Q7.60P - Introduction to Electrodynamics

Question

Question: Suppose $\mathit{j}\mathbf{\left(}\mathit{r}\mathbf{\right)}$is constant in time but $\mathit{\rho }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}$is not-conditions that 

might prevail, for instance, during the charging of a capacitor. 

(a) Show that the charge density at any particular point is a linear function of time: 

                                $\mathit{\rho }\left(r,t\right)\mathbf{=}\mathit{\rho }\left(r,0\right)\mathbf{+}\mathit{\rho }\left(r,0\right)\mathit{t}$           

where $\mathit{\rho }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{0}\mathbf{)}$ is the time derivative of at . [Hint: Use the continuity equation.] 

This is not an electrostatic or magnetostatic configuration: nevertheless, rather surprisingly, both Coulomb's law (Eq. 2.8) and the Biot-Savart law (Eq. 5.42) hold, as you can confirm by showing that they satisfy Maxwell's equations. In particular:

(b) Show that 

                          $\mathit{B}\left(r\right)\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{J}\left(r\text{'}\right)\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathit{\tau }\mathbf{\text{'}}$                      

obeys Ampere's law with Maxwell's displacement current term.

Step-by-Step Solution

Answer

Answer

(a) The required equation is proved that is $\rho \left(\mathrm{t}\right)=\rho (\mathrm{r},0)t+\rho \left(t\right)$

(b) The equation for Ampere’s law with Maxwell’s displacement current term is obtained that is $\nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$ .

Step 1: Write the given data from the question.

The charge density is $\rho \left(r,t\right)$ .

The current density is $j\left(r\right)$ .

Step 2: Determine the equation to calculate the equation for charge density and for magnetic field.

The expression for the continuity equation is given as follows.

 $\frac{\mathbf{\partial }\mathbf{\rho }}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{\times }\mathit{J}$

The expression for Ampere’s law with Maxwell’s displacement current is given as follows.

$\mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{J}\mathbf{+}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{t}}$

Here, B is the magnetic field, E is the electric field, ${\mu }_0$is the free space permeability and is the permittivity.

Step 3: Determine the equation for the charge density.

(a)

Consider the expression for the continuity equation.

                $\rho \left(t\right)=\left(-\nabla ·J\right)t+\text{constant}$                                       ……. (1)

Here the constant should be a function of but not the .

Therefore,  $\rho \left(r,0\right)$

Substitute  $\rho \left(r,0\right)$ for constant$\rho \left(r,0\right)$ and for data-custom-editor="chemistry" $-\nabla ·J$into equation (1).

$\rho \left(\mathrm{t}\right)=\rho (\mathrm{r},0)t+\rho \left(t\right)$

Hence the required equation is proved.

Step 4: Determine the equation for the magnetic field.

(b)

The equation of ampere’s law with the Maxwell’s displacement term is given by,

     $$\begin{gathered} \nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t} \\ \nabla \times B={\mu }_{0}J-\frac{{\mu }_0}{4\pi }\int J·\nabla \frac{\hat{r}}{r^2}d\tau \end{gathered}$$                                      …… (2)

 

Consider the condition,

 $-\left(J·\nabla \right)\frac{\hat{r}}{r^2}=\left(J·\nabla \right)\frac{\hat{r}}{r^2}d\tau$

Substitute$-\left(J·\nabla \right)\frac{\hat{r}}{r^2}$ for $-\left(J·\nabla \right)\frac{\hat{r}}{r^2}$into equation (2).

$$\begin{gathered} \nabla \times B={\mu }_{0}J+\frac{{\mu }_0}{4\pi }\int \left(J·\nabla \text{'}\right)\frac{\hat{r}}{r^2}d\tau \\ \nabla \times B={\mu }_{0}J-\frac{{\mu }_0}{4\pi }\int \left(J·\nabla \text{'}\right)\frac{\hat{r}}{r^2}d\tau \end{gathered}$$

Multiply and divide into the second terms of the right side in the above equation.

  $$\begin{gathered} \nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{1}{4\pi {\epsilon }_0}\frac{\partial }{\partial t}\int \frac{\rho \hat{r}}{r^2}d\tau \\ \nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\int \frac{1}{4\pi {\epsilon }_0}\frac{\rho \hat{r}}{r^2}d\tau \end{gathered}$$

The term $\int \frac{1}{4\pi {\epsilon }_0}\frac{\rho \hat{r}}{r^2}d\tau$ represent the electric filed into above equation.

 Therefore,

 $\nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$

Hence, the equation for Ampere’s law with Maxwell’s displacement current term is obtained.