Q5P - Fundamentals Of Physics

Question

The induced magnetic field at a radial distance 6.0 mm from the central axis of a circular parallel-plate capacitor is $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ }\mathit{T}$. The plates have a radius 3.0 mm. At what rate $\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{/}\mathit{d}\mathit{t}$ is the electric field between the plates changing?

Step-by-Step Solution

Answer

The value of $\frac{d\stackrel{\rightharpoonup }{E}}{dt}$ between the charging plates is $2.4\times {10}^{13}V/m·s$.

Step 1: Identification of the given data

The given data is listed as follows,

 

• The radial distance of induced magnetic field from the central axis of a circular parallel plate capacitor is, $r=6.0mm\times \frac{1m}{1000 mm}=6.0\times {10}^{-3}m$
• The induced magnetic field is, $B=2.0\times {10}^{-7}T$
• The radius of the plate is, $R=3.0mm\times \frac{1m}{1000mm}=3.0\times {10}^{-3}m$

Step 2: Expression for the Maxwell’s law of induction

The expression for Maxwell’s law of induction is as follows,

$\mathbf{\int }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{A}\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}$

Here, B is the induced magnetic field, A is the area enclosed which is given by the following expression $A=\pi R^2$ (Here, R is the radius of the plate), ${\mu }_0$ is the magnetic permeability of the free space with the value of $\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^2\right), {\epsilon }_0$,  is the electric permeability of the free space with the value of $\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)$, and $\frac{dE}{dt}$ is then the rate of change of electric field.

Step 3: Determination of the value of d E ⇀ / d t between the charging plates

Substitute the $2\mathrm{\pi r}$ for dA, and ${\mathrm{\pi R}}^2$ for A in the Maxwell’s law of induction and rearrange the expression.

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

Substitute all the values in the above expression.

$\begin{array}{rcl}\frac{dE}{dt}& =& \frac{2\times 2.0\times {10}^{-7}T\times 6.0\times {10}^{-3}m}{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^2\right)\left(8.85\times {10}^{-12}{C}^2/N·m^2\right){\left(3.0\times {10}^{-3}m\right)}^2}\\ & =& \frac{24.0\times {10}^{-10}}{1001\times {10}^{-25}}T·m·A^2/C^2\times \frac{1N/A·m}{1 T}\\ & =& 2.3975\times {10}^{13}N·A/C^2\times \frac{1C/s·m}{1 A}\\ & =& 2.4\times {10}^{13}N/C·s\times \frac{1V/m}{1N/C}\\ & =& 2.4\times {10}^{13}V/m·s\end{array}$

Thus, the value of $\frac{d\stackrel{\rightharpoonup }{E}}{dt}$ between the charging plates is $2.4\times {10}^{13}V/m·s$.