di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Optics

- 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: Max finds himself trapped in the center of a mirror walled equilateral triangular room. What minimum beam angle must his flashlight have so that any point of illumination in the room can be traced back to his flashlight with at most 1 bounce? (Answer in degrees.) Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant.

The point of illumination refers to any point in the room that is lit.

Solution:
Each time light hits a mirror, we can reflect the entire equilateral triangle about that mirror and continue to trace the straight-line path of the light. For a maximum of 1 bounce, we can reflect our triangle about each of its initial sides. In order for the light to hit every part of the triangle (or an image of that part), by symmetry we require a $120^{\circ}$ angle. In this case, we would just shine the flashlight so that light directly reaches the entirety of one side, and reflections of light will fully reach the other two sides of the triangle.

Final answer: $120$ $^{\circ}$
Subfield: Mechanics


Question: For his art project, Weishaupt cut out $N=20$ wooden equilateral triangular blocks with a side length of $\ell=10 \mathrm{~cm}$ and a thickness of $t=2 \mathrm{~cm}$, each with the same mass and uniform density. He wishes to stack one on top of the other overhanging the edge of his table. In centimeters, what is the maximum overhang? Round to the nearest centimeter. A side view is shown below. Assume that all triangles are parallel to each other.

Note: This diagram is not to scale.

![](https://cdn.mathpix.com/cropped/2023_12_21_145bbb911bbed723cc96g-1.jpg?height=567&width=940&top_left_y=562&top_left_x=587)

Solution:
Let us consider $N=1$ equilateral triangles. From inspection, we need to place the triangle such that the center of mass lies at the edge of the table. The maximum overhang in this case is $(1-f) h$ where $h=\frac{\ell \sqrt{3}}{2}$ is the height of the triangle and $f h=\frac{h}{3}$ is the location of the center of mass.
If we wish to place a second triangle on top, we want to maximize the center of mass to be as far right as possible without the top block toppling. Placing the second block such that its center of mass is at the tip of the first triangle accomplishes this. However, the center of mass of the two triangles combined is now past the edge. Their center of mass is:

$$
x_{\mathrm{cm}}=\frac{f h+h}{2}=\frac{f+1}{2} h
$$

Thus the maximum overhang of the first block is now:

$$
h-\frac{f+1}{2} h=\frac{1-f}{2} h
$$

Now, we will place a third block such that it has the maximum overhang with respect to the top block and then shift the entire setup so that the center of mass of the system lies at the edge of the table. Following the same procedures, we find that the maximum overhang of the first block is:

$$
\frac{1-f}{3} h
$$

The overhang of the top two blocks are $(1-f) h$ and $\frac{1-f}{2} h$, unchanged from earlier. You can show via induction that the maximum overhang of the $n^{\text {th }}$ block (counting from the top downwards) is:

$$
\frac{1-f}{n} h
$$

so if there are 20 such blocks, then the total overhang (summing over all the blocks) is:

$$
\sum_{k=0}^{20} \frac{1-f}{k} h=(1-f) h H_{20}=\frac{2}{3} \frac{\ell \sqrt{3}}{2} H_{20}=\frac{\ell \sqrt{3}}{3} H_{20}=20.77 \mathrm{~cm} \approx 21 \mathrm{~cm}
$$

where $H_{N}$ is the $N^{\text {th }}$ harmonic number.

Final answer: 21 cm
Subfield: Optics

- 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: Kushal finds himself trapped in a large room with mirrors as walls. Being scared of the dark, he has a powerful flashlight to light the room. All references to "percent" refer to area. Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant. The point of illumination refers to any point in the room that is lit.

What percent of a large circular room can be lit up using a flashlight with a 20 degree beam angle if Kushal stands in the center?

Solution:
Each ray emitted follows a straight line, even when it is reflected since it originates from the center of a circle. Thus, the light rays in total trace out two circular sectors with an angle of $\theta=20^{\circ}$ each. Thus, the total percent of the room illuminated is:

$$
f=\frac{2 \theta}{360}=11.1 \% \text {. }
$$

Final answer: 11.1 %
Subfield: Mechanics


Question: The graph provided plots the $y$-component of the velocity against the $x$-component of the velocity of a kiddie roller coaster at an amusement park for a certain duration of time. The ride takes place entirely in a two dimensional plane.

Some students made a remark that at one time, the acceleration was perpendicular to the velocity. Using this graph, what is the minimum x-velocity the ride could be travelling at for this to be true? Round to the nearest integer and answer in meters per second. The diagram is drawn to scale, and you may print this page out and make measurements.

