[User] Kaks ühesugust kuulikest, millest kumbki kannab laengut $q$, asuvad vertikaalsihis kaugusel $H$ üksteisest. Alumine kuulike on jäigalt kinnitatud, ülemine aga hakkab liikuma vertikaalselt alla suunatud algkiirusega $v$. Kui suur on minimaalne kaugus $h$ alumise kuulikeseni, millele suudab läheneda ülemine kuulike? Ülemise kuulikese mass on $m$. Raskuskiirendus on $g$.

Solution:
Energia jäävuse seaduse kohaselt
\[
mgH + \frac{mv^2}{2} + \frac{kq^2}{H} = mgh + \frac{kq^2}{h},
\]
kus peale gravitatsioonilise potentsiaalse energia arvestasime ka elektrilise potentsiaaliga. Niisiis
\[
h-\frac{v^{2}}{2 g}-H-\frac{k q^{2}}{m g H}+\frac{k q^{2}}{m g h}=0,
\]
\[
h^{2}-\left(\frac{v^{2}}{2 g}+\frac{k q^{2}}{m g H}+H\right) h+\frac{k q^{2}}{m g}=0.
\]
Lahendades saadud ruutvõrrandi, saame
\[
h=\frac{1}{2}\left(\frac{v^{2}}{2 g}+\frac{k q^{2}}{m g H}+H\right)-\sqrt{\frac{1}{4}\left(\frac{v^{2}}{2 g}+\frac{k q^{2}}{m g H}+H\right)^{2}-\frac{k q^{2}}{m g}}.
\]
Lahend \enquote{$+$}-märgiga ruutjuure ees oleks kaugusest $H$ suurem ja vastaks maksimaalsele kõrgusele, mille saavutaks kuulike, kui ta saaks samasuguse kuid ülespoole suunatud algkiiruse.

Kuulikesed

Kaks ühesugust kuulikest, millest kumbki kannab laengut $q$, asuvad vertikaalsihis kaugusel $H$ üksteisest. Alumine kuulike on jäigalt kinnitatud, ülemine aga hakkab liikuma vertikaalselt alla suunatud algkiirusega $v$. Kui suur on minimaalne kaugus $h$ alumise kuulikeseni, millele suudab läheneda ülemine kuulike? Ülemise kuulikese mass on $m$. Raskuskiirendus on $g$.

Kehtib energia jäävuse seadus, kus peame arvestama nii gravitatsioonilise kui ka elektrilise potentsiaalse energiaga.


[User]
In this problem we estimate the maximum firing speed of a human neuron. Model
    a human cell simply as a sphere of radius 10−6 m.

      (a) It has been measured that 1 cm2 of cell membrane has a resistance of 1000 Ω. Estimate the
          resistance of a single human cell.

     (b) Estimate the capacitance of a single human cell, treating the two sides of the membrane as
         capacitor plates. You will have to estimate the thickness of the cell membrane.

      (c) By modeling the cell as an RC circuit, estimate the maximum firing speed of a human neuron.
          Is this a reasonable result? If yes, how do you know? If not, how could this model be refined?

Solution:
In this problem we estimate the maximum firing speed of a human neuron. Model
    a human cell simply as a sphere of radius 10−6 m.

      (a) It has been measured that 1 cm2 of cell membrane has a resistance of 1000 Ω. Estimate the
          resistance of a single human cell.

     (b) Estimate the capacitance of a single human cell, treating the two sides of the membrane as
         capacitor plates. You will have to estimate the thickness of the cell membrane.

      (c) By modeling the cell as an RC circuit, estimate the maximum firing speed of a human neuron.
          Is this a reasonable result? If yes, how do you know? If not, how could this model be refined?

    Answer. (c) The RC timescale is on the order of 10 ms, which is reasonable.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
USAPhO 2015, problem A1. (The use of the WKB approximation in this problem
    is technically incorrect: the problem takes α = π when it actually should take α = 3π/2. Often,
    people will apply the WKB approximation in a sloppy way because they’re only after rough estimates
    at small n, or the limiting behavior at n ≫ 1. In the previous problems, we treated α properly

                   01T
    because doing so will give the exact correct answer in those cases.)

Solution:
USAPhO 2015, problem A1. (The use of the WKB approximation in this problem
    is technically incorrect: the problem takes α = π when it actually should take α = 3π/2. Often,
    people will apply the WKB approximation in a sloppy way because they’re only after rough estimates
    at small n, or the limiting behavior at n ≫ 1. In the previous problems, we treated α properly

                   01T
    because doing so will give the exact correct answer in those cases.)


[User]
Theoretical Question – T2
                                                                                Page 1 of 4

                          The Expanding Universe                               (Total Marks : 20)

The most outstanding fact in cosmology is that our universe is expanding. Space is
continuously created as time lapses. The expansion of space indicates that, when the
universe expands, the distance between objects in our universe also expands. It is
convenient to use “comoving” coordinate system 𝑟⃗ = (𝑥, 𝑦, 𝑧) to label points in our
expanding       universe,      in      which       the      coordinate       distance
Δ𝑟 = |𝑟⃗2 − 𝑟⃗1 | = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2 between objects 1 and 2 does not
change. (Here we assume no peculiar motion, i.e. no additional motion of those objects
other than the motion following the expansion of the universe.) The situation is illustrated
in the figure below (the figure has two space dimensions, but our universe actually has three
space dimensions).

The modern theory of cosmology is built upon Einstein’s general relativity. However, under
proper assumptions, a simplified understanding under the framework of Newton’s theory of
gravity is also possible. In the following questions, we shall work in the framework of
Newton’s gravity.

To measure the physical distance, a “scale factor” 𝑎(𝑡) is introduced such that the physical
distance Δ𝑟p between the comoving points 𝑟⃗1 and 𝑟⃗2 is

                                       Δ𝑟p = 𝑎(𝑡)Δ𝑟,

The expansion of the universe implies that 𝑎(𝑡) is an increasing function of time.

On large scales – scales much larger than galaxies and their clusters – our universe is
approximately homogeneous and isotropic. So let us consider a toy model of our universe,
which is filled with uniformly distributed particles. There are so many particles, such that
we model them as a continuous fluid. Furthermore, we assume the number of particles is

                                            Theoretical Question – T2
                                                                                Page 2 of 4

    conserved.

    Currently, our universe is dominated by non-relativistic matter, whose kinetic energy is
    negligible compared to its mass energy. Let 𝜌m (𝑡) be the physical energy density (i.e.
    energyper unit physical volume, which is dominated by mass energy for non-relativistic
    matter and the gravitational potential energy is not counted as part of the “physical energy
    density”) of non-relativistic matter at time 𝑡. We use 𝑡0 to denote the present time.

A   Derive the expression of 𝜌m (𝑡) at time 𝑡 in terms of 𝑎(𝑡), 𝑎(𝑡0 ) and 𝜌m (𝑡0 ).                 2 points

    Besides non-relativistic matter, there is also a small amount of radiation in our current
    universe, which is made of massless particles, for example, photons. The physical
    wavelength of massless particles increases with the universe expansion as 𝜆p ∝ 𝑎(𝑡). Let
    the physical energy density of radiation be 𝜌r (𝑡).

