di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Modern Physics

Problem T3. Protostar formation

Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
Context question:
i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
Context answer:
\boxed{$n=8$}


Context question:
ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
Context answer:
\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}


Context question:
iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.
Context answer:
\boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$}


Context question:
iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.
Context answer:
\boxed{$Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$}


Question: v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r<r_{3}$.

Solution:
The collapse continues adiabatically.

$$
\begin{aligned}
& p V^{\gamma}=\text { const } \Longrightarrow T V^{\gamma-1}=\text { const. } \\
& \therefore T \propto V^{1-\gamma} \propto r^{3-3 \gamma} \\
& \therefore T=T_{0}\left(\frac{r_{3}}{r}\right)^{3 \gamma-3} .
\end{aligned}
$$

Final answer: $T(r)=T_{0}\left(\frac{r_{3}}{r}\right)^{3 \gamma-3}$
Subfield: Thermodynamics

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$.

As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$.

Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain.

<img_4393>

Hints and Data 

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$.

Specific heat capacity of the atmosphere in the pertaining temperature range:

$\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$

Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$

Specific latent heat of vaporisation of the water within the volume of the cloud:

$\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$

$\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$

Question: 1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.

Solution:
1. Temperature $T_{1}$ where the cloud ceiling forms

$$
\mathrm{T}_{1}=\mathrm{T}_{0} \cdot\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{1-\frac{1}{x}}=279 \mathrm{~K}
\tag{1}
$$

Final answer: 279 K
Subfield: Thermodynamics

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$.

As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$.

Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain.

<img_4393>

Hints and Data 

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$.

Specific heat capacity of the atmosphere in the pertaining temperature range:

$\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$

Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$

Specific latent heat of vaporisation of the water within the volume of the cloud:

$\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$

$\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$
Context question:
1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.
Context answer:
\boxed{279}


Question: 2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density?

Solution:
2. Height $h_{1}$ of the cloud ceiling:

$$
p_{0}-p_{1}=\frac{\rho_{0}+\rho_{1}}{2} \cdot g \cdot h_{1}, with \rho_{1}=\rho_{0} \cdot \frac{p_{1}}{p_{0}} \cdot \frac{T_{0}}{T_{1}}
$$

$$
\mathrm{h}_{1}=1410 \mathrm{~m}
\tag{2}
$$

Final answer: 1410 m
Subfield: Thermodynamics

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$.

As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$.

Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain.

<img_4393>

Hints and Data 

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$.

Specific heat capacity of the atmosphere in the pertaining temperature range:

$\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$

Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$

Specific latent heat of vaporisation of the water within the volume of the cloud:

$\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$

$\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$
Context question:
1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.
Context answer:
\boxed{279}


Context question:
2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density?
Context answer:
\boxed{1410}


Question: 3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range?

Solution:
3. Temperature $\mathrm{T}_{2}$ at the ridge of the mountain.

The temperature difference when the air is ascending from the cloud ceiling to the mountain ridge is caused by two processes:

- adiabatic cooling to temperature $\mathrm{T}_{\mathrm{x}}$,



- heating by $\Delta \mathrm{T}$ by condensation.

$$
\mathrm{T}_{2}  =\mathrm{T}_{\mathrm{x}}+\Delta \mathrm{T}
\tag{3}
$$
$$
\mathrm{T}_{\mathrm{x}}  =\mathrm{T}_{1} \cdot\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{1-\frac{1}{x}}=265 \mathrm{~K}
\tag{4}
$$

For each $\mathrm{kg}$ of air the heat produced by condensation is $\mathrm{L}_{\mathrm{v}} \cdot 2.45 \mathrm{~g}=6.125 \mathrm{~kJ}$.

$$
\Delta \mathrm{T}=\frac{6.125}{\mathrm{c}_{\mathrm{p}}} \cdot \frac{\mathrm{kJ}}{\mathrm{kg}}=6.1 \mathrm{~K}
\tag{5}
$$

$$
\mathrm{T}_{2}=271 \mathrm{~K}
\tag{6}
$$

Final answer: 271 K
Subfield: Thermodynamics

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$.

As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$.

Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain.

<img_4393>

Hints and Data 

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$.

Specific heat capacity of the atmosphere in the pertaining temperature range:

$\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$

Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$

Specific latent heat of vaporisation of the water within the volume of the cloud:

$\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$

$\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$
Context question:
1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.
Context answer:
\boxed{279}


Context question:
2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density?
Context answer:
\boxed{1410}


Context question:
3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range?
Context answer:
\boxed{271}


Question: 4. Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points $M_{1}$ and $M_{2}$.

Solution:
4. Height of precipitated water column

$$
\mathrm{h}=35 \mathrm{~mm}
\tag{7}
$$

Final answer: 35 mm
Subfield: Thermodynamics

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$.

As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$.

Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain.

<img_4393>

Hints and Data 

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$.

Specific heat capacity of the atmosphere in the pertaining temperature range:

$\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$

Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$

Specific latent heat of vaporisation of the water within the volume of the cloud:

$\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$

$\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$
Context question:
1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.
Context answer:
\boxed{279}


Context question:
2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density?
Context answer:
\boxed{1410}


Context question:
3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range?
Context answer:
\boxed{271}


Context question:
4. Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points $M_{1}$ and $M_{2}$.
Context answer:
\boxed{35}


Question: 5. What temperature $\mathrm{T}_{3}$ is measured in the back of the mountain range at station $\mathrm{M}_{3}$ ?

Discuss the state of the atmosphere at station $\mathrm{M}_{3}$ in comparison with that at station $\mathrm{M}_{0}$.

Solution:
5. Temperature $\mathrm{T}_{3}$ behind the mountain

$$
\mathrm{T}_{3}=\mathrm{T}_{2} \cdot\left(\frac{\mathrm{p}_{3}}{\mathrm{p}_{2}}\right)^{1-\frac{1}{x}}=300 \mathrm{~K}
\tag{8}
$$

The air has become warmer and dryer. The temperature gain is caused by condensation of vapour.

Final answer: 300 K
Subfield: Electromagnetism

A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$.

Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant.

<img_4363>


Data:

$\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$

Question: 1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus.

Solution:
1. Determination of $\mathrm{B}$ :

The vector of the velocity of any electron is divided into components parallel with and perpendicular to the magnetic field $\overrightarrow{\mathrm{B}}$ :

$$
\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{v}}_{\|}+\overrightarrow{\mathrm{v}}_{\perp}
\tag{1}
$$

The Lorentz force $\overrightarrow{\mathrm{F}}=-\mathrm{e} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ influences only the perpendicular component, it acts as a radial force:

$$
\mathrm{m} \cdot \frac{\mathrm{v}_{\perp}^{2}}{\mathrm{r}}=\mathrm{e} \cdot \mathrm{v}_{\perp} \cdot \mathrm{B}
\tag{2}
$$

Hence the radius of the circular path that has been travelled is

$$
\mathrm{r}=\frac{\mathrm{m}}{\mathrm{e}} \cdot \frac{\mathrm{v}_{\perp}}{\mathrm{B}}
\tag{3}
$$



and the period of rotation which is independent of $\mathrm{v}_{\perp}$ is

$$
\mathrm{T}=\frac{2 \cdot \pi \cdot \mathrm{r}}{\mathrm{v}_{\perp}}=\frac{2 \cdot \pi \cdot \mathrm{m}}{\mathrm{B} \cdot \mathrm{e}}
\tag{4}
$$

The parallel component of the velocity does not vary. Because of $\alpha_{0}<<1$ it is approximately equal for all electrons:

$$
\mathrm{V}_{\| 0}=\mathrm{V}_{0} \cdot \cos \alpha_{0} \approx \mathrm{v}_{0}
\tag{5}
$$

Hence the distance $b$ between the focusing points, using eq. (5), is

$$
\mathrm{b}=\mathrm{v}_{\| 0} \cdot \mathrm{T} \approx \mathrm{v}_{0} \cdot \mathrm{T}
\tag{6}
$$

