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

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Extra Supplementary Reading Materials:

The electric potential at the center of a cube with uniform charge density $\rho$ and side length $a$ is



$$

\Phi \approx \frac{0.1894 \rho a^{2}}{\epsilon_{0}}

$$



You do not need to derive this. ${ }^{1}$



For the entirety of this problem, any computed numerical constants should be written to three significant figures.

Question: a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.

Solution:
By dimensional analysis, the answer takes the form



$$

\Phi_{c}(a, \rho) \approx \frac{C \rho a^{2}}{\epsilon_{0}}

$$



for a dimensionless constant $C$. Note that a cube of side length $a$ consists of 8 cubes of side length $a / 2$, each with a corner at the center of the larger cube. Then



$$

\frac{0.1894 \rho a^{2}}{\epsilon_{0}}=8 \frac{C \rho(a / 2)^{2}}{\epsilon_{0}}

$$



so $C=0.1894 / 2=0.0947$.



Final answer: $\Phi_{c}(a, \rho) \approx \frac{C \rho a^{2}}{\epsilon_{0}}$, $C=0.0947$
Subfield: Electromagnetism


Question: b. What is the electric potential at the tip of a pyramid with a square base of side length $a$, height $a / 2$, and uniform charge density $\rho$ ? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.

Solution:
A cube of side length $a$ consists of 6 such pyramids. Then we simply compute $0.1894 / 6$ for



$$

\Phi_{p}(a, \rho) \approx \frac{0.0316 \rho a^{2}}{\epsilon_{0}}

$$

Final answer: $\Phi_{p}(a, \rho) \approx \frac{0.0316 \rho a^{2}}{\epsilon_{0}}$
Subfield: Electromagnetism


Question: c. What is the electric potential due to a square plate with side length $a$ of uniform charge density $\sigma$ at a height $a / 2$ above its center? Write your answer in terms of $\sigma, a, \epsilon_{0}$, and any necessary numerical constants.

Solution:
Let the potential due to such a square be $\Phi_{s}(a, \sigma)$. Note that adding a square plate of infinitesimal thickness $d z$ and side length $a$ to a square pyramid with base side length $a$ and height $a / 2$ yields a square pyramid with base side length $a+2 d z$ and height $a / 2+d z$.



The surface charge density of a square plate with thickness $d z$ and volume charge density $\rho$ is $\sigma=\rho d z$. Then by the principle of superposition,



$$

\Phi_{s}(a, \rho d z)=\Phi_{p}(a+2 d z, \rho)-\Phi_{p}(a, \rho) \approx \frac{0.0316 \rho\left((a+2 d z)^{2}-a^{2}\right)}{\epsilon_{0}}=\frac{0.126 a \rho d z}{\epsilon_{0}}

$$



so we have



$$

\Phi_{s}(a, \sigma) \approx \frac{0.126 a \sigma}{\epsilon_{0}}

$$

Final answer: $\Phi_{s}(a, \sigma) \approx \frac{0.126 a \sigma}{\epsilon_{0}}$
Subfield: Electromagnetism


Question: d. Let $E(z)$ be the electric field at a height $z$ above the center of a square with charge density $\sigma$ and side length $a$. If the electric potential at the center of the square is approximately $\frac{0.281 a \sigma}{\epsilon_{0}}$, estimate $E(a / 2)$ by assuming that $E(z)$ is linear in $z$ for $0<z<a / 2$. Write your answer in terms of $\sigma, a, \epsilon_{0}$, and any necessary numerical constants.

Solution:
The potential difference between height 0 and $a / 2$ is



$$

\Delta \Phi=(0.281-0.126) \frac{a \sigma}{\epsilon_{0}}=\frac{0.155 a \sigma}{\epsilon_{0}}

$$



On the other hand, we have



$$

\Delta \Phi=\int_{0}^{a / 2} E(z) d z \approx \frac{a}{2} \frac{E(0)+E(a / 2)}{2}

$$



where we approximated $E(z)$ as linear, and $E(0)=\sigma / 2 \epsilon_{0}$ by Gauss's law. Solving for $E(a / 2)$,



$$

E(a / 2) \approx \frac{0.119 \sigma}{\epsilon_{0}}

$$



where the last significant digit is not important. Incidentally, the actual value is exactly $\sigma / 6 \epsilon_{0}$, and this fact has a slick calculation-free proof.