    Derive the physical energy density for radiation 𝜌r (𝑡) at time 𝑡 in terms of 𝑎(𝑡), 𝑎(𝑡0 ) and   2 points
B
    𝜌r (𝑡0 ).

    Consider a gas of non-interacting photons which has thermal equilibrium distribution. In
    this situation, the temperature of the photon depends on time as 𝑇(𝑡) ∝ [𝑎(𝑡)]𝛾 .

C   Calculate the numerical value of 𝛾.                                                              2 points

    Consider the thermodynamics of one type of non-interacting particle X. Note that the space
    expansion is slow enough and thermally isolated such that the entropy of X is a constant in
    time. Let the physical energy density of X be 𝜌X (𝑡), which includes mass energy and
    internal energy. Let the physical pressure be 𝑝X (𝑡).

D   Derive d𝜌X (𝑡)/d𝑡 in terms of 𝑎(𝑡), d𝑎(𝑡)/d𝑡,𝜌X (𝑡),and 𝑝X (𝑡).                                  4 points

                                           Theoretical Question – T2
                                                                                Page 3 of 4

    Consider a star S. At the present time 𝑡0 , the star is at a physical distance 𝑟p = 𝑎(𝑡0 )𝑟 away
    from us, where 𝑟 is the comoving distance. Here we ignore the peculiar motion, i.e. assume
    that both the star and us just follow the expansion of the universe without additional
    motion.

    The star is emitting energy in the form of light at power 𝑃e , which is isotropic in every
    direction. We use a telescope to observe its starlight. For simplicity, assume the telescope
    can observe all frequencies of light with 100% efficiency. Let the area of the telescope lens
    be 𝐴.

    Derive the power received by the telescope 𝑃r from the star S, as a function of 𝑟, 𝐴, 𝑃e , the
E   scale factor 𝑎(𝑡e ) at the starlight emission time 𝑡e , and the present (i.e. at the observation   4 points
    time) scale factor 𝑎(𝑡0 ).

    If there were no gravity, the expansion speed of the universe should be a constant. In
    Newton’s framework, this can be understood as that, without force, matter just moves
    away from each other with constant speed and thus d𝑎(𝑡)/d𝑡 is a constant depending on
    the initial condition.

    Let us now consider how Newton’s gravity affects the scale factor 𝑎(𝑡), in a universe filled
    with non-relativistic matter in a homogeneous and isotropic way.

    As illustrated in the above figure, let us assume C is the center of our universe (this
    assumption can be removed in Einstein’s general relativity, which is beyond the scope of
    this question). We slice matter into thin shells around C. Let us focus on one thin shell (the
    sphere in the above figure) whose comoving distance from the center is 𝑟 (recall that this
    comoving distance is a constant in time).

                                            Theoretical Question – T2
                                                                                Page 4 of 4

      Use the motion of the shell to find a relation between d𝑎(𝑡)/d𝑡, 𝑎(𝑡) and the density of
F     mass energy 𝜌(𝑡). (In the final relation, if you encounter a constant depending on the        5 points
      initial condition, it can be kept as it is.)

      Based on the model described in Part (F), is the expansion of the universe
G                                                                                1 points
      (a) accelerating or (b) decelerating? Choose from (a) or (b).

For your information, in 1998, a new type of energy component of our universe is discovered. It actually
changes the conclusion in Part (G).

Solution:
Marking Scheme – T2
                                                                                Page 1 of 4

                                                                                (Full Marks: 20)
Part                                              Model Answer                                           Marks

 A     The physical volume is                                                                              2

                                         𝑉p = 𝑎3 (𝑡)𝑉. (0.5 points).

       The comoving number density is a constant, thus the physical number density is

                                        𝑛(𝑡)        𝑎(𝑡 ) 3
                                              = ( 𝑎(𝑡)0 ) . (0.5 points)
                                        𝑛(𝑡 )
                                            0

       The kinetic energy for non-relativistic particles are negligible, thus the energy density is

                                       𝜌m (𝑡) = 𝑚 𝑛(𝑡), (0.5 points)

       where 𝑚 is the mass of a particle.

       Thus

                                                         𝑎(𝑡 ) 3
                                  𝜌m (𝑡) = 𝜌m (𝑡0 ) ( 𝑎(𝑡)0 ) (0.5 points)

       [Remarks: It is acceptable if the student just writes 𝝆𝐦 ∝ 𝟏/𝒂𝟑 and full points will be given.]

 B     The Einstein’s energy relation for a massless particle is                                           2

                                                𝐸 = 𝑝𝑐. (0.5 points)

       From de Brogile’s relation:

                                      𝑝 ∝ 1/𝜆P ∝1/a(t). (0.5 points)

       [Remarks: No point if only 𝝀𝐩 ∝ a(t) is written because already given.]

       Thus

                                            𝐸 ∝1/a(t). (0.5 points)

       Physical number density is 𝑛 ∝ 1/𝑎3 .

       Energy density is 𝑛 𝐸.

       Thus

                                                         𝑎(𝑡 ) 4
                                   𝜌r (𝑡) = 𝜌r (𝑡0 ) ( 𝑎(𝑡)0 ) (0.5 points)

       [Remarks: It is acceptable if the student just write 𝝆𝐫 ∝ 𝟏/𝒂𝟒 .]

                                            Marking Scheme – T2
                                                                                Page 2 of 4

C   The photons in thermal equilibrium satisfy Boltzmann distribution                               2
                                                        𝐸(𝑎)
                                                    −
                                   𝑛(𝐸(𝑎)) ∝ 𝑒 𝑘B 𝑇(𝑎) , (1 point)

    where 𝐸 ∝1/a(t).

    Condition of being non-interacting implies that there is no energy transfer. Thus the energy
    distribution must be stable.

    To be explicit, for two different comoving wavelengths,

                             𝑛(𝐸1 (𝑎))
                                         = 𝑒 [𝐸2 (𝑎)−𝐸1 (𝑎)]/[𝑘B 𝑇(𝑎)] = const.
                             𝑛(𝐸2 (𝑎))

    [Remarks: All the above steps can be replaced by the intuition of 𝑬 ∝ 𝑻, if the students
    realize it, the above 1 point can be given.]

    Thus

                                𝑇(𝑎) ∝ 1/𝑎, i.e. 𝛾 = −1. (1 point)

D   The 1st law of thermodynamics is                                                                4

                                         d𝐸X = −𝑝X d𝑉p . (1 point)

    Here no entropy term appears, because 𝑆 =const. No chemical potential appears, because
    of particle number conservation.

    Here 𝑉p = 𝑎3 𝑉.

                                         d𝑉p = 3𝑎2 𝑉d𝑎. (1 point)

                                         𝐸X = 𝜌X 𝑉p . (0.5 points)

                       d𝐸X = d(𝜌X 𝑉p ) = 𝑎3 𝑉d𝜌X + 3𝜌X 𝑎2 𝑉d𝑎. (0.5 points)

    Thus

                                            d𝑎
                             d𝜌X + 3 ( 𝑎 )(𝜌X + 𝑝X ) = 0. (0.5 points)

                                           𝑎̇
                               𝜌̇ X + 3 (𝑎)(𝜌X + 𝑝X ) = 0. (0.5 points)

    [Remarks: 0.5 point for relating variation and time derivative no matter in which step it is
    being used.]

