Q11P - Fundamentals Of Physics

Question

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  $\mathbf{0}\mathbf{<}\mathit{r}\mathbf{<}\mathbf{10}\mathbf{ }\mathbf{\text{cm}}$.

Step-by-Step Solution

Answer

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

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,  $f=60 \mathrm{Hz}$

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

 

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-

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

The electric field is given as-

 $E=\frac{V}{d}$

Where, B  is the magnetic field,   is the area enclosed by the 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, 

 style="max-width: none; vertical-align: -15px;" $\overrightarrow{B}.d\overrightarrow{A}={\mu }_0{\mathcal{E}}_{0 }A\frac{dE}{dt}$

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

 

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 $r=R=0.03 \mathrm{m}$ 

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

 

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

Step 4: (b) Determining the required plot

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

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

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.