![](https://cdn.mathpix.com/cropped/2023_12_21_e7e16de1e9199580e484g-1.jpg?height=1900&width=1106&top_left_y=655&top_left_x=496)

Solution:
The solution revolves around the idea that when the acceleration is perpendicular to the velocity, the work done is 0 , and thus, the instantaneous rate of change of the magnitude of velocity $v_{x}^{2}+v_{y}^{2}$ is 0 . Thus, at such points, when the vertical velocity is plotted against the horizontal velocity, the curve will be tangent to a circle centered at the origin because $v_{y}^{2}+v_{x}^{2}$ is nonchanging at that instant.
This is equivalent to stating that the line from the origin to the curve is perpendicular to the curve. Drawing such lines to the curve, the first time this occurs is at $v_{x}=1 \mathrm{~m} / \mathrm{s}$.

Final answer: 1 m/s
Subfield: Modern Physics


Question: While exploring outer space, Darth Vader comes upon a purely reflective spherical planet with radius $R_{p}=$ $40,000 \mathrm{~m}$ and mass $M_{p}=8.128 \times 10^{24} \mathrm{~kg}$. Around the planet is a strange moon of orbital radius $R_{s}=$ $6,400,000 \mathrm{~m}\left(R_{s} \gg R_{p}\right)$ and mass $M_{s}=9.346 \times 10^{19} \mathrm{~kg}\left(M_{s} \ll M_{p}\right)$. The moon can be modelled as a blackbody and absorbs light perfectly. Darth Vader is in the same plane that the planet orbits in. Startled, Darth Vader shoots a laser with constant intensity and power $P_{0}=2 \times 10^{32} \mathrm{~W}$ at the reflective planet and hits the planet a distance of $\frac{R_{p}}{2}$ away from the line from him to the center of the planet. Upon hitting the reflective planet, the light from the laser is plane polarized. The angle of the planet's polarizer is always the same as the angle of reflection. After reflectance, the laser lands a direct hit on the insulator planet. Darth Vader locks the laser in on the planet until it moves right in front of him, when he turns the laser off. Determine the energy absorbed by the satellite. Assume the reflective planet remains stationary and that the reflective planet is a perfect polarizer of light.

![](https://cdn.mathpix.com/cropped/2023_12_21_cc3e4746f35af35f4e0bg-1.jpg?height=776&width=1008&top_left_y=764&top_left_x=556)

Solution:
The original angle of reflectance can be found with some optical geometry to be $\theta_{0}=\sin ^{-1}\left(\frac{1}{2}\right)=$ $\frac{\pi}{6}$. By Malus' Law, when the laser hits the planet with an angle $\theta$, the final power after reflection is $P_{0} \cos ^{2}(\theta)$. Also, note that the satellite has to be at an angle $2 \theta$ due to the law of reflection. Therefore, by Kepler's Third law, the time for the satellite to reach $\theta=0$ is
$$
t=2 \theta \sqrt{\frac{r^{3}}{G M_{p}}} \Longrightarrow d t=2 \sqrt{\frac{r^{3}}{G M_{p}}} d \theta
$$

Thus, the total power absorbed by the satellite is

$$
\int_{0}^{\theta_{0}} P_{0} \cos ^{2}(\theta) d t=2 P_{0} \sqrt{\frac{r^{3}}{G M_{p}}} \int_{0}^{\theta_{0}} \cos ^{2}(\theta) d \theta=P_{0} \sqrt{\frac{r^{3}}{G M_{p}}}\left(\theta_{0}+\sin \left(\theta_{0}\right) \cos \left(\theta_{0}\right)\right)=1.33 \cdot 10^{35} \mathrm{~J} .
$$

Final answer: $1.33 \cdot 10^{35}$ J
Subfield: Modern Physics

- 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: A particle of rest mass $m$ moving at a speed $v=0.7 c$ decomposes into two photons which fly off at a separated angle $\theta$. What is the minimum value of the angle of separation assuming that the two photons have equal wavelength. (Answer in degrees)

Solution:
Conservation of momentum and energy gives:
$$
\begin{gathered}
p_{m}=2 E_{\gamma} \cos (\theta / 2) \\
E_{m}=2 E_{\gamma}
\end{gathered}
$$

Relativity demands that:

$$
E_{m}^{2}=m^{2}+p_{m}^{2}
$$

Solving this system of three equations gives:

$$
\begin{aligned}
4 E_{\gamma}^{2} & =m^{2}+p_{m}^{2} \\
\frac{\gamma^{2} m^{2} v^{2}}{\cos ^{2}(\theta / 2)} & =m^{2}+\gamma^{2} m^{2} v^{2} \\
\gamma^{2} v^{2}\left(-1+\frac{1}{\cos ^{2}(\theta / 2)}\right) & =1 \\
\frac{1}{\cos ^{2}(\theta / 2)} & =\frac{1}{\gamma^{2} v^{2}}+1 \\
\cos (\theta / 2) & =v \\
\theta & =91.1^{\circ}
\end{aligned}
$$

