Q61P - Introduction to Electrodynamics

Question

Problem 7.61 The magnetic field of an infinite straight wire carrying a steady current I can be obtained from the displacement current term in the Ampere/Maxwell law, as follows: Picture the current as consisting of a uniform line charge $\mathit{\lambda }$ moving along the z axis at speed v (so that $\mathbf{I}\mathbf{=}\mathbf{\lambda v}$), with a tiny gap of length E , which reaches the origin at time $\mathbf{t}\mathbf{=}\mathbf{0}$. In the next instant (up to $\mathbf{t}\mathbf{=}\mathbf{E}\mathbf{/}\mathbf{v}$) there is no real current passing through a circular Amperian loop in the xy plane, but there is a displacement current, due to the "missing" charge in the gap.

(a) Use Coulomb's law to calculate the z component of the electric field, for points in the xy plane a distances from the origin, due to a segment of wire with uniform density -$\mathit{\lambda }$ . extending from to ${\mathbf{z}}_{\mathbf{1}}\mathbf{=}\mathbf{vt}\mathbf{-}\mathbf{E}$ to ${\mathbf{z}}_{\mathbf{2}}\mathbf{=}\mathbf{vt}$ . 

(b) Determine the flux of this electric field through a circle of radius a in the xy plane. 

(c) Find the displacement current through this circle. Show that ${\mathbf{I}}_{\mathbf{d}}$ is equal to I , in the limit as the gap width $\left(E\right)$ goes to zero.${}^{\mathbf{35}}$

Step-by-Step Solution

Answer

(a) The value of electric field due to small element parallel to z-axis is $E_2=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right\}$.

(b) The value of flux of this electric field through a circle of radius a in the xy plane is ${\varphi }_E=\frac{\lambda }{2{\epsilon }_0}\left[\sqrt{{\left(vt+\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\left(\epsilon -vt\right)+\left(vt\right)\right]$.

(c) The value of displacement current $I_d$ through this circle is $I_d=I$.

Step 1: Write the given data from the question.

The value of current as consisting of a uniform line charge $\lambda$ moving along the z axis at speed $v$ is $I=\lambda v$.

The value of time with a tiny gap of lengthE, which reaches the origin at time $t=0$

$t=E/v.$

Step 2: Determine the formula for component of electric field, flux of this electric field through a circle of radius a in the xy plane and displacement current I d .

Write the formula of electric field due to small element parallel to z-axis.

 

${\mathbf{dE}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{-}\mathbf{\lambda dz}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{sin\theta }$                                                                   …… (1)

Here, $\mathit{\lambda }$ is uniform density, $\mathbf{r}$ is radius and ${\mathbf{\epsilon }}_{\mathbf{0}}$ is permittivity. 

Write the formula of flux of this electric field through a circle of radius a in the plane.

${\mathbf{\varphi }}_{\mathbf{E}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{a}}{\mathbf{E}}_{\mathbf{2}}\mathbf{.}\mathbf{da}$                                                                                      …… (2)

Here, ${\mathbf{E}}_{\mathbf{2}}$ is electric charge and da is the radius of the circle.

Write the formula of displacement current through this circle.

${\mathbf{I}}_{\mathbf{d}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\frac{{\mathbf{d\varphi }}_{\mathbf{E}}}{\mathbf{dt}}$                                                                                        …… (3)

Here, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is permittivity, $\mathit{\varphi }$ is flux of this electric field in a circle.

Step 3: (a) Determine the value of electric field due to small element parallel to z-axis.

Draw the figure of given provide information.

                           

Here, a wire with a uniform line charge of $\lambda$ is travelling down the z-axis at a speed of v .

Then current  $I=zv$

Determine the electric field component resulting from a small element parallel to the z-axis is

Substitute $\frac{z}{r}$ for $\sin \theta$ and $\sqrt{z^2+s^2}$ for r  into equation (1).

$$\begin{gathered} E_2=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\int \frac{zdz}{{\left(z^2+s^2\right)}^{\frac{3}{2}}} \\ =\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}{\left[\frac{-1}{\left(z^2+s^2\right)}\right]}_{vt-\epsilon }^{vt} \\ =\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right\} \end{gathered}$$

Therefore, the value of electric field due to small element parallel to -axis is $E_2=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right\}$.

Step 4: (b) Determine the value of flux of this electric field through a circle of radius a in the xy plane.

Determine the flux due to $E_2$ through a circle of radius 'a' in xy- plane is

Substitute $\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right]$ for $E_2$ and $\left(2\mathrm{\pi s}\right)$ ds  for da  into equation (2).

 

$$\begin{gathered} {\varphi }_E=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}{\int }_0^a\left[\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right]2\mathrm{\pi }s ds \\ =\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}{\left[\sqrt{{\left(vt+\epsilon \right)}^2+s^2}-\sqrt{{\left(vt\right)}^2+s^2}\right]}_0^a \\ =\frac{\lambda }{2{\epsilon }_0}\left[\sqrt{{\left(vt+\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\sqrt{{\left(\epsilon -vt\right)}^2+0+\sqrt{{\left(vt\right)}^2+0}}\right] \\ =\frac{\lambda }{2{\epsilon }_0}\left[\sqrt{{\left(vt-\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\left(\epsilon -vt\right)+\left(vt\right)\right] \end{gathered}$$

Therefore, the value of flux of this electric field through a circle of radius a in the xy plane is ${\varphi }_E=\frac{\lambda }{2{\epsilon }_0}\left[\sqrt{{\left(vt-\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\left(\epsilon -vt\right)+\left(vt\right)\right]$ .

Step 5: (c) Determine the value of displacement current I d through this circle.

Determine the displacement current $I_d$ .

Substitute $\left[\frac{\lambda }{2{\epsilon }_0}\left(\sqrt{{\left(vt-\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\left(\epsilon -vt\right)+\left(vt\right)\right)\right]$ for ${\varphi }_E$ into equation (3).

$$\begin{gathered} I_d={\epsilon }_0\frac{d}{dt}\left[\frac{\lambda }{2{\epsilon }_0}\left(\sqrt{{\left(vt-\epsilon \right)}^2+a^2}-\sqrt{{\left(vt\right)}^2+a^2}-\left(\epsilon -vt\right)+\left(vt\right)\right)\right] \\ =\frac{\lambda }{2}\left\{\frac{v\left(vt-\epsilon \right)}{{\left(vt-\epsilon \right)}^2+a^2}-\frac{v\left(vt\right)}{{\left(vt\right)}^2+a^2}+2v\right\} \end{gathered}$$ 

As, $\epsilon \to 0$ then $vt<\epsilon$ then  $vt\to 0$

Then, 

 

$$\begin{gathered} I_d=\frac{\lambda }{2}\left(2v\right) \\ =\lambda v \\ =I \end{gathered}$$

Therefore, the value of displacement current $I_d$ through this circle is $I_d=I$ .