                                              Marking Scheme – T2
                                                                                Page 3 of 4

E   With lens area 𝐴, we only receive part of the starlight. The area ratio is                       4

                                     𝐴/(4𝜋𝑎2 (𝑡0 )𝑟 2 ). (1 point)

    The wavelength of each photon emitted from the star gets stretched. Thus energy per
    photon is lowered, contributing a ratio

                                          𝑎(𝑡e )/𝑎(𝑡0 ). (1 point)

    The separation among the photons also increases due to cosmic expansion, contributing a
    ratio

                                          𝑎(𝑡e )/𝑎(𝑡0 ). (1 point)

    As a result, the power that the telescope receives is

                                               𝐴 𝑎2 (𝑡 )
                                   𝑃r = 4𝜋𝑎4 (𝑡 e)𝑟 2 × 𝑃e . (1 point)
                                                         0

F   The kinetic energy and gravitational energy of the shell adds up to a constant:                  5
                                          1                        𝐺𝑀𝑚
                                 𝐸 = 2 𝑚 (𝑟̇p )2 −                           , (2 points)
                                                                        𝑟p

    where
                                                    4𝜋         𝜌
                                         𝑀 = 3 𝑟p 3 𝑐 2, (1 point)

    (Note: energy conservation without evolving pressure requires the assumption of non-
    relativistic matter.)

                                           𝑟p = 𝑎(𝑡)𝑟, (1 point)

    [Remarks: The point is given because the student understand that the shell is not pulled
    gravitationally from the outside, because the force due to the mass outside cancels.]

    Thus
                                    2𝐸                       8𝜋𝐺
                                          = 𝑎̇ 2 − 3𝑐 2 𝜌𝑎2 . (1 point)
                                   𝑚𝑟 2

    Alternative Solution:

    For the gravitational force due to the mass inside:
                                          𝐺𝑀𝑚                      4𝜋
                            𝑚 𝑟̈p = −         𝑟p2
                                                     = − 3𝑐 2 𝐺𝑚𝜌𝑟p , (2 points)

    where 𝑚 is mass of shell.

                                             Marking Scheme – T2
                                                                                Page 4 of 4

                                             𝑟p = 𝑎(𝑡)𝑟, (1 point)

    [Remarks: The point is given because the student understand that the shell is not pulled
    gravitationally from the outside, because the force due to the mass outside cancels.]

    and 𝜌 = 𝜌(𝑡0 )𝑎3 (𝑡0 )/𝑎3 (𝑡).

    Thus
                                            4𝜋
                                𝑎̈ = − 3𝑐 2 𝐺𝜌(𝑡0 )𝑎3 (𝑡0 )𝑎−2 . (1 point)

    Integrate the above equation. One gets

                            1         4𝜋𝐺                1     4𝜋𝐺
                       𝑐 = 2 𝑎̇ 2 − 3𝑐 2 𝜌(𝑡0 )𝑎−1 = 2 𝑎̇ 2 − 3𝑐 2 𝜌𝑎2 , (1 point)

    where 𝑐 is an integration constant.

    (b) decelerating. This is because gravity is attractive for the matter that we are considering
G   here. As a result, d𝑎(𝑡)/d𝑡 is a decreasing function of 𝑡.                                        1

    Appendix: Notes about the physics behind this set of problems:                                   N/A

    To reduce students’ reading load, we have not mentioned in the problems, that those
    problems set up the framework of researches in modern cosmology:

    A theory of gravity (especially Einstein’s general relativity) contains two aspects: Gravity
    tells matter how to move (kinematics of matter motion in a gravitational field); and matter
    determines the gravitational field (dynamics of the gravitational field). Parts (A)-(E) are
    about kinematics and part (F) is about dynamics in this sense. The two key equations in
    cosmology are derived in part (D) (this is known as the continuity equation, containing
    parts (A) and (B) as special cases) and (F), upon which the whole theory of modern
    cosmology is built.

    The equation derived in part (F) is known as the Friedmann equation, which is
                                     𝑎̇ 2    𝑘    8𝜋𝐺
    conventionally written as (𝑎) − 𝑎2 =                𝜌 . This equation governs the dynamics of
                                                   3
    cosmic expansion and actually not only applies for non-relativistic matter but also for
    general matter components (which needs general relativity to derive). The constant 𝑘 is
    related to the curvature of 3-dimensional space, which is observed to be vanishingly small.

    Part (C) indicates that the universe was hotter at earlier ages. The hot universe in local
    thermal equilibrium determines the whole thermal history of our universe, which answers
    questions such as where the light elements come from, and when the universe becomes
    transparent for light. Part (E) defines the luminosity distance, which relates the telescope
    observations to the cosmic reality.

APhO 2016 Theory T2


[User]
The acceleration due to gravity can be measured by measuring the time period of a
    simple pendulum. However, it can be challenging to get an accurate result.

     (a) Suppose you constructed a pendulum using regular household materials. Name at least five
         sources of possible experimental error in your calculated value of g. How would you make the
         pendulum and perform the measurements to minimize these sources of error?

     (b) Make an actual pendulum yourself and carry out the measurement. Describe your experimental
         procedure and show your data. Estimate as many of the sources of experimental error identified
         in part (a) as you can, and using them, give a value of g with a reasonable uncertainty.

        (c) If you had $1,000 and a week to do plenty of measurements, how would you go about it?
            How precise a result do you think you could get? What would be the dominant sources of
            uncertainty remaining?

Solution:
The acceleration due to gravity can be measured by measuring the time period of a
    simple pendulum. However, it can be challenging to get an accurate result.

      (a) Suppose you constructed a pendulum using regular household materials. Name at least five
          sources of possible experimental error in your calculated value of g. How would you make the
          pendulum and perform the measurements to minimize these sources of error?

     (b) Make an actual pendulum yourself and carry out the measurement. Describe your experimental
         procedure and show your data. Estimate as many of the sources of experimental error identified
         in part (a) as you can, and using them, give a value of g with a reasonable uncertainty.

        (c) If you had $1,000 and a week to do plenty of measurements, how would you go about it?
            How precise a result do you think you could get? What would be the dominant sources of
            uncertainty remaining?


[User]
01W   USAPhO 1997, problem A1.

      Idea 4
      When a musical instrument plays a note, typically multiple standing waves are excited, so
      the resulting sound is composed of multiple frequencies. As you saw in problem 9, often the
      standing wave frequencies are all multiples of a single, lowest frequency. This frequency f0 is
      called the fundamental, or first harmonic, while the multiple nf0 is called the nth harmonic.
      The fundamental frequency determines the pitch we perceive, while the distribution of energy
      among the harmonics determines the timbre, or tonal quality, of the instrument.

Solution:
01W   USAPhO 1997, problem A1.

      Idea 4
      When a musical instrument plays a note, typically multiple standing waves are excited, so
      the resulting sound is composed of multiple frequencies. As you saw in problem 9, often the
      standing wave frequencies are all multiples of a single, lowest frequency. This frequency f0 is
      called the fundamental, or first harmonic, while the multiple nf0 is called the nth harmonic.
      The fundamental frequency determines the pitch we perceive, while the distribution of energy
      among the harmonics determines the timbre, or tonal quality, of the instrument.