Final answer: $E(a / 2) \approx \frac{0.119 \sigma}{\epsilon_{0}}$



where the last significant digit is not important. Incidentally, the actual value is exactly $\sigma / 6 \epsilon_{0}$, and this fact has a slick calculation-free proof.
Subfield: Optics


Question: b. Use the uncertainty principle, $\Delta r \Delta p_{r} \geq \hbar / 2$, to place a bound on how accurately we can know the angle of the photon from the axis of the telescope. Give your answer in terms of $R$ and $\lambda$. If you were unable to solve part a, you may also give your answer in terms of $\Delta r$.

Solution:
We have



$$

\Delta p_{r} \approx p \Delta \theta=\frac{h}{\lambda} \Delta \theta

$$







Applying the uncertainty principle, we have



$$

\Delta \theta \geq \frac{\sqrt{18} \lambda}{4 \pi R}

$$



Since the factors of $\hbar$ canceled out, this is really a classical calculation; one could get a similar result by considering classical diffraction from a circular aperture.

Final answer: $$

\Delta \theta \geq \frac{\sqrt{18} \lambda}{4 \pi R}

$$
Subfield: Modern Physics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Extra Supplementary Reading Materials:

Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.


a. The force of radiation on a spherical particle of radius $r$ is given by



$$

F=P Q \pi r^{2}

$$



where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.

Question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.

Solution:
We will assume that light from the sun is completely absorbed, and then re-radiated as blackbody isotropically. In that case,



$$

P=\frac{I}{c}=\frac{L \odot}{4 \pi R^{2} c}

$$



The relationship between the pressure $P$ and the energy density $I$ can be derived from the equation $p=E / c$ for photons, or simply postulated by dimensional analysis, since there is no other relevant speed.



Alternatively, this relation can be derived using classical electromagnetism, though this was not required. For a plane wave, the momentum density (i.e. pressure) of the electromagnetic field is $P=\epsilon_{0}|\mathbf{E} \times \mathbf{B}|$, while the energy density is $I=|\mathbf{S}| / c$ where $\mathbf{S}=\mathbf{E} \times \mathbf{B} / \mu_{0}$. Then $P=I / c$ since $c^{2}=1 / \epsilon_{0} \mu_{0}$.

Final answer: $P=\frac{L \odot}{4 \pi R^{2} c}$
Subfield: Modern Physics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Extra Supplementary Reading Materials:

Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.


a. The force of radiation on a spherical particle of radius $r$ is given by



$$

F=P Q \pi r^{2}

$$



where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
Context question:
i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
Context answer:
\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}


Question: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.

Solution:
We have



$$

F_{\text {gravity }}=\frac{G M_{\odot}}{R^{2}} \frac{4}{3} \pi \rho r^{3}, \quad F_{\text {radiation }}=\frac{L_{\odot}}{4 \pi R^{2} c} Q \pi r^{2}=\frac{L_{\odot}}{4 R^{2} c} Q r^{2}

$$



which gives



$$

\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}

$$

Final answer: $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$
Subfield: Modern Physics


Question: iii. The quality factor is given by one of the following



- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$

- If $r \sim \lambda, Q \sim 1$.

- If $r \gg \lambda, Q=1$



Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?

Solution:
In order to be blown away, the ratio should be greater than one. Since it is independent of distance from the sun, if it is blown away, it will be blown away at any distance. For $r \gg \lambda$, the ratio is proportional to $1 / r$, so smaller particles are more likely to be blown away. For $r \ll \lambda$, the ratio is proportional to $r$, so larger particles are more likely to be blown away. Thus particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.

