Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Consider an infinite square grid of equal resistors where the nodes are exactly the lattice points in the 2D Cartesian plane. A current $I=2.7 \mathrm{~A}$ enters the grid at the origin $(0,0)$. Find the current in Amps through the resistor connecting the nodes $(N, 0)$ and $(N, 1)$, where $N=38$ can be assumed to be much larger than 1.

Solution:
WLOG, let each resistor have unit resistance.

Kirchoff's current law says that the total current entering the node $(x, y) \neq(0,0)$ is zero:

$$
\begin{array}{r}
{[U(x+1, y)-U(x, y)]+[U(x-1, y)-U(x, y)]+[U(x, y+1)-U(x, y)]+[U(x, y-1)-U(x, y)]=0} \\
{[U(x+1, y)-2 U(x, y)+U(x-1, y)]+[U(x, y+1)-2 U(x, y)+U(x, y-1)]=0}
\end{array}
$$

This is an approximation of the equation

$$
\frac{\partial^{2} U}{\partial x^{2}}+\frac{\partial^{2} U}{\partial y^{2}}=0
$$

which is Laplace's equation $\nabla^{2} U=0$ in 2D. Given $U \rightarrow 0$ as $r \rightarrow \infty$, this implies that the potential



field approximates that of a point charge at $O$ in 2D. This approximation is valid far from the origin where changes in $U$ over unit length are small.

The electric field corresponding to this potential field is

$$
\mathbf{E}=-\nabla U=\left(-\partial_{x} U,-\partial_{y} U\right) \approx(U(x-1, y)-U(x, y), U(x, y-1)-U(x, y))=\left(i_{x}(x, y), i_{y}(x, y)\right)
$$

far from the origin, where $i_{x}(x, y), i_{y}(x, y)$ are the horizontal and vertical currents passing through node $(x, y)$. (Note that the current horizontal current is different to the left vs. to the right of the node, and similarly for the vertical current, but the difference is negligible for $N \gg 1$.)

The current $i=\sqrt{i_{x}^{2}+i_{y}^{2}}$ at a distance $r \gg 1$ away from the origin is given by Gauss's law:

$$
2 \pi r i(r) \approx I
$$

so at $(N, 0)$ we have

$$
i_{x}(N, 0)=i(N) \approx \frac{I}{2 \pi N}
$$

The difference between the entering and exiting horizontal currents at $(N, 0)$ is approximately

$$
-\partial_{x} i_{x}(N, 0)=\frac{I}{2 \pi N^{2}}
$$

This difference is directed equally into the vertical resistors adjacent to $(N, 0)$, so the final answer is

$$
\frac{I}{4 \pi N^{2}}=1.488 \times 10^{-4} \mathrm{~A}
$$

Final answer: $1.488 \times 10^{-4}$ A