[User]
6. Two blocks, say B and C, each of mass m are connected by a light spring of force
   constant k and natural length L. The whole system is resting on a frictionless table
   such that xB = 0 and xC = L, where xB and xC are the coordinates of the blocks B
   and C respectively. Another block (named A) of mass M , which is travelling at speed
   V0 collides head-on with the block B at an instant t = 0 (see Fig. (3)).
                                                               [ 3 + 2 + 2 + 3 = 10]

  CS
                             V0

                            A                B            C

                                      Figure 3:

   (a) Obtain the velocities of blocks A, B and C just after the collision at t = 0? Ex-
       press the velocities in terms of V0 and γ = m/M . Assume that the collision is
       elastic.

                                       Page 16

INPhO - 2011                                         Roll Number:P 1 1 0 0 0

        VA =

        VB =

        VC =

  (b) Draw free body diagrams for the blocks B and C after the collison and write down
      the equation of motion.
       Free body diagram:

                                                   E
HB
  CS    Equation of motion:

   (c) For t > 0 the positions of the blocks are given by

                                     xB = αt + β sin(ωt)
                                 xC = L + αt − β sin(ωt)
                                                                                (1)
                                                                                (2)

       Find ω in terms of m and k. Express α and β in terms of V0 , γ and ω. (For this
       part, ignore the motion of block A.)

                                       Page 17

INPhO - 2011                                     Roll Number:P 1 1 0 0 0

                                               E
HB
       ω=

       α=

       β=

  (d) Obtain the condition on γ such that the block A will collide with the block B
      again at some time t > 0?

  CS
                                    Page 18

INPhO - 2011                                          Roll Number:P 1 1 0 0 0

                                                   E
HB       Condition on γ:

Solution:
6. (a) VA =         V0
              1+γ
                2V0
         VB =
              1+γ
         VC = 0
   (b) See Fig. (2) for free body diagrams.
       Equations of motion:
               2                                2

         m ddtx2B =        −FB and m ddtxC = FC     2

                                                             Page 2

                                                    N                   N

                                     FB             B                   C             FC

                                                                                E
                                                    mg                  mg

                                                    (a)                 (b)

Figure 2: (a) Free body diagram for the block B . Here FB = k (L                           − (xC − xB )) (b) Free body
diagram for the block C . Here FC = k (L (xC xB )).       −       −
                   

                                                            S
                 2k
      (c) ω =
                 m
                V0
          α=
              1+γ
                  V0
          β=
              ω(1 + γ )
      (d) 0 < γ < 2/3π

INPhO 2011 Problem 6


[User]
Here we’ll introduce a simple model for a ferromagnetic phase transition. Consider
    N electrons, which may have spins si = ±1. The energy of a configuration is
                                                X              J X
                                       E = −B          si −        si sj .
                                                              2N
                                                  i               i̸=j

    The first term represents the effect of an external magnetic field B, while the second term represents
    an interaction, with strength described by the constant J, which tries to make the spins parallel.
    (In this simple model, we suppose all distinct pairs of spins interact equally. We could also make
    spins only interact with their neighbors, but this would complicate the analysis.)
                                                      P
      (a) Define the average magnetization as m = i si /N . Find E(m), the energy in terms of m and
          the other constants in the problem.

     (b) For a fixed value of m, write down the number of states Ω(m) with that magnetization.

      (c) The probability of having a given value of m is proportional to e−βE(m) Ω(m). Argue that this
          probability is maximized for the value of m that minimizes the free energy

                                                      F = E − T S.

          Hence the equilibrium configuration minimizes the free energy. This is the statistical mechan-
          ical way to argue that F is minimized; the thermodynamic way was covered in T2.

     (d) Assuming that N ≫ 1 and using the approximation for log N ! mentioned in T2, show that
         the free energy F (m) is minimized when

                                       m = tanh(βB + βJm),               β=        .
                                                                              kB T

        (e) For a fixed B > 0, plot m(T ). This should match with Curie’s law, which you proved in T1.
        (f) Now let B = 0. Show that there exists a critical temperature Tc , above which m(T ) is exactly
            zero and below which it is nonzero; also find an approximate expression for the magnetization
            just below Tc .
   This is a phase transition where the material spontaneously becomes magnetized, and the simplest

                      01h
   example of a phase transition which can be understood analytically.

Solution:
(a) The first term is easy,
                                                         X
                                                 −B           si = −N mB.
                                                         i

     To do the second term, we note that
                                                                            !2
                  X                XX                X             X                 X
                         si sj =           si sj −       si si =       si        −       s2i = N 2 m2 − N.
                  i̸=j             i   j             i             i                 i

     We hence have
                                                       1       1
                                            E = −N mB − JN m2 + J.
                                                       2       2
 (b) Of all N electrons, N+ have si = +1 and N− have si = −1, where N+ + N− = N and
     mN = N+ − N− . Thus N+ = N/2 + mN/2 and N− = N/2 − mN/2. The number of states is
     N choose N+ .
                                                 N!
                               Ω(m) = N           N         .
                                        2 (1 + m) ! 2 (1 − m) !

 (c) The probability P (m) = e−βE(m) Ω(m) is found with Boltzmann’s theorem, where each state
     has a probability of e−E/kB T , so β = 1/kB T . Maximizing P (m) is the same as maximizing
     log(P (m)), which gives
                                                     E
                                    log(P (m)) = −      + log(Ω(m)).
                                                   kB T
     Since we’re optimizing with respect to m, we can multiply both sides by −kB T and maximize
     P (m) by minimizing −kB T log(P (m)).

                              −kB T log(P (m)) = E − T kB log(Ω(m)) = E − T S.

     This is just the expression for free energy, thus minimizing free energy for a given value of m
     will maximize the probability.

 (d) Using log N ! ≈ N log N − N gives
                                    !
                      N!                      N          N (1 + m) N          N (1 − m)
     log   N
                      N          ≈ N log N − (1+m) log          − (1−m) log           .
            2 (1 + m) ! 2 (1 − m) !
                                              2              2     2              2

    To simplify this, we can expand the logarithms to get

                           1+m                1+m       1−m              1−m
             N log N −            log N + log         −      log N + log
                             2                  2         2               2

    and various factors cancel to give

                                   1+m     1+m 1−m     1−m
                            −N         log    +    log       .
                                     2      2   2       2

    Setting dF/dm to zero gives
                                                              ∂S
                                      0 = −N B − N Jm − T
                                                              ∂m
    which is equivalent to

                                   ∂ log Ω                  1     1+m 1    1−m
                N (B + Jm) = −kB T         = N kB T           log    − log             .
                                     ∂m                     2      2  2     2

    Then we have
                                              1     1+m
                               βB + βJm =       log     = tanh−1 m
                                              2     1−m
    which gives the desired result.