Final answer: particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.
Subfield: Thermodynamics

A hollow insulated cylinder of height $2 \mathrm{H}$ and volume $2 \mathrm{~V}$ is closed from below by an insulating piston. The cylinder is divided into two initially identical chambers by an insulating diaphragm of mass $m$. The diaphragm rests on a circular ledge and a gasket between them provides tight contact. Both chambers are filled with gaseous helium at pressure $p$ and temperature $T$. A force is applied to the piston, so that it moves upwards slowly.

<img_4481>

Question: a. Find the volume of the lower chamber $V_{0}$ when the gas starts to leak between the chambers.

Solution:
Let $p_{1}, V_{1}, T_{1}$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_{2}, V_{2}$, $T_{2}$ - those in the lower one. Note that $V_{1} \equiv V$ does not change.

Consider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_{2}$ and $T_{2}$. The energy conservation for the parcel is

$$
0=p_{2} \mathrm{~d} v+\mathrm{d}\left(\frac{3}{2} n_{v} R T_{2}\right)=\frac{5}{2} p_{2} \mathrm{~d} v+\frac{3}{2} v \mathrm{~d} p_{2}
$$

This gives

$$
v^{5} p_{2}^{3} =\text { const }
\tag{1}
$$

$$
T_{2}^{5} p_{2}^{-2} =\text { const }
\tag{2}
$$

The leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\Delta p \equiv$ $p_{0}-p=m g H / V=p x$, where

$$
x=\frac{m g H}{p V} \tag{(a., b., c.)}
$$



a. We may let $v=V_{2}$ before that.

$$
\begin{gathered}
V^{5} p^{3}=V_{0}^{5}(p+\Delta p)^{3}=V_{0}^{5} p^{3}(1+x)^{3} \\
V_{0}=V(1+x)^{-3 / 5}
\end{gathered}
$$

Final answer: $V(1+x)^{-3 / 5}$
Subfield: Electromagnetism

A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$.

<img_4344>

This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}<V<V_{2}$, the temperature of the resistor starts oscillating. Assume that (i) the heat flux $P$ from the resistor to the ambient medium is given by $P=\alpha\left(T-T_{0}\right)$, where $\alpha$ is a constant, $T$ denotes the temperature of the resistor, and $T_{0}$ is the ambient temperature; (ii) the geometrical size of the resistor is so small that it will reach a thermal equilibrium much faster than the characteristic time $L / R_{2}$.

Question: (a) Express $V_{1}$ and $V_{2}$ in terms of the other parameters defined above.

Solution:
The power heating the resistor is $P_{\mathrm{el}}=V^{2} / R_{j}$. The thermal equilibrium is reached when $P_{\mathrm{el}}=P=\alpha\left(T_{\mathrm{eq}}-T_{0}\right)$. To avoid oscillations, the equilibrium temperature $T_{\text {eq }}$ must satisfy $T_{\text {eq }}<T_{c}$ if $R=R_{1}$ and $T_{\text {eq }}>T_{c}$ if $R=R_{2}$. Solving for $V$, we have

$$
V=\sqrt{R_{j} \alpha\left(T_{\mathrm{eq}}-T_{0}\right)}
\tag{1}
$$

The critical values therefore are

$$
V_{1}=\sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)} \quad \text { and } \quad V_{2}=\sqrt{R_{2} \alpha\left(T_{c}-T_{0}\right)}
\tag{2}
$$

Final answer: $V_{1}=\sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)}$ , $V_{2}=\sqrt{R_{2} \alpha\left(T_{c}-T_{0}\right)}$
Subfield: Electromagnetism

A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$.

