Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
Context question:
1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
Context answer:
$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$


Context question:
2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.
Context answer:
\boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$}


Context question:
3.  Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces.
Context answer:
<img_4501>


Extra Supplementary Reading Materials:

For the following questions, ignore any edge effects.
Context question:
4.  Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin.
Context answer:
the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$

The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$.

The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$.


Context question:
5.  Calculate the capacitance $C$ of the system.
Context answer:
\boxed{$\frac{l \pi \epsilon_{0}}{\ln 3}$}


Context question:
6.  Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$. Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon=\epsilon_{0}$.
Context answer:
\boxed{$I=\frac{V_{0} \pi \sigma l}{\ln 3}$}


Context question:
7.  Calculate the resistance $R$ of the system. Calculate $R C$ of the system.
Context answer:
\boxed{$R C=\frac{\epsilon_{0}}{\sigma}$}


Question: 8.  Calculate the magnetic field due to the current in question 6. Assume that the permeability of the liquid is equal to that of vacuum $\mu=\mu_{0}$.

Notes $\int \frac{\alpha d x}{\alpha^{2}+x^{2}}=\arctan \frac{x}{a}+$ const

Solution:
Since the system has a high symmetry, we may use Ampere's law. The magnetic field should not have any $z$ dependence, since the current has no $z$ dependence.

Figure 4 shows the current density $\mathbf{J}$ flow from one cylinder to the other cylinder. Choose an Ampere loop on a constant $x$ plane in a symmetrical way, so that the first path is pointing in the positive $z$ direction with constant $y$ coordinate, the second path is pointing to the negative $y$ direction with constant $z$ coordinate. The third path is pointing to the negative $z$ direction, but with constant $-y$ coordinate. The fourth path is pointing in the positive $y$ direction with constant $-z$ coordinate.

Having this path, we need to calculate the current that flow through the loop

$$
\begin{aligned}
I & =\int \mathbf{J} \cdot \mathbf{d} \mathbf{A} \\
& =\int J_{x} l d y \\
& =\frac{V_{0} \sigma l}{2 \ln 3} \int_{-y}^{y}\left(\frac{4 a+x}{(4 a+x)^{2}+y^{2}}+\frac{4 a-x}{(4 a-x)^{2}+y^{2}}\right) d y
\end{aligned}
$$

<img_4435>

Figure 4: The Ampere loop

$$
I=\frac{V_{0} \sigma l}{\ln 3}\left(\arctan \frac{y}{4 a+x}+\arctan \frac{y}{4 a-x}\right)
\tag{29}
$$

Using the Ampere's law

$$
\oint \mathbf{B} \cdot \mathbf{d} \mathbf{l}=\mu_{0} I
\tag{30}
$$

$$
\begin{aligned}
2 B_{z} l & =\frac{\mu_{0} V_{0} \sigma l}{\ln 3}\left(\arctan \frac{y}{4 a+x}+\arctan \frac{y}{4 a-x}\right) \\
B_{z} & =\mu_{0} \frac{V_{0} \sigma}{2 \ln 3}\left(\arctan \frac{y}{4 a+x}+\arctan \frac{y}{4 a-x}\right)
\end{aligned}
\tag{31}
$$



therefore

$$
\mathbf{B}=\hat{z} \frac{\mu_{0} V_{0} \sigma}{2 \ln 3}\left(\arctan \frac{y}{4 a+x}+\arctan \frac{y}{4 a-x}\right)
\tag{32}
$$

Final answer: $\mathbf{B}=\hat{z} \frac{\mu_{0} V_{0} \sigma}{2 \ln 3}\left(\arctan \frac{y}{4 a+x}+\arctan \frac{y}{4 a-x}\right)$