 (e) The graph should look something like this:

 (f) With B = 0, for the magnetization to be nonzero, we need a non-zero solution to

                                                    Jm
                                        m = tanh          .
                                                   kB T

    Since the second derivative of tanh(ax), −2a2 sech2 (ax) tanh(ax) is always negative for x > 0,
    that means that the slope of tanh(ax) will always be decreasing for x > 0. In order for the
    graphs of x and tanh(ax) to intersect again after x = 0, the slope of tanh(ax) must be greater
    than the slope of x, which is 1. Thus the critical value is where a = 1, so
                                                      J
                                              Tc =      .
                                                     kB

          To find what happens just below Tc , we Taylor expand the hyperbolic tangent about zero,
          which is useful because m will be small,
                                                            x3
                                            tanh(x) = x −      + O(x5 ).

          Letting T = (1 − ϵ)Tc , we have Jm/kB T = m/(1 − ϵ), giving

                                                     m    1 m3
                                              m≈        −
                                                   1 − ϵ 3 (1 − ϵ)3
          which means that at lowest order at epsilon,
                                              √        3(Tc − T ) 1/2

                                        m ≈ 3ϵ =                      .
                                                          Tc


[User]
A cue ball is a uniform sphere of radius R.

     (a) Find the height at which the cue ball must be hit horizontally so that it immediately begins
         rolling without slipping.

     (b) Skillful players can hit the cue ball so that it begins moving forwards, but then ends up moving
         backwards. Model the hit as an instantaneous impulse applied at an arbitrary point on the
         back half of the cue ball, in an arbitrary direction. For what impulses will this trick work?
         You can treat the situation as two-dimensional; justify your answer carefully.

Solution:
A cue ball is a uniform sphere of radius R.
     (a) Find the height at which the cue ball must be hit horizontally so that it immediately begins
         rolling without slipping.
     (b) Skillful players can hit the cue ball so that it begins moving forwards, but then ends up moving
         backwards. Model the hit as an instantaneous impulse applied at an arbitrary point on the
         back half of the cue ball, in an arbitrary direction. For what impulses will this trick work?
         You can treat the situation as two-dimensional; justify your answer carefully.
    Answer. (a) 2R/5, (b) the line of the impulse passes under the contact point with the ground


[User]
3. Consider a bicycle in vertical position accelerating forward without slipping on a
   straight horizontal road. The combined mass of the bicycle and the rider is M and
   the magnitude of the accelerating torque applied to the rear wheel by the pedal and
   gear system is τ . The radius and the moment of inertia of each wheel is R and I
   (with respect to axis) respectively. The acceleration due to gravity is g.    [Marks: 8]
   (a) Draw the free body diagram of the system (bicycle and rider).                       [2]

                          R
                                         E           R

                    CS
                       Wheel 2                  Wheel 1

   (b) Obtain the acceleration a in terms of the above mentioned quantities.               [2]

HB      a=

   (c) For simplicity assume that the centre of mass of the system is at height R from
       the ground and equidistant at 2R from the center of each of the two wheels. Let
       µ be the coefficient of friction (both static and dynamic) between the wheels and
       the ground. Consider M ≫ I/R2 and no slipping. Obtain the conditions for the
       maximum acceleration am of the bike.                                                [3]

                                       Page 5

INPhO - 2013                                Roll Number:P 1 3 0 0 0

                                       E
                       CS
       am =

  HB
  (d) For µ = 1.0 calculate am .

       am =
                                                                      [1]

                                   Page 6

INPhO - 2013                                          Roll Number:P 1 3 0 0 0

Solution:
3. (a) Here f1 , f2 are frictional forces and N1 , N2 are normal reactions.

                        R                                R

                    N                                N
                2                       mg           1
           Wheel 2                  f       f            Wheel 1

  HB
                                    2        1
                                τ
     (b) a =                            R
                    M R2 + 2I
               µg/2
      (c) a ≤    µ
              1−
                  4
     (d) am = 2g/3

             R       R

INPhO 2013 Problem 3


[User]
USAPhO 2016, problem B2.

       Remark
       We noted in M7 that clouds are visible because the radiation scattered by a small droplet of
       n water molecules grows as n2 . To understand why, note that each of the molecules performs
       independent Rayleigh scattering, as computed above. For separated molecules, the energy
       scattered just adds. However, for nearby molecules the electromagnetic waves scattered
       interfere constructively, so the amplitude grows as n and hence the energy scattered as n2 .

       This quadratic enhancement breaks down once the droplets exceed the wavelength λ of the
       light. This means the maximum possible enhancement is larger for larger wavelengths, acting
       against the ω 4 dependence of Rayleigh scattering. This is why clouds are white, not blue.

    Radiation pressure can also have mechanical effects.

       Example 5: NBPhO 2018.6
                                                                                √
       A laser pointer of power P is directed at a glass cube, with refractive index n > 2. The
       surface of the cube has an anti-reflective coating, so there is no partial reflection when light
       enters or exits it; the laser pointer only refracts. What is the maximum force the laser pointer
       can exert on the cube?

       Solution
       The force is due to a change in momentum of the light. The greatest possible force is attained
       if the direction of the light is reversed, which can occur as shown, in the limit α → 90◦ .

                      √
       Assuming n > 2, we then have θr < 45◦ , and then the laser internally reflects when it hits
       the top surface of the cube. It exits in the opposite direction it came in.

       If the laser pointer has power P , then the momentum of the laser beam per time is P/c. The
       momentum is reversed, so the force is 2P/c.

[3] Problem 24 (IZhO 2022). In 2018, the Nobel Prize in physics was awarded to Arthur Ashkin
    for the creation of the “laser tweezer”, a device that allows one to hold and move transparent
    microscopic objects with the help of light. In one such device, a parallel beam of light from a laser
    passes through a converging lens L and hits a microparticle M , which can also be considered a
    converging lens. Point F is the common focus of L and M .

    The light intensity in the beam is I = 1.00 µW/cm2 , the beam radius is R = 1.00 cm, and the focal

    length of the lens L is F = 10.0 cm. Ignore the absorption and reflection of light.

      (a) Calculate the force acting on the microparticle, in the setup shown at left above.

      (b) Next, the left half of the lens L is covered by a diaphragm, as shown at right above. Calculate
          the force acting on the microparticle in the transverse direction of the beam.

[3] Problem 25 (Feynman). In one proposed means of space propulsion, a spaceship of mass 103 kg
    carries a thin sheet of area 100 m2 . The sheet is made of highly reflective plastic film, and can be
    used as a solar radiation pressure “sail”. The spaceship travels in a circular orbit of radius r, which
    is initially equal to the Earth’s orbit radius, where the intensity of sunlight is 1400 W/m2 . Assume
    the spaceship is moving nonrelativistically and the gravitational effect of the Earth is negligible.

      (a) Find the angle at which the sail should be pointed to maximize dr/dt.

      (b) Assuming the sail is pointed this way, find the numeric value of dr/dt.

      (c) If this continues for a very long time, then r will grow as r ∝ tn . Find the value of n.

    Finally, we’ll consider electromagnetic wave propagation in transmission lines.

[4] Problem 26 (Griffiths 7.62, Crawford 4.8). A certain transmission line is constructed from two
    thin metal ribbons, of width w, a very small distance h ≪ w apart. The current travels down one
    strip and back along the other. In each case it spreads out uniformly over the surface of the ribbon.

