$A$ long straight cable of length $l$ is placed symmetrically along $z-$ axis and has radius $a (a << l)$. The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I(t) = I_0 \sin(2\pi \nu t)$ 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 $\vec{E}(s,t) = \mu_0 I_0 \nu \cos(2\pi \nu t) \ln(s/a) \hat{k}$.
$(i)$ 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 $I_d$.
$(iii)$ Compare the conduction current $I_0$ with the displacement current $I_d$.

(N/A) $(i)$ The displacement current density is given by $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.
Given $\vec{E}(s,t) = \mu_0 I_0 \nu \cos(2\pi \nu t) \ln(s/a) \hat{k}$.
$\vec{J}_d = \epsilon_0 \frac{\partial}{\partial t} [\mu_0 I_0 \nu \cos(2\pi \nu t) \ln(s/a) \hat{k}] = \epsilon_0 \mu_0 I_0 \nu \ln(s/a) \hat{k} \cdot \frac{d}{dt} [\cos(2\pi \nu t)]$.
Since $\epsilon_0 \mu_0 = 1/c^2$,we have $\vec{J}_d = \frac{1}{c^2} I_0 \nu \ln(s/a) \hat{k} \cdot (-2\pi \nu \sin(2\pi \nu t)) = -\frac{2\pi \nu^2 I_0}{c^2} \ln(s/a) \sin(2\pi \nu t) \hat{k}$.
Using $\lambda = c/\nu$,$\vec{J}_d = \frac{2\pi I_0}{\lambda^2} \ln(a/s) \sin(2\pi \nu t) \hat{k}$.
$(ii)$ The total displacement current $I_d = \int \vec{J}_d \cdot d\vec{A} = \int_0^a J_d (2\pi s ds)$.
$I_d = \int_0^a \frac{2\pi I_0}{\lambda^2} \ln(a/s) \sin(2\pi \nu t) (2\pi s ds) = \frac{4\pi^2 I_0 \sin(2\pi \nu t)}{\lambda^2} \int_0^a s \ln(a/s) ds$.
Let $x = s/a$,then $ds = a dx$. The integral becomes $a^2 \int_0^1 x \ln(1/x) dx = a^2 \int_0^1 -x \ln x dx = a^2 [1/4] = a^2/4$.
Thus,$I_d = \frac{4\pi^2 I_0 \sin(2\pi \nu t)}{\lambda^2} \cdot \frac{a^2}{4} = I_0 \sin(2\pi \nu t) (\frac{\pi a}{\lambda})^2$.
$(iii)$ Comparing $I_d$ with $I(t) = I_0 \sin(2\pi \nu t)$,we find $I_d = I(t) (\frac{\pi a}{\lambda})^2$. Since $a << \lambda$,the displacement current is much smaller than the conduction current.