcal{E}}_0{R}^2}{2rd}\frac{dV}{dt}\right)}_{max}\\ B_{max}& =& {\left(\frac{{\mu }_0{\mathcal{E}}_0{R}^2}{2rd}{V}_{max}\omega \cos \omega t\right)}_{max}\\ B_{max}& =& {\left(\frac{{\mu }_0{\mathcal{E}}_0{R}^2}{2rd}{V}_{max}\omega \right)}_{max}\end{array}$

So, all values are constant except r.

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

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

Here, r varies from 0 to 0.1.

Hence, plotted the graph $B_{max}$vs $r$, for varying from 0 to 0.1.