      (a) Find the capacitance per unit length C, and the inductance per unit length L.

      (b) Argue that the√ speed of propagation of electromagnetic waves through this transmission line
          is of order 1/ LC, and evaluate this quantity.

      (c) Repeat the first two parts for a coaxial transmission line, consisting of two cylinders of radii
          a < b with the same axis of symmetry.

     (d) Repeat the first two parts for a parallel-wire transmission line, consisting of two wires of radius
         r whose axes are a distance D ≫ r apart.
                                         √
    You should find that in all cases, 1/ LC is the same, yielding the same speed for electromagnetic
    waves. This isn’t a coincidence, and applies for transmission lines with conductors of any shape,
    though the general proof requires some elaborate vector calculus, as you can see here.

Solution:
See the official solutions here.

[3] Problem 25 (Feynman). In one proposed means of space propulsion, a spaceship of mass 103 kg
    carries a thin sheet of area 100 m2 . The sheet is made of highly reflective plastic film, and can be
    used as a solar radiation pressure “sail”. The spaceship travels in a circular orbit of radius r, which
    is initially equal to the Earth’s orbit radius, where the intensity of sunlight is 1400 W/m2 . Assume
    the spaceship is moving nonrelativistically and the gravitational effect of the Earth is negligible.

      (a) Find the angle at which the sail should be pointed to maximize dr/dt.

      (b) Assuming the sail is pointed this way, find the numeric value of dr/dt.

      (c) If this continues for a very long time, then r will grow as r ∝ tn . Find the value of n.

    Solution. (a) We have that E = −GM m/2r, so increasing radius as fast as possible is the same
         as imparting the most energy to the spaceship; in other words, we want to maximize F · v. The
         velocity is almost purely tangential, so only the tangential component of the force contributes
         to the power.
          Let the normal vector of the sail be at an angle θ to the radial direction. To find the net
          force on the sail, we can think of the reflection process in two steps: absorbing the light and
          reemitting it. Absorbing the light yields an outward radial force F . Reemitting it yields a
          force F of the same magnitude, directed at an angle 2θ to the radial direction.
          Only the second force can contribute to the power, so

                                               P = F · v = F v sin 2θ.

          Finally, since the radial area covered by the sail is A cos θ, we have F = I(A cos θ)/c, so
                                                       2IAv
                                                 P =        sin θ cos2 θ.
                                                         c

                                                                         √
          Setting the derivative to zero, the maximum is at θ = sin−1 (1/ 3) = 35.3◦ , giving power

                                                   IAv         4
                                        Pmax = β       ,   β = √ ≈ 0.77.
                                                    c         3 3
          Strictly speaking, the answer is very slightly different, by corrections of order v/c, since the
          spaceship is moving, but we’ll neglect this here.

     (b) The power is
                                                     dE   GM m dr
                                               P =      =         .
                                                     dt    2r2 dt
          We also know from force balance that
                                                   GM m   mv 2
                                                        =
                                                    r2     r
          since the radial force from the sail is negligible, and combining these results gives
                                              dr   βIA 2r   2βIA
                                                 =        =
                                              dt    c mv    mωc
          where ω = 2π/(1 year) is the angular velocity of the Earth. Plugging in the numbers gives
          dr/dt = 4 m/s.

      (c) We have
                                            dr   Ir   (1/r2 ) r 1
                                               ∝    ∝    √ =√ .
                                            dt   v     1/ r      r
          Separating and integrating gives t ∝ r3/2 in the long run (i.e. when the contribution from the
          initial condition is negligible), so n = 2/3.

    Finally, we’ll consider electromagnetic wave propagation in transmission lines.

[4] Problem 26 (Griffiths 7.62, Crawford 4.8). A certain transmission line is constructed from two
    thin metal ribbons, of width w, a very small distance h ≪ w apart. The current travels down one
    strip and back along the other. In each case it spreads out uniformly over the surface of the ribbon.

      (a) Find the capacitance per unit length C, and the inductance per unit length L.

     (b) Argue that the√ speed of propagation of electromagnetic waves through this transmission line
         is of order 1/ LC, and evaluate this quantity.

      (c) Repeat the first two parts for a coaxial transmission line, consisting of two cylinders of radii
          a < b with the same axis of symmetry.

     (d) Repeat the first two parts for a parallel-wire transmission line, consisting of two wires of radius
         r whose axes are a distance D ≫ r apart.
                                         √
    You should find that in all cases, 1/ LC is the same, yielding the same speed for electromagnetic
    waves. This isn’t a coincidence, and applies for transmission lines with conductors of any shape,
    though the general proof requires some elaborate vector calculus, as you can see here.

    Solution. Suppose the length is ℓ.

     (a) We have C = ϵ0 wℓ/h, so C = ϵ0 w/h. Similarly, L = µw0 h ℓ, so L = µ0 h/w.
                                                        √
     (b) Note that the
                    √  time scale
                              √   of this LC circuit is   LC. Thus, the transmission velocity is on the
         order of ℓ/ LC = 1/ LC = c. Alternatively, this is easily implied by dimensional analysis.

     (c) The capacitance of two coaxial cylinders can be found by giving the inner cylinder a charge
         Q and using C = Q/V , where
                                        Z            Z b
                                                         (Q/ℓ)       Q/ℓ
                                  V =       E ds =              dr =      log(b/a)
                                                       a 2πϵ0 r      2πϵ0

         and therefore
                                                             2πϵ0
                                                     C=             .
                                                           log(b/a)
         To find the inductance for a transmission line setup, the current will flow parallel to the axis,
         so by Ampere’s law the field inside the region is B = µ0 I/2πr, and using L = Φ/I where Φ
         will be the flux going around the inner cylinder will give
                                            Z b
                                                  µ0 I      µ0 Iℓ      b
                                       Φ=       ℓ      dr =       log
                                             a    2πr        2π        a

         and therefore
                                                      µ0         b
                                                   L=    log        .
                                                      2π         a
                       √
         Thus we get 1/ LC = c.

     (d) The capacitance can be found similarly (factors of 2 appear because the negative charge/opposing
         current will contribute to the E and B fields too),
                                             Z D
                                                   2Q/ℓ                    πϵ0
                                      V =                 dr,    C=              .
                                              r    2πϵ0 r               log(D/r)
                                            Z D
                                                   2µ0 I           µ0            D
                                    ΦB =                 dr,    L=    log                .
                                             r     2πr             π             r
                      √
         This gives 1/ LC = c.


[User]
Kaubaauto, pikkusega $l = \SI{10}{m}$, möödub kiirusega $v = \SI{92}{km/h}$ paigalseisvast pilukatikuga varustatud fotoaparaadist. Fotoaparaat teeb kaubaautost pildi, kusjuures säriaeg on $t_s = 1/1500\si{s}$. Negatiivil on kaubaauto kujutise pikkus $d = \SI{32}{mm}$. Negatiivi suurus on $36\times\SI{24}{mm}$. Kui palju lühem või pikem oleks kaubaauto kujutis, kui oleks pildistatud paigalseisvat autot? Aeg, mille jooksul katik läbib kaadri, on $t_k = 1/30\si{s}$. Kui suure pildi saab sellest negatiivist valmistada, kui me tahame, et häguste piirjoonte laius pildil ei ületaks $\delta = \SI{0,1}{mm}$?

