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A long straight cable of length $l$ is placed symmetrically along z-axis and has radius $a(< < l)$. The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I(t)=I_0 \sin (2 \pi v t)$ flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is
$$
E(s, t)=\mu_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{k} \text {. }
$$
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current $\mathrm{I}^{\mathrm{d}}$.
(iii) Compare the conduction current $I_0$ with the displacement current $\mathrm{I}_0^{\mathrm{d}}$
$$
E(s, t)=\mu_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{k} \text {. }
$$
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current $\mathrm{I}^{\mathrm{d}}$.
(iii) Compare the conduction current $I_0$ with the displacement current $\mathrm{I}_0^{\mathrm{d}}$
Solution:
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Verified Answer
We know that, Displacement current density
$$
J_d=\varepsilon_0 \frac{d E}{d t}
$$
(i) The induced electric field $\mathrm{E}(\mathrm{s}, \mathrm{t})$ at a distance $(\mathrm{s} < $ radius of co-axial cable) from the wire inside the cable is given as :
$$
E(s, t)=\mu_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{k}
$$
Now, displacement current density, $\left(\mathrm{J}_{\mathrm{d}}\right)$ is given by,
$$
\begin{aligned}
&\mathrm{J}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{dE}}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{d}}{\mathrm{dt}}\left[\mu_0 \mathrm{I}_0 \mathrm{v} \cos (2 \pi v \mathrm{t}) \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\right] \\
&=\varepsilon_0 \mu_0 \mathrm{I}_0 \mathrm{v} \frac{\mathrm{d}}{\mathrm{dt}}[\cos 2 \pi \mathrm{vt}] \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\left[\because \mathrm{c}^2=\frac{1}{\mu_0 \varepsilon_0}\right] \\
&=\frac{1}{\mathrm{c}^2} \mathrm{I}_0 \mathrm{v}^2 2 \pi[-\sin 2 \pi v \mathrm{t}] \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\left[\because l_4 \frac{\mathrm{s}}{\mathrm{a}}=-l_4 \frac{\mathrm{a}}{\mathrm{s}}\right] \\
&=+\frac{\mathrm{v}^2}{\mathrm{c}^2} 2 \pi \mathrm{I}_0 \sin 2 \pi v t \ln \left(\frac{\mathrm{a}}{\mathrm{S}}\right) \hat{\mathrm{k}} \quad\left[\because \lambda=\frac{\mathrm{C}^2}{\mathrm{~V}^2}\right] \\
&=\frac{1}{\lambda^2} 2 \pi \mathrm{I}_0 \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \sin 2 \pi v t \hat{\mathrm{k}} \\
&\mathrm{J}_{\mathrm{d}}=\frac{2 \pi \mathrm{I}_0}{\lambda^2} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
(ii) $I_d=\int J_d s d s d \theta=\int^a J_d \operatorname{sds} \int^{2 \pi} d \theta=\int^a J_d \operatorname{sds} \times[2 \pi]$
$$
\begin{aligned}
&=\int_{s=0}^a\left[\frac{2 \pi}{\lambda^2} I_0 \log _e\left(\frac{\mathrm{a}}{\mathrm{s}}\right) \operatorname{sds} \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}\right] \times[2 \pi] \\
&=\left(\frac{2 \pi}{\lambda}\right)^2\left(\mathrm{I}_0 \int_{\mathrm{s}=0}^{\mathrm{a}} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \mathrm{sds}\right) \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0\left[\int_{\mathrm{s}=0}^{\mathrm{a}} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \frac{1}{2} \mathrm{~d}\left(\mathrm{~s}^2\right)\right] \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}} \\
&=\frac{\mathrm{a}^2}{\lambda}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0[\sin 2 \pi v t \hat{\mathrm{k}}]\left[\int_{\mathrm{s}=0}^{\mathrm{a}}\left[\ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \cdot \mathrm{d}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)^2\right] \mathrm{ds}\right] \\
&=\frac{\mathrm{a}^2}{2}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi v t \hat{\mathrm{k}} \times\left(\frac{-\mathrm{a}}{4}\right) \\
&\therefore \quad\left[\because \int_{\mathrm{s}=0}^{\mathrm{a}}\left[\ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \mathrm{d}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)^2\right] \mathrm{ds}=\frac{-\mathrm{a}}{4}\right] \\
&\quad \mathrm{I}_{\mathrm{d}}=\frac{-\mathrm{a}^3}{8}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi v t(\hat{\mathrm{k}}) \\
&=\frac{-\mathrm{a}}{2}-\frac{\mathrm{a}^2}{4}\left[\frac{2 \pi}{\lambda}\right]^2 \mathrm{I}_0 \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
The negative sign shows that the displacement current $\mathrm{I}_{\mathrm{d}}$ is oposite to the conduction current $\mathrm{I}_{\mathrm{c}}$
So, $\frac{\mathrm{a}}{2}\left(\frac{2 \pi \mathrm{a}}{2 \lambda}\right)^4 \mathrm{I}_0 \sin 2 \pi v t(\hat{\mathrm{k}})$
$\mathrm{I}_{\mathrm{d}}$ is in z-direction as $\mathrm{I}_{\mathrm{c}}$ is in $+\mathrm{z}$ direction.