From the law of conservation of energy follows the relation between the acceleration voltage $\mathrm{V}_{0}$ and the velocity $\mathrm{v}_{0}$ :

$$
\frac{\mathrm{m}}{2} \cdot \mathrm{v}_{0}^{2}=\mathrm{e} \cdot \mathrm{V}_{0}
\tag{7}
$$

Using eq. (7) and eq. (4) one obtains from eq. (6)

$$
\mathrm{b}=\frac{2 \cdot \pi}{\mathrm{B}} \cdot \sqrt{2 \cdot \frac{\mathrm{m}}{\mathrm{e}} \cdot \mathrm{V}_{0}}
\tag{8}
$$

and because of $b=\frac{2 \cdot \pi \cdot R}{4}$ one obtains

$$
\mathrm{B}=\frac{4}{\mathrm{R}} \cdot \sqrt{2 \cdot \frac{\mathrm{m}}{\mathrm{e}} \cdot \mathrm{V}_{0}}=1.48 \cdot 10^{-2} \frac{\mathrm{Vs}}{\mathrm{m}^{2}}
\tag{9}
$$

Final answer: $1.48 \cdot 10^{-2}$ $\frac{\mathrm{Vs}}{\mathrm{m}^{2}}$
Subfield: Electromagnetism

A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$.

Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant.

<img_4363>


Data:

$\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$
Context question:
1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus.
Context answer:
\boxed{$1.48 \cdot 10^{-2}$}


Question: 2. Determine the value of $\vec{B}$ which gives four focussing points separated by $\pi / 2$ as indicated in the diagram.

Note: When considering the electron paths you may disregard the curvature of the magnetic field.

Solution:
2. Determination of $\mathrm{B}_{1}$ :

Analogous to eq. (2)

$$
\mathrm{m} \cdot \frac{\mathrm{v}_{0}^{2}}{\mathrm{R}}=\mathrm{e} \cdot \mathrm{v}_{0} \cdot \mathrm{B}_{1}
\tag{10}
$$

must hold.

From eq. (7) follows

$$
\mathrm{B}_{1}=\frac{1}{\mathrm{R}} \cdot \sqrt{2 \cdot \frac{\mathrm{m}}{\mathrm{e}} \cdot \mathrm{V}_{0}}=0.37 \cdot 10^{-2} \frac{\mathrm{Vs}}{\mathrm{m}^{2}}
\tag{11}
$$

Final answer: $0.37 \cdot 10^{-2}$ $\frac{\mathrm{Vs}}{\mathrm{m}^{2}}$
Subfield: Electromagnetism

When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$.

<img_4530>


Formulae:

$\cos \alpha-\cos \beta=-2 \cdot \sin \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$

$\sin \alpha-\sin \beta=2 \cdot \cos \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$
Context question:
a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave).
Context answer:
$\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)$ with $0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$


Question: b) Determine the velocity of propagation of the waves if the length of each unit is $\ell$.

Solution:
b) The distance $\ell$ is covered in the time $\Delta \mathrm{t}$ thus the propagation velocity is

$$
\mathrm{v}=\frac{\ell}{\Delta \mathrm{t}}=\frac{\omega \cdot \ell}{\varphi} \quad \text { or } \quad \mathrm{v}=\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)}
\tag{10}
$$

Final answer: $\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)}$
Subfield: Electromagnetism

When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$.

<img_4530>


Formulae:

$\cos \alpha-\cos \beta=-2 \cdot \sin \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$

$\sin \alpha-\sin \beta=2 \cdot \cos \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$
Context question:
a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave).
Context answer:
$\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)$ with $0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$


Context question:
b) Determine the velocity of propagation of the waves if the length of each unit is $\ell$.
Context answer:
\boxed{$\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)}$}


Question: c) State under what conditions the propagation velocity of the waves is almost independent of $\omega$. Determine the velocity in this case.

Solution:
c)

<img_4451>

Slightly dependent means arc sin $\left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right) \sim \omega$, since $\mathrm{v}$ is constant in that case.

This is true only for small values of $\omega$. That means $\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2} \ll 1$ and therefore

$$
\mathrm{v}_{0}=\frac{\ell}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}
\tag{11}
$$

Final answer: $\frac{\ell}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$