<img_4344>

This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}<V<V_{2}$, the temperature of the resistor starts oscillating. Assume that (i) the heat flux $P$ from the resistor to the ambient medium is given by $P=\alpha\left(T-T_{0}\right)$, where $\alpha$ is a constant, $T$ denotes the temperature of the resistor, and $T_{0}$ is the ambient temperature; (ii) the geometrical size of the resistor is so small that it will reach a thermal equilibrium much faster than the characteristic time $L / R_{2}$.
Context question:
(a) Express $V_{1}$ and $V_{2}$ in terms of the other parameters defined above.
Context answer:
\boxed{$V_{1}=\sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)}$ , $V_{2}=\sqrt{R_{2} \alpha\left(T_{c}-T_{0}\right)}$}


Question: (b) Assuming that $V_{1}<V<V_{2}$, sketch qualitatively how the temperature of the resistor $T$ depends on time $t$, and find the ratio $\left(T_{\max }-T_{0}\right) /\left(T_{\min }-T_{0}\right)$, where $T_{\max }$ and $T_{\min }$ denote the maximal and minimal values of $T$, respectively.

Solution:
In the oscillating regime, we have a time-dependent current $I(t)$. The power dissipated over the resistor is $P_{\mathrm{el}}(t)=$ $R(t) I(t)^{2}$. By assumption (ii), we may assume that the thermal equilibrium is reached very fast, i.e. $P_{\mathrm{el}}(t)=P(t)$. The temperature $T(t)$ is therefore determined by the current via

$$
T(t)=T_{0}+\frac{R(t) I(t)^{2}}{\alpha}
\tag{3}
$$

If the resistance has value $R_{1}$, the current will increase, trying to reach $J_{1}=V / R_{1}$. The difference $I(t)-V / R_{1}$ will decay exponentially, with characteristic time $L / R_{1}$. The phase transition occurs once the critical current

$$
I_{1}=\sqrt{\frac{\alpha\left(T_{c}-T_{0}\right)}{R_{1}}}
$$

is reached. After the phase transition, the current will decrease, approaching the new equilibrium value $J_{2}=V / R_{2}$. Again, $I(t)-V / R_{2}$ will decay exponentially with characteristic time $L / R_{2}$, until the critical current

$$
I_{2}=\sqrt{\frac{\alpha\left(T_{c}-T_{0}\right)}{R_{2}}}
$$

is reached. This behaviour is shown in Fig. 1.

<img_4391>

Fig. 1

Together with (3), we see that the temperature behaves like in Figure 2.

<img_4335>

Fig. 2

The maximum and minimum temperatures will be attained just after the phase transitions occur. We obtain that

$$
\frac{T_{\max }-T_{0}}{T_{\min }-T_{0}}=\frac{R_{2} I_{1}^{2}}{R_{1} I_{2}^{2}}=\frac{R_{2}^{2}}{R_{1}^{2}}
\tag{4}
$$

Final answer: $\frac{R_{2}^{2}}{R_{1}^{2}}$
Subfield: Electromagnetism

A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$.

<img_4344>

This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}<V<V_{2}$, the temperature of the resistor starts oscillating. Assume that (i) the heat flux $P$ from the resistor to the ambient medium is given by $P=\alpha\left(T-T_{0}\right)$, where $\alpha$ is a constant, $T$ denotes the temperature of the resistor, and $T_{0}$ is the ambient temperature; (ii) the geometrical size of the resistor is so small that it will reach a thermal equilibrium much faster than the characteristic time $L / R_{2}$.
Context question:
(a) Express $V_{1}$ and $V_{2}$ in terms of the other parameters defined above.
Context answer:
\boxed{$V_{1}=\sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)}$ , $V_{2}=\sqrt{R_{2} \alpha\left(T_{c}-T_{0}\right)}$}


Context question:
(b) Assuming that $V_{1}<V<V_{2}$, sketch qualitatively how the temperature of the resistor $T$ depends on time $t$, and find the ratio $\left(T_{\max }-T_{0}\right) /\left(T_{\min }-T_{0}\right)$, where $T_{\max }$ and $T_{\min }$ denote the maximal and minimal values of $T$, respectively.
Context answer:
\boxed{$\frac{R_{2}^{2}}{R_{1}^{2}}$}


Question: (c) Find the period of oscillations if $V=\sqrt{V_{1} V_{2}}$ and $R_{2}=16 R_{1}$.

