Q10P - Fundamentals Of Physics

Question

The figure shows an electric field is directed out of the page within a circular region of radius $\mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$. The field magnitude is $\mathit{E}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{ }\mathbf{\text{V/ms}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{R}\mathbf{)}\mathit{t}$, where t is in seconds and r is the radial distance $\left(\mathbf{r}\le \mathbf{R}\right)$. What is the magnitude of the induced magnetic field at a radial distance $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$? What is the magnitude of the induced magnetic field at a radial distance $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{cm}}$?

Step-by-Step Solution

Answer

1. The magnitude of an induced magnetic field at a given radial distance is $B=3.09\times {10}^{-20} \mathrm{T}.$
2. The magnitude of an induced magnetic field at a given radial distance is $B=1.67\times {10}^{-20} \mathrm{T}.$

Step 1: Given

$\begin{array}{rcl}R& =& 3.00 \mathrm{cm}=0.03 \mathrm{m}\\ \mathrm{r}& =& 0.02 \mathrm{m}\\ \mathrm{E}& =& \left(0.500\frac{\mathrm{V}}{\mathrm{m}.\mathrm{s}}\right)\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)\mathrm{t}\\ & & \end{array}$

Step 2: Determining the concept

For a non-uniform electric field, first, find the electric flux for the region inside and outside the circular the region. Then calculate the magnetic field by using the Maxwell equation for a non-uniform electric field.  

Formulae are as follows:

 $\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_{0 }\frac{d\varphi }{dt}$

Where, $\overrightarrow{B}$ is the magnetic field, $\varphi$ is the flux.  

Step 3: (a) Determining the magnitude of an induced magnetic field at a given radial distance 2.00 cm

By using the formula, find the magnetic field inside the circle as,

$\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_{0 }\frac{d{\varphi }_E}{dt}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where, $\varphi$ electric flux for a non-uniform field can be defined as,

$\begin{array}{rcl}{\varphi }_E& =& \int EdA\\ {\varphi }_E& =& {\int }_0^{r}EdA\\ {\varphi }_E& =& {\int }_0^{r}E\left(2\pi rdr\right)\end{array}$

By the given value, 

$\begin{array}{rcl}{\varphi }_E& =& {\int }_0^r\left(0.500\right)\left(1-\frac{r}{R}\right)t2\pi rdr\\ {\varphi }_E& =& \left(0.500\right)t2\pi {\int }_0^r\left(1-\frac{r}{R}\right)rdr\\ {\varphi }_E& =& \left(0.500\right)t.2\pi \left\{{{\left[\frac{r^2}{2}\right]}_0}^r-{\left[\frac{r^3}{3R}\right]}_0^r\right\}\\ {\varphi }_E& =& \left(0.500\right)t.2\pi \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ & & \end{array}$

 Simplify further.

$\begin{array}{rcl}{\varphi }_E& =& t.\pi \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ \frac{{\varphi }_E}{t}& =& \pi \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\end{array}$

Then equation (1) can be written as,

$\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{s}& =& {\mu }_0{\mathcal{E}}_{0 } \pi \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B\left(2\pi r\right)& =& {\mu }_0{\mathcal{E}}_{0 } \pi \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2r} \left(\frac{r^2}{2}-\frac{r^3}{3R}\right)\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2} \left(\frac{r}{2}-\frac{r^2}{3R}\right)\\ & & \end{array}$

By substituting the given value, 

$\begin{array}{rcl}\mathrm{B}& =& \frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)}{2} \left(\frac{\mathrm{r}}{2}-\frac{{\mathrm{r}}^2}{3\mathrm{R}}\right)\\ \mathrm{B}& =& \frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)}{2} \left(\frac{0.02}{2}-\frac{{\left(0.02\right)}^2}{3\times 0.03}\right)\\ \mathrm{B}& =& 5.56\times {10}^{-18}\times \left(5.6\times {10}^{-3}\right)\\ \mathrm{B}& =& 3.09\times {10}^{-20} \mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a given radial distance is $B=3.09\times {10}^{-20} \mathrm{T}.$

Step 4: (b) Determining the magnitude of an induced magnetic field at a radial distance 5.00 cm

For the r > R in the above equation, the limit of integration is 0 to R, i.e.,

For $r=0.05 \mathrm{m}$,

$\begin{array}{rcl}{\varphi }_E& =& \int EdA\\ {\varphi }_E& =& {\int }_0^{R}EdA\\ {\varphi }_E& =& {\int }_0^{R}E\left(2\pi rdr\right)\end{array}$

By the given value,  

$\begin{array}{rcl}{\varphi }_E& =& {\int }_0^R\left(0.500\right)\left(1-\frac{r}{R}\right)t2\pi rdr\\ {\varphi }_E& =& \left(0.500\right)t2\pi {\int }_0^R\left(1-\frac{r}{R}\right)rdr\\ {\varphi }_E& =& \left(0.500\right)t.2\pi \left\{{{\left[\frac{r^2}{2}\right]}_0}^R-{\left[\frac{r^3}{3R}\right]}_0^R\right\}\\ {\varphi }_E& =& \left(0.500\right)t.2\pi \left(\frac{R^2}{2}-\frac{R^3}{3R}\right)\end{array}$

 Simplify further. 

$\begin{array}{rcl}{\varphi }_E& =& t.\pi \left(\frac{R^2}{2}-\frac{R^2}{3}\right)\\ \frac{{\varphi }_E}{t}& =& \pi \left(\frac{R^2}{6}\right)\end{array}$

Then equation (1) can be written as,

$\begin{array}{rcl}\oint \overrightarrow{B}·d\overrightarrow{s}& =& {\mu }_0{\mathcal{E}}_{0 } \pi \left(\frac{R^2}{6}\right)\\ B\left(2\pi r\right)& =& {\mu }_0{\mathcal{E}}_{0 } \pi \left(\frac{R^2}{6}\right)\\ B& =& \frac{{\mu }_0{\mathcal{E}}_{0 }}{2r} \left(\frac{R^2}{6}\right)\end{array}$

By substituting the given value, 

$\begin{array}{rcl}B& =& \frac{\left(4\pi \times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)}{2\times 0.05} \left(\frac{{0.03}^2}{6}\right)\\ B& =& \frac{\left(4\pi \times {10}^{-7}\right)\left(8.85\times {10}^{-12}\right)}{2} \left(1.5\times {10}^{-4}\right)\\ B& =& 1.12\times {10}^{-16}\times \left(1.5\times {10}^{-4}\right)\\ B& =& 1.67\times {10}^{-20} \mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a given radial distance is $\begin{array}{rcl}B& =& 1.67\times {10}^{-20} \mathrm{T}\end{array}$.

By using the relation between the non-uniform electric field and electric flux, the magnetic field for outside and inside the given circular region can be found.