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Found in: Page 766

### Fundamentals Of Physics

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

# The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by ${\mathbit{J}}{\mathbf{=}}\left(3.00×{10}^{8}\right){{\mathbit{r}}}^{{\mathbf{2}}}$ , with J in amperes per square meter and radial distance r in meters. What is the current through the outer section bounded by r = 0.900 R and r = R ?

Current through the outer section bounded by radial distance r = 0.900 R and r = R is $2.59×{10}^{-3}A$ .

See the step by step solution

## Step 1: The given data

a) Current density, $J\left(r\right)=\left(3×{10}^{8}\right){r}^{2}$

b) Radius of the wire,$R=2mm$

c) Bounded radial distances, $r=0.9R$ and $r=R$

## Step 2: Understanding the concept of the flow of current

The current is the rate of flow of charges per unit of time. The current density is the current per unit cross-section area of the rate of flow of charges per unit time per unit area.

We can use the relation between the current and the current density to find the current through the outer section bounded between the two radii.

Formulae:

The equation of the current flowing through a small area, $i=\int \stackrel{\to }{J}.\stackrel{\to }{dA}$ ...(i)

Here, i is current, $\stackrel{\to }{J}$ is current density, $\stackrel{\to }{dA}$ is the area of cross-section.

The cross-sectional area of the circle, $A={\mathrm{\pi r}}^{2}$ ...(ii)

## Step 3: Calculation of the current through the bounded outer section

We have, the given value of the current density as: $J\left(r\right)=\left(3×{10}^{8}\right){r}^{2}$

The differential cross-sectional area value using equation (ii) can be given as follows:

$dA=2\mathrm{\pi r}\mathrm{dr}$

Substituting these above values in the equation (i), we can get the contained current within the width of the concentric ring assuming is directed along the wire for the radial distance varying from $r=0.900R$ to r = R as follows:

$i=\int \left(3×{10}^{8}\right){r}^{2}2\mathrm{\pi rdr}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }\left(3×{10}^{8}\right){\int }_{\mathrm{r}=0.900}^{\mathrm{r}=\mathrm{R}}{\mathrm{r}}^{3}\mathrm{dr}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }\left(3×{10}^{8}\right){\left[\frac{{\mathrm{r}}^{4}}{4}\right]}_{0.900\mathrm{R}}^{\mathrm{R}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }\left(3×{10}^{8}\right)}{2}{\left[{\mathrm{r}}^{4}\right]}_{0.900\mathrm{R}}^{\mathrm{R}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(3×{10}^{8}\right)\left[{\mathrm{R}}^{4}-{\left(0.900\mathrm{R}\right)}^{4}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(3×{10}^{8}\right)\left[{\mathrm{R}}^{4}-0.6561{\mathrm{R}}^{4}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(3×{10}^{8}\right)\left[0.3439\right]{\mathrm{R}}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(3×{10}^{8}\right)\left[0.3439\right]{\left(0.002\right)}^{4}\mathrm{A}\left(\because \mathrm{R}=0.002\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}=2.59×{10}^{-3}\mathrm{A}$

Hence, the value of the current is $2.59×{10}^{-3}\mathrm{A}$ .