Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The figure shows an electric field is directed out of the page within a circular region of radius $R = 3.00 cm$ . The field magnitude is $E = (0.500 V/ms) (1 - r / R) t$ , where t is in seconds and r is the radial distance $(r \leq R)$ . What is the magnitude of the induced magnetic field at a radial distance $2.00 cm$ ? What is the magnitude of the induced magnetic field at a radial distance $5.00 cm$ ?

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

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

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:

$\oint \vec{B} \cdot d \vec{s} = μ_{0} E_{0} \frac{d ϕ}{d t}$

Where, $\vec{B}$ is the magnetic field, $ϕ$ 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,

$\oint \vec{B} \cdot d \vec{s} = μ_{0} E_{0} \frac{d ϕ_{E}}{d t} \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots (1)$

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

localid="1663156293618" $\begin{array}{rcl} ϕ_{E} & = & \int E d A \\ ϕ_{E} & = & \int_{0}^{r} E d A \\ ϕ_{E} & = & \int_{0}^{r} E (2 π r d r) \end{array}$

By the given value,

localid="1663156368547" $\begin{array}{rcl} ϕ_{E} & = & \int_{0}^{r} (0.500) (1 - \frac{r}{R}) t 2 π r d r \\ ϕ_{E} & = & (0.500) t 2 π \int_{0}^{r} (1 - \frac{r}{R}) r d r \\ ϕ_{E} & = & (0.500) t . 2 π \{{[\frac{r^{2}}{2}]}_{0}^{r} - {[\frac{r^{3}}{3 R}]}_{0}^{r}\} \\ ϕ_{E} & = & (0.500) t . 2 π (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \end{array}$

Simplify further.

localid="1663156414306" $\begin{array}{rcl} ϕ_{E} & = & t . π (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \\ \frac{ϕ_{E}}{t} & = & π (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \end{array}$

Then equation (1) can be written as,

localid="1663156492531" $\begin{array}{rcl} \oint \vec{B} \cdot d \vec{s} & = & μ_{0} E_{0} π (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \\ B (2 π r) & = & μ_{0} E_{0} π (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \\ B & = & \frac{μ_{0} E_{0}}{2 r} (\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) \\ B & = & \frac{μ_{0} E_{0}}{2} (\frac{r}{2} - \frac{r^{2}}{3 R}) \end{array}$

By substituting the given value,

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

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

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 $r = 0.05 m$ ,

$\begin{array}{rcl} ϕ_{E} & = & \int E d A \\ ϕ_{E} & = & \int_{0}^{R} E d A \\ ϕ_{E} & = & \int_{0}^{R} E (2 π r d r) \end{array}$

By the given value,

localid="1663157413775" $\begin{array}{rcl} ϕ_{E} & = & \int_{0}^{R} (0.500) (1 - \frac{r}{R}) t 2 π r d r \\ ϕ_{E} & = & (0.500) t 2 π \int_{0}^{R} (1 - \frac{r}{R}) r d r \\ ϕ_{E} & = & (0.500) t . 2 π \{{[\frac{r^{2}}{2}]}_{0}^{R} - {[\frac{r^{3}}{3 R}]}_{0}^{R}\} \\ ϕ_{E} & = & (0.500) t . 2 π (\frac{R^{2}}{2} - \frac{R^{3}}{3 R}) \end{array}$

Simplify further.

localid="1663157442965" $\begin{array}{rcl} ϕ_{E} & = & t . π (\frac{R^{2}}{2} - \frac{R^{2}}{3}) \\ \frac{ϕ_{E}}{t} & = & π (\frac{R^{2}}{6}) \end{array}$

Then equation (1) can be written as,

localid="1663157466171" $\begin{array}{rcl} \oint \vec{B} \cdot d \vec{s} & = & μ_{0} E_{0} π (\frac{R^{2}}{6}) \\ B (2 π r) & = & μ_{0} E_{0} π (\frac{R^{2}}{6}) \\ B & = & \frac{μ_{0} E_{0}}{2 r} (\frac{R^{2}}{6}) \end{array}$

By substituting the given value,

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

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

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.

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Step 1: Given

Step 2: Determining the concept

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

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

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

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