\emph{Vihje.} pilukatikuks nimetatakse varje-ekraani, milles on kindla laiusega (kitsas) pilu (kasutatakse peegelkaamerates). See pilu liigub filmi pinna vahetus läheduses kujutise eest läbi. Seega langeb negatiivi igale punktile valgust ainult siis, kui pilu on antud punktiga kohakuti. Säriaeg (aeg, mille jooksul antud punkt saab valgust) sõltub pilu laiusest ja liikumiskiirusest. Eeldagem, et tegemist on horisontaalse pilukatikuga, s.t. pilu liigub kas vasakult paremale või paremalt vasakule

Solution:
Tähistame $L=\SI{36}{mm}$. Aeg, mille jooksul katik läbib kujutise pikkuse on ligikaudu $t=d t_{k} / L$. Auto kujutis negatiivil liigub kiirusega $(d / l) v$. Aja $t$ jooksul jõuab kujutis edasi nihkuda
$$
\Delta d=\frac{d}{l} v t=\frac{d^{2} v t_{k}}{l L} \approx \SI{2,4}{mm}
$$

võrra. Liikuva auto kujutis on $\Delta d$ võrra lühem või pikem sõltuvalt katiku liikumise suunast.
Filmi iga punkti valgustatakse $t_{s}$ jooksul, seega kujutise teravus on
$$
\delta_{0}=\frac{d}{l} v t_{s}=\SI{0,054}{mm}.
$$
Negatiivi võib suurendada $\delta / \delta_{0}=\num{1,8}$ korda.

Kaubaauto

Kaubaauto, pikkusega $l = \SI{10}{m}$, möödub kiirusega $v = \SI{92}{km/h}$ paigalseisvast pilukatikuga varustatud fotoaparaadist. Fotoaparaat teeb kaubaautost pildi, kusjuures säriaeg on $t_s = 1/1500\si{s}$. Negatiivil on kaubaauto kujutise pikkus $d = \SI{32}{mm}$. Negatiivi suurus on $36\times\SI{24}{mm}$. Kui palju lühem või pikem oleks kaubaauto kujutis, kui oleks pildistatud paigalseisvat autot? Aeg, mille jooksul katik läbib kaadri, on $t_k = 1/30\si{s}$. Kui suure pildi saab sellest negatiivist valmistada, kui me tahame, et häguste piirjoonte laius pildil ei ületaks $\delta = \SI{0,1}{mm}$?

\emph{Vihje.} pilukatikuks nimetatakse varje-ekraani, milles on kindla laiusega (kitsas) pilu (kasutatakse peegelkaamerates). See pilu liigub filmi pinna vahetus läheduses kujutise eest läbi. Seega langeb negatiivi igale punktile valgust ainult siis, kui pilu on antud punktiga kohakuti. Säriaeg (aeg, mille jooksul antud punkt saab valgust) sõltub pilu laiusest ja liikumiskiirusest. Eeldagem, et tegemist on horisontaalse pilukatikuga, s.t. pilu liigub kas vasakult paremale või paremalt vasakule

Auto liigub negatiivil kiirusega $(d/l)v$ ning katik läbib auto kujutise ajaga $d t_k / \SI{36}{mm}$.


[User]
EFPhO 2012, problem 6.

                    01W
    The following problems concern nuclear fusion processes in stars, an important topic.

Solution:
See the official solutions here.

    The following problems concern nuclear fusion processes in stars, an important topic.


[User]
36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

                   Th 2    ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES

       The technological and scientific transformations underwent during the XIX century produced a compelling need of
universally accepted standards for the electrical quantities. It was thought the new absolute units should only rely on the
standards of length, mass and time established after the French Revolution. An intensive experimental work to settle the
values of these units was developed from 1861 until 1912. We propose here three case studies.
Marks are indicated at the beginning of each subquestion, in parenthesis.

Determination of the ohm (Kelvin)                                                                                Z
                                                                                ω
        A closed circular coil of N turns, radius a and total resistance R is rotated with
uniform angular velocity ω about a vertical diameter in a horizontal magnetic
      r       r
field B0 = B0 i .

         1. (0.5+1.0) Compute the electromotive force ε induced in the coil, and also the
             mean power1 P required for maintaining the coil in motion. Neglect the coil                             r
                                                                                B0                     θ
             self inductance.
                                                                                X
        A small magnetic needle is placed at the center of the coil, as shown in Figure F-1. It                                                         F-1
is free to turn slowly around the Z axis in a horizontal plane, but it cannot follow the rapid
rotation of the coil.
                                                                                r
         2. (2.0) Once the stationary regime is reached, the needle will set at a direction making a small angle θ with B0 .
             Compute the resistance R of the coil in terms of this angle and the other parameters of the system.

      Lord Kelvin used this method in the 1860s to set the absolute standard for the ohm. To avoid the rotating coil,
Lorenz devised an alternative method used by Lord Rayleigh and Ms. Sidgwick, that we analyze in the next paragraphs.

Determination of the ohm (Rayleigh, Sidgwick).
        The experimental setup is shown in Figure
                                                                                R
F-2. It consists of two identical metal disks D and D'                        ω
of radius b mounted on the conducting shaft SS'. A                                       4                           G
motor rotates the set at an angular velocity ω , which                                       3
                                                                               S
can be adjusted for measuring R. Two identical coils                                                                                1
C and C' (of radius a and with N turns each)                                                 D
                                                                                C                                                    2
surround the disks. They are connected in such a
                                                                                D'                     S'
form that the current I flows through them in                                                                            C'
                                                                                I
opposite directions. The whole apparatus serves to
measure the resistance R.
                                                                                F-2

                                                                                T
                                                                                ∫ X (t ) dt
                                                                                1
                          of a quantity X (t ) in a periodic system of period T is
1
    The mean value X                                                                   X =
                                                                                T    0

    You may need one or more of these integrals:
              2π             2π               2π                        2π               2π
            ∫ sin x dx = ∫ cos x dx = ∫ sin x cos x dx = 0 , ∫ sin x dx = ∫ cos x dx = π , and later ∫ x dx = n + 1 x
                                                                               2                  2                                 n               1       n +1
             0               0               0                          0                0

                                                                                Th 2 Page 1 of 3

                                        36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

         3. (2.0) Assume that the current I flowing through the coils C and C' creates a uniform magnetic field B around D
            and D', equal to the one at the centre of the coil. Compute1 the electromotive force ε induced between the rims 1
            and 4, assuming that the distance between the coils is much larger than the radius of the coils and that a >> b.
       The disks are connected to the circuit by brush contacts at their rims 1 and 4. The galvanometer G detects the flow
of current through the circuit 1-2-3-4.
         4. (0.5) The resistance R is measured when G reads zero. Give R in terms of the physical parameters of the system.

