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Found in: Page 967

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The figure shows an electric field is directed out of the page within a circular region of radius ${\mathbit{R}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{cm}}}$. The field magnitude is ${\mathbit{E}}{\mathbf{=}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{\text{V/ms}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{R}\mathbf{\right)}{\mathbit{t}}$, 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 ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{cm}}}$?

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

## 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:

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

Where, $\stackrel{\to }{B}$ 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,

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

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

localid="1663156293618" $\begin{array}{rcl}{\varphi }_{E}& =& \int EdA\\ {\varphi }_{E}& =& {\int }_{0}^{r}EdA\\ {\varphi }_{E}& =& {\int }_{0}^{r}E\left(2\pi rdr\right)\end{array}$

By the given value,

localid="1663156368547" $\begin{array}{rcl}{\varphi }_{E}& =& {\int }_{0}^{r}\left(0.500\right)\left(1-\frac{r}{R}\right)t2\pi rdr\\ {\varphi }_{E}& =& \left(0.500\right)t2\pi {\int }_{0}^{r}\left(1-\frac{r}{R}\right)rdr\\ {\varphi }_{E}& =& \left(0.500\right)t.2\pi \left\{{{\left[\frac{{r}^{2}}{2}\right]}_{0}}^{r}-{\left[\frac{{r}^{3}}{3R}\right]}_{0}^{r}\right\}\\ {\varphi }_{E}& =& \left(0.500\right)t.2\pi \left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\\ & & \end{array}$

Simplify further.

localid="1663156414306" $\begin{array}{rcl}{\varphi }_{E}& =& t.\pi \left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\\ \frac{{\varphi }_{E}}{t}& =& \pi \left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\end{array}$

Then equation (1) can be written as,

localid="1663156492531" $\begin{array}{rcl}\oint \stackrel{\to }{B}·d\stackrel{\to }{s}& =& {\mu }_{0}{\mathcal{E}}_{0}\pi \left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\\ B\left(2\pi r\right)& =& {\mu }_{0}{\mathcal{E}}_{0}\pi \left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}}{2r}\left(\frac{{r}^{2}}{2}-\frac{{r}^{3}}{3R}\right)\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}}{2}\left(\frac{r}{2}-\frac{{r}^{2}}{3R}\right)\\ & & \end{array}$

By substituting the given value,

localid="1663156562796" $\begin{array}{rcl}\mathrm{B}& =& \frac{\left(4\mathrm{\pi }×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(\frac{\mathrm{r}}{2}-\frac{{\mathrm{r}}^{2}}{3\mathrm{R}}\right)\\ \mathrm{B}& =& \frac{\left(4\mathrm{\pi }×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(\frac{0.02}{2}-\frac{{\left(0.02\right)}^{2}}{3×0.03}\right)\\ \mathrm{B}& =& 5.56×{10}^{-18}×\left(5.6×{10}^{-3}\right)\\ \mathrm{B}& =& 3.09×{10}^{-20}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a given radial distance is localid="1663156604486" $B=3.09×{10}^{-20}\mathrm{T}.$

## 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,

localid="1663157413775" $\begin{array}{rcl}{\varphi }_{E}& =& {\int }_{0}^{R}\left(0.500\right)\left(1-\frac{r}{R}\right)t2\pi rdr\\ {\varphi }_{E}& =& \left(0.500\right)t2\pi {\int }_{0}^{R}\left(1-\frac{r}{R}\right)rdr\\ {\varphi }_{E}& =& \left(0.500\right)t.2\pi \left\{{{\left[\frac{{r}^{2}}{2}\right]}_{0}}^{R}-{\left[\frac{{r}^{3}}{3R}\right]}_{0}^{R}\right\}\\ {\varphi }_{E}& =& \left(0.500\right)t.2\pi \left(\frac{{R}^{2}}{2}-\frac{{R}^{3}}{3R}\right)\end{array}$

Simplify further.

localid="1663157442965" $\begin{array}{rcl}{\varphi }_{E}& =& t.\pi \left(\frac{{R}^{2}}{2}-\frac{{R}^{2}}{3}\right)\\ \frac{{\varphi }_{E}}{t}& =& \pi \left(\frac{{R}^{2}}{6}\right)\end{array}$

Then equation (1) can be written as,

localid="1663157466171" $\begin{array}{rcl}\oint \stackrel{\to }{B}·d\stackrel{\to }{s}& =& {\mu }_{0}{\mathcal{E}}_{0}\pi \left(\frac{{R}^{2}}{6}\right)\\ B\left(2\pi r\right)& =& {\mu }_{0}{\mathcal{E}}_{0}\pi \left(\frac{{R}^{2}}{6}\right)\\ B& =& \frac{{\mu }_{0}{\mathcal{E}}_{0}}{2r}\left(\frac{{R}^{2}}{6}\right)\end{array}$

By substituting the given value,

localid="1663157500012" $\begin{array}{rcl}B& =& \frac{\left(4\pi ×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2×0.05}\left(\frac{{0.03}^{2}}{6}\right)\\ B& =& \frac{\left(4\pi ×{10}^{-7}\right)\left(8.85×{10}^{-12}\right)}{2}\left(1.5×{10}^{-4}\right)\\ B& =& 1.12×{10}^{-16}×\left(1.5×{10}^{-4}\right)\\ B& =& 1.67×{10}^{-20}\mathrm{T}\end{array}$

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

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.