(iii) The displacement current.
$$
\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{a}}{2}\left(\frac{\pi \mathrm{a}}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi \mathrm{vt}=\mathrm{I}_0^{\mathrm{d}} \sin 2 \pi \mathrm{vt}
$$
So, $\mathrm{I}_0^{\mathrm{d}}=\frac{\mathrm{a}}{2}\left(\frac{\pi \mathrm{a}}{\lambda}\right)^2 \mathrm{I}_0=\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2 \mathrm{I}_0 \times\left(\frac{\mathrm{a}}{2}\right)$
$$
\therefore \quad \frac{\mathrm{I}_0^{\mathrm{d}}}{\mathrm{I}_0}=\left[\frac{\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2 \mathrm{I}_0 \times\left(\frac{\mathrm{a}}{2}\right)}{\mathrm{I}_0}\right]=\frac{\mathrm{a}}{2}\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2=\frac{\mathrm{a}^3 \pi^2}{2 \lambda^2}
$$
So, the required ratio is $\frac{\mathrm{I}_0^{\mathrm{d}}}{\mathrm{I}_0}=\frac{\mathrm{a}^3 \pi^2}{2 \lambda^2}$
$$
J_d=\varepsilon_0 \frac{d E}{d t}
$$
(i) The induced electric field $\mathrm{E}(\mathrm{s}, \mathrm{t})$ at a distance $(\mathrm{s} < $ radius of co-axial cable) from the wire inside the cable is given as :
$$
E(s, t)=\mu_0 I_0 v \cos (2 \pi v t) \ln \left(\frac{s}{a}\right) \hat{k}
$$
Now, displacement current density, $\left(\mathrm{J}_{\mathrm{d}}\right)$ is given by,
$$
\begin{aligned}
&\mathrm{J}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{dE}}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{d}}{\mathrm{dt}}\left[\mu_0 \mathrm{I}_0 \mathrm{v} \cos (2 \pi v \mathrm{t}) \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\right] \\
&=\varepsilon_0 \mu_0 \mathrm{I}_0 \mathrm{v} \frac{\mathrm{d}}{\mathrm{dt}}[\cos 2 \pi \mathrm{vt}] \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\left[\because \mathrm{c}^2=\frac{1}{\mu_0 \varepsilon_0}\right] \\
&=\frac{1}{\mathrm{c}^2} \mathrm{I}_0 \mathrm{v}^2 2 \pi[-\sin 2 \pi v \mathrm{t}] \ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \hat{\mathrm{k}}\left[\because l_4 \frac{\mathrm{s}}{\mathrm{a}}=-l_4 \frac{\mathrm{a}}{\mathrm{s}}\right] \\
&=+\frac{\mathrm{v}^2}{\mathrm{c}^2} 2 \pi \mathrm{I}_0 \sin 2 \pi v t \ln \left(\frac{\mathrm{a}}{\mathrm{S}}\right) \hat{\mathrm{k}} \quad\left[\because \lambda=\frac{\mathrm{C}^2}{\mathrm{~V}^2}\right] \\
&=\frac{1}{\lambda^2} 2 \pi \mathrm{I}_0 \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \sin 2 \pi v t \hat{\mathrm{k}} \\
&\mathrm{J}_{\mathrm{d}}=\frac{2 \pi \mathrm{I}_0}{\lambda^2} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
(ii) $I_d=\int J_d s d s d \theta=\int^a J_d \operatorname{sds} \int^{2 \pi} d \theta=\int^a J_d \operatorname{sds} \times[2 \pi]$
$$
\begin{aligned}
&=\int_{s=0}^a\left[\frac{2 \pi}{\lambda^2} I_0 \log _e\left(\frac{\mathrm{a}}{\mathrm{s}}\right) \operatorname{sds} \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}\right] \times[2 \pi] \\
