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.${}^{\mathbf{35}}$

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 $t=0$

localid="1657531425699" $t=E/v.$

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, role="math" localid="1657530384359" $\mathbf{r}$is radius and${\mathbf{\epsilon }}_{\mathbf{0}}$ is permittivity.

Write the formula offlux 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 ofdisplacement 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.

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 localid="1657531329091" $E_2=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{\sqrt{{\left(vt+\epsilon \right)}^2+s^2}}-\frac{1}{\sqrt{{\left(vt\right)}^2+s^2}}\right\}$.

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).

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

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

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$I_d=I$ .