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### Fundamentals Of Physics

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

# The figure 32-20 shows a circular region of radius${\mathbit{R}}{\mathbf{=}}{\mathbf{3}}{\mathbf{cm}}$ in which a displacement current is directed out of the page. The magnitude of the density of this displacement current is ${{\mathbit{J}}}_{{\mathbf{d}}}{\mathbf{=}}{\mathbf{\left(}}\mathbf{4}\mathbf{}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{\right)}}{\mathbf{\left(}\mathbf{1}\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{R}\mathbf{\right)}}$, where is the radial distance $\left(r\le R\right)$.(a) What is the magnitude of the magnetic field due to displacement current at ${\mathbf{2}}{\mathbf{cm}}$?(b) What is the magnitude of the magnetic field due to displacement current at ${\mathbf{5}}{\mathbf{cm}}$ ?

1. The magnitude of the magnetic field due to displacement current at a radial distance at$2\mathrm{cm}$ is $B=27.9\mathrm{nT}$.
2. The magnitude of the magnetic field due to displacement current at a radial distance at $5\mathrm{cm}$ is $B=15.1\mathrm{nT}$.
See the step by step solution

## The given data

1. Displacement current density ${J}_{d}=\left(4\mathrm{A}/{\mathrm{m}}^{2}}{\left(1-\mathrm{r}/\mathrm{R}\right)}\right)$,
2. The radius of the circular region $R=3\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}=3×{10}^{-2}\mathrm{m}$,
3. Radial distances at which the magnetic field is induced ${r}_{1}=2\mathrm{cm}x\frac{1}{100\mathrm{m}}=0.02\mathrm{m}$ ${r}_{2}=5\mathrm{cm}x\frac{1}{100\mathrm{m}}=0.05\mathrm{m}$,

## Understanding the concept of induced magnetic field

When a conductor is placed in a region of changing magnetic field, it induces a displacement current that starts flowing through it as it causes the case of an electric field produced in the conductor region. According to Lenz law, the current flows through the conductor such that it opposes the change in magnetic flux through the area enclosed by the loop or the conductor. The magnitude of the magnetic field is due to the displacement current using the displacement current density which is non-uniformly distributed.

Formulae:

The magnetic field at a point inside the capacitor,$B=\left(\frac{{\mu }_{0}{i}_{d}r}{2\pi {R}^{2}}\right)$ (i)

where, $B$ is the magnetic field, ${\mu }_{0}=\left(4\pi x{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)$ is the magnetic permittivity constant, $r$ is the radial distance, ${i}_{d}$ is the displacement current, $R$ is the radius of the circular region.

The magnetic field at a point outside the capacitor, $B=\left(\frac{{\mu }_{0}{i}_{d}}{2\pi r}\right)$ (ii)

Where, ${\mu }_{0}=\left(4\pi x{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)$ is the magnetic permittivity constant , $r$ is the radial distance, ${i}_{d}$ is the displacement current.

The current flowing in a given region for non-uniform electric field, ${i}_{d,enc}={\int }_{0}^{r}J\left(2\pi r\right)dr$ (iii)

Where, $J$ is the current density of the material, $r$ is the radial distance of the circular region, $dr$ is the differential form of the radial distance.

## (a) Determining the magnitude of the magnetic field due to displacement current at a radial distance R=2cm.

The displacement current density is non-uniform. Hence, the displacement current is determined by taking the integration over the closed path of radius $r,{r}_{1}=0.02\mathrm{m},{\mathrm{r}}_{1}<\mathrm{R}$ and that is given using the given data in equation (i) as follows:

${i}_{d,enc}={\int }_{0}^{r}\left(4.00\mathrm{A}/{\mathrm{m}}^{2}\right)\left(1-r/R\right)\left(2\pi r\right)dr\phantom{\rule{0ex}{0ex}}{i}_{d,enc}=\left(8\pi \mathrm{A}/{\mathrm{m}}^{2}\right){\int }_{0}^{\mathrm{r}}\left(\mathrm{r}-{\mathrm{r}}^{2}/\mathrm{R}\right)\mathrm{dr}\phantom{\rule{0ex}{0ex}}{\mathrm{i}}_{\mathrm{d},\mathrm{enc}}=\left(8\mathrm{\pi A}/{\mathrm{m}}^{2}\right)\left(\frac{{\mathrm{r}}^{2}}{2}-\frac{{\mathrm{r}}^{3}}{3\mathrm{R}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{i}}_{\mathrm{d},\mathrm{enc}}=\left(8\mathrm{\pi A}/{\mathrm{m}}^{2}\right)\left(\frac{{\left(0.02\mathrm{m}\right)}^{2}}{2}-\frac{{\left(0.02\mathrm{m}\right)}^{3}}{3×0.03\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{i}}_{\mathrm{d},\mathrm{enc}}=2.79×{10}^{-3}\mathrm{A}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ …………………………….. (I)

The integral is limited to . Hence, by taking in equation (i), the magnetic field can be determined as follows:

$B=\frac{{\mu }_{0}{i}_{d}{r}_{1}}{2\pi {r}_{1}^{2}}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{0}{i}_{d}}{2\pi {r}_{1}}\phantom{\rule{0ex}{0ex}}B=\frac{\left(4\pi ×{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(2.79×{10}^{-3}\mathrm{A}\right)}{2\pi ×\left(0.02\mathrm{m}\right)}\phantom{\rule{0ex}{0ex}}B=2.79×{10}^{-7}\mathrm{T}\phantom{\rule{0ex}{0ex}}\mathrm{B}=2.79\mathrm{nT}$

Therefore, the magnitude of the magnetic field due to displacement current at a radial distance $R=2.0\mathrm{cm}$ is$B=27.9\mathrm{nT}$ .