&=\left(\frac{2 \pi}{\lambda}\right)^2\left(\mathrm{I}_0 \int_{\mathrm{s}=0}^{\mathrm{a}} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \mathrm{sds}\right) \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0\left[\int_{\mathrm{s}=0}^{\mathrm{a}} \ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \frac{1}{2} \mathrm{~d}\left(\mathrm{~s}^2\right)\right] \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}} \\
&=\frac{\mathrm{a}^2}{\lambda}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0[\sin 2 \pi v t \hat{\mathrm{k}}]\left[\int_{\mathrm{s}=0}^{\mathrm{a}}\left[\ln \left(\frac{\mathrm{a}}{\mathrm{s}}\right) \cdot \mathrm{d}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)^2\right] \mathrm{ds}\right] \\
&=\frac{\mathrm{a}^2}{2}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi v t \hat{\mathrm{k}} \times\left(\frac{-\mathrm{a}}{4}\right) \\
&\therefore \quad\left[\because \int_{\mathrm{s}=0}^{\mathrm{a}}\left[\ln \left(\frac{\mathrm{s}}{\mathrm{a}}\right) \mathrm{d}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)^2\right] \mathrm{ds}=\frac{-\mathrm{a}}{4}\right] \\
&\quad \mathrm{I}_{\mathrm{d}}=\frac{-\mathrm{a}^3}{8}\left(\frac{2 \pi}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi v t(\hat{\mathrm{k}}) \\
&=\frac{-\mathrm{a}}{2}-\frac{\mathrm{a}^2}{4}\left[\frac{2 \pi}{\lambda}\right]^2 \mathrm{I}_0 \sin (2 \pi v \mathrm{t}) \hat{\mathrm{k}}
\end{aligned}
$$
The negative sign shows that the displacement current $\mathrm{I}_{\mathrm{d}}$ is oposite to the conduction current $\mathrm{I}_{\mathrm{c}}$
So, $\frac{\mathrm{a}}{2}\left(\frac{2 \pi \mathrm{a}}{2 \lambda}\right)^4 \mathrm{I}_0 \sin 2 \pi v t(\hat{\mathrm{k}})$
$\mathrm{I}_{\mathrm{d}}$ is in z-direction as $\mathrm{I}_{\mathrm{c}}$ is in $+\mathrm{z}$ direction.
(iii) The displacement current.
$$
\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{a}}{2}\left(\frac{\pi \mathrm{a}}{\lambda}\right)^2 \mathrm{I}_0 \sin 2 \pi \mathrm{vt}=\mathrm{I}_0^{\mathrm{d}} \sin 2 \pi \mathrm{vt}
$$
So, $\mathrm{I}_0^{\mathrm{d}}=\frac{\mathrm{a}}{2}\left(\frac{\pi \mathrm{a}}{\lambda}\right)^2 \mathrm{I}_0=\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2 \mathrm{I}_0 \times\left(\frac{\mathrm{a}}{2}\right)$
$$
\therefore \quad \frac{\mathrm{I}_0^{\mathrm{d}}}{\mathrm{I}_0}=\left[\frac{\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2 \mathrm{I}_0 \times\left(\frac{\mathrm{a}}{2}\right)}{\mathrm{I}_0}\right]=\frac{\mathrm{a}}{2}\left(\frac{\mathrm{a} \pi}{\lambda}\right)^2=\frac{\mathrm{a}^3 \pi^2}{2 \lambda^2}
$$
So, the required ratio is $\frac{\mathrm{I}_0^{\mathrm{d}}}{\mathrm{I}_0}=\frac{\mathrm{a}^3 \pi^2}{2 \lambda^2}$
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