Question

Uniform electric flux.Figure 32-30shows a uniform 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 given by, $\mathit{E}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{\text{V/ms}}\mathbf{)}\mathit{t}$where t is in seconds. (a)What is the magnitude of the induced magnetic field at a radial distance 2.00 cm? (b) What is the magnitude of the induced magnetic field at aradial distance 5.00 cm?

Short answer

1. The magnitude of an induced magnetic field at a given radial distance is$B=5.01\times {10}^{-22}\mathrm{T}.$
2. The magnitude of an induced magnetic field at a given radial distance is$B=4.51\times {10}^{-22}\mathrm{T}.$

Step-by-step solution

Step 1: The given data

1. The radius of the circular region,$R=3.00\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.03\mathrm{m}$
2. The magnitude of the electric field,$E=\left(4.50\times {10}^{-3}\mathrm{V}/\mathrm{m}.\mathrm{s}\right)t$
3. Radial distances at which the magnetic field is induced,$r_1=2\mathrm{cm}\times \frac{1}{100\mathrm{m}}=0.02\mathrm{m}$and$r_2=5\mathrm{cm}\times \frac{1}{100\mathrm{m}}=0.05\mathrm{m}$.

Understanding the concept of the magnetic field

Maxwell's equations describe the wave propagation of a magnetic field induced due to the changing electric flux in addition to Ampere's law. When a body is placed inside a changing electric field, the magnetic field induced is considered using an Amperian loop such that within the loop the magnetic field flux is equal to the rate of change of electric flux within the region. For the regions with radial distances smaller than the radius of the governing region, an Amperian loop of radius equal to radial distance is considered, while for a radial distance equal to or more than the radius the electric flux change is similar to that change in the region.

Formulae:

The Maxwell equation for the magnetic field,

$\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_0\frac{d\varphi }{dt}$ (i)

Where, $\overrightarrow{B}$is the magnetic field, $\varphi$is the magnetic flux, $d\overrightarrow{s}$is the length vector, ${\mathcal{E}}_0=\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)$is the permittivity in a vacuum, ${\mu }_0=\left(4\pi \times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)$is the magnetic permittivity constant.

The amount of electric flux passing through the region, $\varphi =EA$ (ii)

Where, E is the magnitude of the electric field, A is the area of the region.

(a) Determining the magnitude of an induced magnetic field at a radial distance.

The given radial distance $r=0.02\mathrm{m}$is less than the radius of the region, thus substituting the value of electric flux of equation (ii) in equation (i), the Maxwell equation of the magnetic field can be given as follows: (For $r_1<R$)

localid="1663152291274" $\begin{array}{c}\oint \overrightarrow{B}·d\overrightarrow{s}={\mu }_0{\mathcal{E}}_{0}A\frac{dE}{dt}\\ B\left(2\pi r_1\right)={\mu }_0{\mathcal{E}}_0\pi {r_1}^2\frac{dE}{dt}\left(\mathrm{Areaoftheameperianloop},A=\pi r^2\right)\\ B=\frac{{\mu }_0{\mathcal{E}}_0}{2}{r}_1\frac{dE}{dt}......\left(\text{iii}\right)\end{array}$

Now, using the electric field value, the rate of the electric field can be given as:

$\frac{E}{t}=\left(4.50\times {10}^{-3}\mathrm{V}/\mathrm{m}.\mathrm{s}\right)$

Using the above value and the given data in equation (iii), the magnitude of the electric field can be given as follows:

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)}{2}{\left(0.02\mathrm{m}\right)}^2\left(4.50\times {10}^{-3}\mathrm{V}/\mathrm{m}.\mathrm{s}\right)\\ =5.01\times {10}^{-22}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a radial distance localid="1663152377692" width="62" style="max-width: none;" $2.00\mathrm{cm}$is$B=5.01\times {10}^{-22}\mathrm{T}.$

(b) Determine the magnitude of an induced magnetic field at a radial distance.

For the $r_2>R,$the above equation can be changed only in the Amperian loop for the area value $A=\pi R^2$so that, the Maxwell equation for the magnetic field can be given using equation (i) as follows:

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

where, A is the area enclosed by the Amperian loop, which is $A=\pi r^2$.

$\begin{array}{c}B\left(2\pi r_2\right)={\mu }_0{\mathcal{E}}_0\pi R^2\frac{dE}{dt}\\ B=\frac{{\mu }_0{\mathcal{E}}_0}{2r_2}{R}^2\frac{dE}{dt}\\ =\frac{\left(4\pi \times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)}{2\times \left(0.02\mathrm{m}\right)}{\left(0.03\mathrm{m}\right)}^2\left(4.50\times {10}^{-3}\mathrm{V}/\mathrm{m}.\mathrm{s}\right)\\ \left(\mathrm{Frompart}\left(\mathrm{a}\right),\frac{E}{t}=\left(4.50\times {10}^{-3}\mathrm{V}/\mathrm{m}.\mathrm{s}\right)\right)\\ =4.51\times {10}^{-22}\mathrm{T}\end{array}$

Therefore, the magnitude of an induced magnetic field at a radial distance $5.00\mathrm{cm}$is$B=4.51\times {10}^{-22}\mathrm{T}.$