Q25P - Fundamentals Of Physics

Question

As a parallel-plate capacitor with circular plates $\mathbf{20}\mathbf{ }\mathbf{\text{cm}}$ in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of $\mathbf{20}\mathbf{ }\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. (a) Calculate the magnitude $\mathit{B}$ of the magnetic field at a distance  $\mathbf{r}\mathbf{=}\mathbf{50}\mathbf{ }\mathbf{mm}$ from the axis of symmetry of this region. (b) Calculate $\mathbf{dE}\mathbf{/}\mathbf{dt}$  in this region.

Step-by-Step Solution

Answer

(a) The magnitude of the magnetic field at distance $r=50 \mathrm{mm}$  is $6.3\times {10}^{-7} \mathrm{T}$.

(b) The rate of change of electric field $\frac{dE}{dT}$ is $2.3\times {10}^{12} \mathrm{V}/\mathrm{m}.\mathrm{s}$.

The given data

a) Diameter of circular plates, $d = 20 \mathrm{cm}$

b) Displacement current density,  $J_d=20{\text{ A/m}}^2$

c) Radial distance where the field is calculated from the axis of symmetry,  $r=50\text{mm}\times \frac{1\text{ m}}{{\text{10}}^{-3}\text{ mm}}=0.05\text{ m}$

Understanding the concept of Ampere’s law

Ampere's law states that an induced magnetic field due to the flow of an electric current in the given current region is directly proportional to the size of the current and constant, that is, the permeability of free space or the magnetic permittivity constant. Again, Ampere's circuital law states that the line integral of the magnetic field within a closed region is the sum of the algebraic sum of the current presence in the loop. 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 current density of the wire,  $J = i/A$ ......(i)

where, $i$ is the amount of the current flow through the wire, $A$ is the cross-sectional area of the wire.

The magnetic flux equation according to Ampere’s law, $\oint B.ds={\mu }_{0}i$...... (ii)

where, $i$  is current, $B$  is magnetic induction, $dS$  is the area of the enclosed region ${\mu }_0=\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)$ is the magnetic permittivity constant.

The displacement current due to the change in the electric field,

$i_d=\frac{{\epsilon }_{0}d{\varphi }_E}{dt}$ ......(iii)

where, ${\mathcal{E}}_0=\left(8.85\times {10}^{-12}{\text{ C}}^2/{\text{N.m}}^2\right)$ is the permittivity in a vacuum, $\frac{d{\varphi }_E}{dt}$ is the rate of change in the electric flux within the enclosed region.

 

(a) Determining the magnitude of the magnetic field at distance r = 50   mm .

Considering an Amperian loop of radius equal to the given radial distance $r=0.05 \mathrm{m}$, the magnitude of the magnetic field within the enclosed region can be given using the given values in equation (ii) as follows:

$$\begin{gathered} B\left(2\pi r\right)={\mu }_0{i}_{enc} \\ B=\frac{{\mu }_0{i}_{enc}}{2\pi r} \end{gathered}$$......(I)

Now, the value of the displacement current within this region can be given using equation (i) as follows:

$$\begin{gathered} i_{enc}\left(=i_d\right)=A{J}_d \\ =\left(\pi r^2{J}_d\right) \end{gathered}$$......(II)

Now, using the above value of current and the given data in equation (I), the value of the magnetic field can be given as follows:

  $$\begin{gathered} B=\frac{{\mu }_0\pi r^2{J}_d}{2\pi r} \\ =\frac{{\mu }_0\pi r{J}_d}{2\pi } \\ =\frac{\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)\pi \left(50\times {10}^{-3}\text{m}\right)\left(20{\text{ A/m}}^2\right)}{2\pi } \\ =6.3\times {10}^{-7}\text{T} \end{gathered}$$

Therefore, the magnitude of the magnetic field at distance, $r=0.05\text{ m}$ is $B=6.3\times {10}^{-7}\text{T}$.

(b) Determining the rate of change of electric field, d E d T

The electric flux flowing through a region can be given as:

$\begin{array}{rcl}d{\varphi }_E& =& EA\\ & =& E\left(\pi r^2\right)\end{array}$

The value of the displacement current using the above value in equation (iii) becomes:

 $i_d={\epsilon }_0\pi r^2\frac{dE}{dt}$......(III)

When displacement current is uniform, the current density is $J_d$, thus the current is given using equation (II).

Now, comparing equations (II) and (III), the value of the rate of electric field can be given as follows:

 $$\begin{gathered} d{\epsilon }_0\pi r^2\frac{dE}{dt}=J_d\pi r^2 \\ \frac{dE}{dt}=\frac{J_d}{{\epsilon }_0} \\ =\frac{20{\text{ A/m}}^2}{\left(8.85\times {10}^{-12}{\text{ C}}^2/{\text{N.m}}^2\right)} \\ =2.3\times {10}^{12}\text{ V/m.s} \end{gathered}$$

Hence, the rate of change of the electric field is $2.3\times {10}^{12}\text{ V/m.s}$.