Solution:
If the phase transition occurs at $t=0$, with the resistance changing from $R_{j^{\prime}}$ to $R_{j}$, the current is given by

$$
I(t)=\frac{V}{R_{j}}+\left(I_{j^{\prime}}-\frac{V}{R_{j}}\right) \mathrm{e}^{-R_{j} t / L}
\tag{5}
$$

until the next phase transition occurs when $I\left(t_{j}\right)=I_{j}$. Hence, the period is

$$
t_{1}+t_{2}=\frac{L}{R_{1}} \ln \left(\frac{I_{2}-V / R_{1}}{I_{1}-V / R_{1}}\right)+\frac{L}{R_{2}} \ln \left(\frac{I_{1}-V / R_{2}}{I_{2}-V / R_{2}}\right)
\tag{6}
$$

Inserting the relations $R_{2}=\eta R_{1}$ and $V=\sqrt{V_{1} V_{2}}=$ $\eta^{1 / 4} \sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)}$, we obtain the period

$$
\begin{aligned}
\frac{L}{R_{1}} \ln \left(\frac{7}{4}\right)+\frac{L}{R_{2}} \ln (7)=\frac{L}{R_{1}}\left(\ln \left(\frac{7}{4}\right)\right. & \left.+\frac{1}{16} \ln (7)\right) \\
& \approx 0.68 \frac{L}{R_{1}} .
\end{aligned}
\tag{7}
$$

Final answer: $\frac{L}{R_{1}}(\ln (\frac{7}{4})+\frac{1}{16} \ln (7))$
$0.68 \frac{L}{R_{1}}$
Subfield: Electromagnetism


Question: (a) Find $\omega_{0}$ and the direction of $\vec{v}_{0}$, so that the center of mass will move with the constant velocity $\vec{v}=\vec{v}_{0}$ ?

Solution:
Lorentz forces acting on the charges:

$$
\begin{gathered}
\vec{F}_{+}=q \vec{v}_{+} \times \vec{B}=q(\vec{v}+\vec{\omega} \times \vec{r}) \times \vec{B} \\
\vec{F}_{-}=(-q) \vec{v}_{-} \times \vec{B}=(-q)(\vec{v}-\vec{\omega} \times \vec{r}) \times \vec{B}
\end{gathered}
$$

where $\vec{r}$ is a vector from the center of mass to the position of the positive charge.

![](https://cdn.mathpix.com/cropped/2023_12_21_dce611b73b4c47125558g-1.jpg?height=305&width=711&top_left_y=750&top_left_x=204)

According to Newton's first law, the center-of-mass $C$ of the dipole will move with constant velocity provided that the net force:

$$
\vec{F}=\vec{F}_{+}+\vec{F}_{-}=q\left(\vec{v}_{+}-\vec{v}_{-}\right) \times \vec{B}
$$

acting on the dipole, is zero. Since $\vec{v}_{+}, \vec{v}_{-}$and $\vec{B}$ are perpendicular, we require $\vec{v}_{+}=\vec{v}_{-}$. It means that dipole does not rotate: $\omega=\omega_{0}=0$.

The pure translation, however, is possible if the pair of forces $\vec{F}_{+}, \vec{F}_{-}$, has zero torque about $C$ :

$$
\begin{aligned}
& \vec{\tau}=\vec{r} \times \vec{F}_{+}-\vec{r} \times \vec{F}_{-}=2 q \vec{r} \times(\vec{v} \times \vec{B})= \\
& 2 q(\vec{v}(\vec{r} \cdot \vec{B})-\vec{B}(\vec{r} \cdot \vec{v}))=-2 q \vec{B}(\vec{r} \cdot \vec{v})
\end{aligned}
$$

We conclude that scalar product is zero only when $\vec{v} \perp \vec{r}$, i.e. the initial velocity should be parallel to $Y$ direction.

In summary, the dipole will move uniformly along $Y$ if, and only if, $\vec{v}_{0} \| Y$ and $\omega_{0}=0$.

Final answer: $\vec{v}_{0} \| Y$ , $\omega_{0}=0$