Final answer: $91.1$ $^{\circ}$
Subfield: Mechanics


Question: Mario is racing with Wario on Moo Moo Meadows when a goomba, ready to avenge all of his friends' deaths, came and hijacked Mario's kart. A graph representing the motion of Mario at any instant is shown below. The velocity acquired by Mario is shown on the x-axis, and the net power of his movement is shown on the $\mathrm{y}$-axis. When Mario's velocity is $6 \mathrm{~m} / \mathrm{s}$, he eats a mushroom which gives him a super boost.

![](https://cdn.mathpix.com/cropped/2023_12_21_0f703e5630a441a05081g-1.jpg?height=1309&width=1272&top_left_y=473&top_left_x=421)

You may need to make measurements. Feel free to print this picture out as the diagram is drawn to scale. Find the total distance from Mario runs from when his velocity is $0 \mathrm{~m} / \mathrm{s}$ to when his velocity just reaches $9 \mathrm{~m} / \mathrm{s}$ given that Mario's mass is $m=89 \mathrm{~kg}$. Answer in meters and round to one significant digit.

Solution:
Our first goal is to find an expression for the power curve $P(v)$. To do this, let us select a few points on the curve. The easiest point to pick is $(0,0)$ since it is fixed at the origin. The next two easiest points to pick are those that are on the lines $x=3$ and $x=6$ which are given as approximately $(3,1.8)$ and $(6,7.2)$. Note that $y$-axis is in units of $100 \mathrm{~W}$ so in reality these two points are given as $(3,180)$ and $(6,720)$. This curve is resemblant of a quadratic in the form of $y=k_{1} x^{2}$ and upon solving for $k_{1}$ we find that the curve is given as $P=20 v^{2}$. Secondly, the next line remains constant with respect to time as a line $P=900$. Therefore, we can write a piecewise function for power defined by
$$
P(v)=\left\{\begin{array}{ll}
20 v^{2} & \text { if } v \geq 0, \quad \text { and } \quad v<6 \\
900 & \text { if } v \geq 6
\end{array} .\right.
$$



We need to find the relationship between power, velocity, and displacement of Mario. Consider dividing the displacement into tiny rectangular pieces with width $\Delta t$ such that

$$
s=\sum_{i \in \mathbb{N}} \Delta s_{i}=\sum_{i} v_{i}(t) \cdot \Delta t
$$

We want to the displacement to be expressed in terms without $\Delta t$. This means that we have to find a relationship for $\Delta t$. Note that

$$
\Delta t=\Delta v \cdot \frac{\Delta t}{\Delta v}=\Delta v \cdot \frac{1}{\Delta v / \Delta t}=\frac{\Delta v}{a}
$$

Therefore, we now know the displacement to be expressed as

$$
s=\sum_{i \rightarrow 0} v_{i}(t) \frac{\Delta v}{a}=\int \frac{v}{a(v)} d v
$$

To find an expression for $a(v)$ in terms of power, we note that

$$
P(v)=F(v) \cdot v \Longrightarrow a(v)=\frac{P(v)}{v m}
$$

which means that upon substituting,

$$
s=\int \frac{v^{2} m}{P(v)} d v \Longrightarrow s=\int_{0}^{6} \frac{89 v^{2}}{20 v^{2}} d v+\int_{6}^{9} \frac{89 v^{2}}{900} d v=43.61 \approx 40 \mathrm{~m}
$$

Final answer: 40 m
Subfield: Thermodynamics


Question: An engineer has access to a tetrahedron building block with side length $\ell=10 \mathrm{~cm}$. The body is made of a thermal insulator but the edges are wrapped with a thin copper wiring with cross sectional area $S=2 \mathrm{~cm}^{2}$. The thermal conductivity of copper is $385.0 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. He stacks these tetrahedrons (all facing the same direction) to form a large lattice such that the copper wires are all in contact. In the diagram, only the front row of a small section is coloured. Assume that the lattice formed is infinitely large.

At some location in the tetrahedral building block, the temperature difference between two adjacent points is $1^{\circ} \mathrm{C}$. What is the heat flow across these two points? Answer in Watts.

Note: Two adjacent points refer to two adjacent points on the tetrahedron.

