Q59P - Fundamentals Of Physics

Question

A parallel-plate capacitor with circular plates of radius $\mathbf{R}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{mm}$ and gap width $\mathbf{d}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mm}$  has a uniform electric field between the plates. Starting at time $\mathbf{t}\mathbf{=}\mathbf{0}$, the potential difference between the two plates is $\mathbf{V}\mathbf{=}\mathbf{\left(}\mathbf{100}\mathbf{V}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{\tau }}}$ , where the time constant $\mathbf{\tau }\mathbf{=}\mathbf{12}\mathbf{ }\mathbf{ms}$. At radial distance $\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.8}\mathbf{ }\mathbf{R}$ from the central axis, what is the magnetic field magnitude (a) as a function of time for  $\mathbf{t}\mathbf{\ge }\mathbf{0}$ and (b) at time $\mathbf{t}\mathbf{=}\mathbf{3}\mathbf{\tau }$?

Step-by-Step Solution

Answer

1. Magnetic field as a function of time for $\mathrm{t}\ge 0$ is $(1.2\times {10}^{-13} \mathrm{T}) {\mathrm{e}}^{-\frac{-\mathrm{t}}{12 \mathrm{ms}}}$
2. Magnetic field as a function of time at $t=3\tau$ is  $5.9\times {10}^{-15} \mathrm{T}$

Step 1: Listing the given quantities

Radius of circular plate is $\mathrm{R}=16 \mathrm{mm}$

Width is $\mathrm{d}=5.0 \mathrm{mm}$

$\mathrm{V}=\left(100\mathrm{V}\right){\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{\tau }}}$

Radius is $\mathrm{r}=0.8 \mathrm{R}$

Time constant, $\mathrm{\tau }=12 \mathrm{ms}$

Step 2: Understanding the concepts of Magnetic and electric field

We have to use the formula for magnetic field induced by electric field, and then we use the formula for electric field in terms of voltage and distance and substitute it in the equation of magnetic field.

Formula:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\mathrm{r}}{2}\times \frac{\mathrm{dE}}{\mathrm{dt}}$

$\mathrm{E}=\mathrm{V}/\mathrm{d}$

Step 3: (a) Calculations of the Magnetic field as a function of time for t ≥ 0

Magnetic field induced by changing electric field is given as

$\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\mathrm{r}}{2}\times \frac{\mathrm{dE}}{\mathrm{dt}}$

We know $\mathrm{E}=\mathrm{V}/\mathrm{d}$.

Differentiating the above with respect to time:

$\begin{array}{c}\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{V}}{\mathrm{d}}\right)\\ =\frac{1}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\left(100\right){\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{\tau }}}\right)\\ =\frac{-100\times {\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{\tau }}}}{\mathrm{\tau d}}\end{array}$

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\mathrm{r}}{2}\times \frac{-100\times {\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{\tau }}}}{\mathrm{\tau d}}\\ =-\frac{4\mathrm{\pi }\times {10}^{-7}\times 8.85\times {10}^{-12}\times 16\times {10}^{-3}\times 100\times 0.8\times {\mathrm{e}}^{-\frac{\mathrm{t}}{12\mathrm{ms}}}}{2\times 12\times {10}^{-3}\times 5\times {10}^{-3}}\\ =-(1.2\times {10}^{-13} \mathrm{T}) {\mathrm{e}}^{-\frac{-\mathrm{t}}{12 \mathrm{ms}}}\end{array}$

The magnitude is $\mathrm{B}=(1.2\times {10}^{-13} \mathrm{T}) {\mathrm{e}}^{-\frac{-\mathrm{t}}{12 \mathrm{ms}}}$

Step 4: (b) Calculations of the Magnetic field as a function of time for t = 3 τ

$\begin{array}{c}\mathrm{B}=(1.2\times {10}^{-13} \mathrm{T}) {\mathrm{e}}^{-\left(\frac{3\mathrm{\tau }}{\mathrm{\tau }}\right)}\\ =5.9\times {10}^{-15} \mathrm{T}\end{array}$

Magnetic field as a function of time at $\mathrm{t}=3\mathrm{\tau }$  is $5.9\times {10}^{-15} \mathrm{T}$