![](https://cdn.mathpix.com/cropped/2023_12_21_fa0c803e66fc3d4839bfg-1.jpg?height=1285&width=1507&top_left_y=756&top_left_x=301)

Solution:
There are many ways to solve this problem. We first identify that this is exactly the same as an infinite lattice resistor problem. To solve these, we can imagine injecting a current at a node and seeing how this current spreads out. However, a faster approach is by applying Foster's Theorem on this lattice.
The resistance of a single wire is:

$$
R=\frac{\ell}{k S}=1.299 \mathrm{~W} / \mathrm{K}
$$

Foster's theorem tells us that

$$
E R=V-1
$$



where $V$ is number of vertices and $E$ is edges. Taking the limit as $E, V \rightarrow \infty$, we get: $E=6 \mathrm{~V}$ (since each vertex is connected to 12 edges, but each edge is shared by two vertices). Therefore:

$$
R_{\mathrm{eff}}=\frac{1}{6} R=0.2165 \mathrm{~W} / \mathrm{K}
$$

From Fourier's Law, we have:

$$
\dot{Q}=\frac{\Delta T}{R_{\mathrm{eff}}}=4.62 \mathrm{~W}
$$

Final answer: 4.62 N
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: Two infinitely long current carrying wires carry constant current $i_{1}=2 \mathrm{~A}$ and $i_{2}=3 \mathrm{~A}$ as shown in the diagram. The equations of the wire curvatures are $y^{2}-8 x-6 y+25=0$ and $x=0$. Find the magnitude of force (in Newtons) acting on one of the wires due to the other.

<img_4264>

Note: The current-carrying wires are rigidly fixed. The units for distances on the graph should be taken in metres.

Solution:
The magnetic field from the wire is given by $B=\frac{\mu_{0} i_{1}}{2 \pi x}$. Let $\theta$ be the direction of a component of force from the vertical. It is then seen that
$$
d F=B i_{2} d \ell \Longrightarrow d F_{x}=B i_{2} d \ell \sin \theta=B i_{2} d y
$$

We only consider the force in the $x$-direction which means that

$$
F_{x}=\int_{\infty}^{\infty} d F_{x}=\frac{\mu_{0} i_{1} i_{2}}{2 \pi} \int_{-\infty}^{\infty} \frac{d y}{x}
$$

Solving the equation in terms of $x$ and then plugging in gives us

$$
F_{x}=\frac{8 \mu_{0} i_{1} i_{2}}{2 \pi} \int_{-\infty}^{\infty} \frac{d y}{y^{2}-6 y+25}=\frac{8 \mu_{0} i_{1} i_{2}}{2 \pi} \cdot \frac{\pi}{4}=\mu_{0} i_{1} i_{2}=7.5398 \cdot 10^{-6} \mathrm{~N} .
$$

Final answer: $7.5398 \cdot 10^{-6}$ N
Subfield: Electromagnetism


Question: Two infinitely long current carrying wires carry constant current $i_{1}=2 \mathrm{~A}$ and $i_{2}=3 \mathrm{~A}$ as shown in the diagram. The equations of the wire curvatures are $y^{2}-8 x-6 y+25=0$ and $x=0$. Find the magnitude of force (in Newtons) acting on one of the wires due to the other.

![](https://cdn.mathpix.com/cropped/2023_12_21_e2bb42caca393489220dg-1.jpg?height=1041&width=1507&top_left_y=1339&top_left_x=301)

Note: The current-carrying wires are rigidly fixed. The units for distances on the graph should be taken in metres.

Solution:
The magnetic field from the wire is given by $B=\frac{\mu_{0} i_{1}}{2 \pi x}$. Let $\theta$ be the direction of a component of force from the vertical. It is then seen that
$$
d F=B i_{2} d \ell \Longrightarrow d F_{x}=B i_{2} d \ell \sin \theta=B i_{2} d y
$$

We only consider the force in the $x$-direction which means that

$$
F_{x}=\int_{\infty}^{\infty} d F_{x}=\frac{\mu_{0} i_{1} i_{2}}{2 \pi} \int_{-\infty}^{\infty} \frac{d y}{x}
$$

Solving the equation in terms of $x$ and then plugging in gives us

$$
F_{x}=\frac{8 \mu_{0} i_{1} i_{2}}{2 \pi} \int_{-\infty}^{\infty} \frac{d y}{y^{2}-6 y+25}=\frac{8 \mu_{0} i_{1} i_{2}}{2 \pi} \cdot \frac{\pi}{4}=\mu_{0} i_{1} i_{2}=7.5398 \cdot 10^{-6} \mathrm{~N} .
$$

Final answer: $7.5398 \cdot 10^{-6}$ N