Determination of the ampere
       Passing a current through two conductors and measuring the force between them provides an absolute determination
of the current itself. The “Current Balance” designed by Lord Kelvin in 1882 exploits this method. It consists of six
identical single turn coils C1... C6 of radius a, connected in series. As shown in Figure F-3, the fixed coils C1, C3, C4, and
C6 are on two horizontal planes separated by a small distance 2h. The coils C2 and C5 are carried on balance arms of length
d, and they are, in equilibrium, equidistant from both planes.
       The current I flows through the various coils in such a direction that the magnetic force on C2 is upwards while that
on C5 is downwards. A mass m at a distance x from the fulcrum O is required to restore the balance to the equilibrium
position described above when the current flows through the circuit.

                                            d                                    d

                                                        x
                                                m
                                                            O
                                                                        l
                                                            G

                                                                        I

                                       C1                                            C6                                     F-3
                             F                                                               F                 h
                                       C2                                            C5
                             F                                                               F                 h
                                       C3                                            C4

         5. (1.0) Compute the force F on C2 due to the magnetic interaction with C1. For simplicity assume that the force per
            unit length is the one corresponding to two long, straight wires carrying parallel currents.
         6. (1.0) The current I is measured when the balance is in equilibrium. Give the value of I in terms of the physical
            parameters of the system. The dimensions of the apparatus are such that we can neglect the mutual effects of the
            coils on the left and on the right.

         Let M be the mass of the balance (except for m and the hanging parts), G its centre of mass and l the distance OG .

         7. (2.0) The balance equilibrium is stable against deviations producing small changes δ z in the height of C2 and
             −δ z in C5. Compute2 the maximum value δ z max so that the balance still returns towards the equilibrium
            position when it is released.

2
    Consider that the coils centres remain approximately aligned.

                                 1                          1
    Use the approximations           ≈ 1 m β + β 2 or               ≈1m β
                                                                            2
                                                                                for β << 1 , and sinθ ≈ tanθ for small θ.
                             1± β                       1± β
                                                                2

                                                                                Th 2 Page 2 of 3

                          36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.
           COUNTRY CODE          STUDENT CODE             PAGE NUMBER           TOTAL No OF PAGES

                                        Th 2      ANSWER SHEET

                                                                                Marking
Question       Basic formulas used                             Analytical results
                                                                                guideline

                                                   ε=
   1                                                                                1.5
                                                    P =

   2                                               R=                                                     2.0

   3                                               ε=                                                     2.0

   4                                               R=                                                     0,5

   5                                               F=                                                     1.0

   6                                               I=                                                     1.0

   7                                               δ z max =                                              2.0

                                                                                Th 2 Page 3 of 3

Solution:
36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

             Th 2        ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES

                                                                  SOLUTION

                                                                                r       r
1. After some time t, the normal to the coil plane makes an angle ω t with the magnetic field B0 = B0 i . Then, the
   magnetic flux through the coil is
                         r r
                 φ = N B0 ⋅ S
                            r
                                                              (                      )
                                          r                 r           r
   where the vector surface S is given by S = π a 2 cos ω t i + sin ω t j

   Therefore      φ = N π a 2 B0 cosω t

   The induced electromotive force is
                            dφ
                   ε =−                  ⇒            ε = N π a 2 B0 ω sin ω t
                            dt

   The instantaneous power is P = ε 2 /R , therefore

                    P =
                        (N π a 2 B0ω )
                                       2

                                  2R

                                       1 T 2
                                         ∫
                                                        1
   where we used < sin 2ω t >=             sin ω t dt =
                                       T 0              2

2. The total field at the center the coil at the instant t is
                   r     r     r
                   B t = B 0 + Bi

                                                                                (              )
          r                                                   r              r           r
   where Bi is the magnetic field due to the induced current Bi = Bi cos ω t i + sin ω t j

                         μ0 N I
   with           Bi =                  and           I=ε/R
                            2a

                         μ 0 N 2π a B0 ω
   Therefore      Bi =                       sinω t
                                 2R
   The mean values of its components are

                            μ 0 N 2π a B0 ω
                   Bix =                        sinω t cosω t = 0
                                   2R
                            μ 0 N 2π a B0 ω                   μ 0 N 2π a B0 ω
                    Biy =                        sin 2ω t =
                                   2R                               4R
   And the mean value of the total magnetic field is
                    r       r μ N 2π a B0 ω r
                    Bt = B0 i + 0           j
                                   4R
   The needle orients along the mean field, therefore

                            μ 0 N 2π aω
                  tan θ =
                                  4R

                                                                                Solution Th 2 Page 1 of 4

                                         36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

   Finally, the resistance of the coil measured by this procedure, in terms of θ , is

                          μ 0 N 2π aω
                    R=
                             4 tanθ

3. The force on a unit positive charge in a disk is radial and its modulus is
                  r r
                  v × B = vB =ωrB

   where B is the magnetic field at the center of the coil
                       μ I
                    B=N 0
                        2a
   Then, the electromotive force (e.m.f.) induced on each disk by the magnetic field B is
                                               b
                    ε D = ε D ' = Bω ∫ r dr = 1 Bω b 2
                                               0    2

   Finally, the induced e.m.f. between 1 and 4 is ε = εD + ε D'

                            μ 0 b 2ω I
                    ε =N
                                 2a

4. When the reading of G vanishes, I G = 0 and Kirchoff laws give an immediate answer. Then we have

                                                            μ 0 b 2ω
                    ε =IR                      ⇒     R=N
                                                              2a

5. The force per unit length f between two indefinite parallel straight wires separated by a distance h is.
                          μ 0 I1 I 2
                    f =
                          2π      h

   for I 1 = I 2 = I and length 2π a , the force F induced on C2 by the neighbor coils C1 is

                          μ0 a
                    F=            I2
                            h

6. In equilibrium
                    m g x = 4F d

   Then
                                4μ 0 a d 2
                    mg x =              I                    (1)
                                   h
   so that
                                        1/ 2
                         ⎛ mghx ⎞
                    I = ⎜⎜          ⎟
                                    ⎟
                         ⎝ 4μ 0 a d ⎠

                                                                                Solution Th 2 Page 2 of 4

                                      36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

7. The balance comes back towards the equilibrium position for a little angular deviation δφ if the gravity torques with
   respect to the fulcrum O are greater than the magnetic torques.

                                                          ⎛ 1        1 ⎞
                  Mg l sinδφ + m g x cosδφ > 2 μ 0 a I 2 ⎜⎜       +       ⎟⎟ d cosδφ
                                                          ⎝ h −δ z h +δ z ⎠

                                                               x         d
                                                  δφ
                                      δz                           O
                                                       mg
                                                                l
                         h - δz                                G δφ
                                                                                h + δz
                      h + δz
                                                                   Mg
                                                                                h - δz

   Therefore, using the suggested approximation

                                                   4 μ 0 a d I 2 ⎛⎜ δ z 2 ⎞⎟
                  Mg l sinδφ + m g x cosδφ >                        1 + 2 cosδφ
                                                         h        ⎜    h ⎟⎠
                                                                  ⎝
   Taking into account the equilibrium condition (1), one obtains

                                           δ z2
                  M g l sin δφ > m g x            cos δφ
                                            h2
                                      δz
   Finally, for tan δ φ ≈ sin δ φ =
                                       d

                        M l h2                                 M lh2
                  δz<                  ⇒           δ z max =
                        m xd                                   m xd

                                                                                Solution Th 2 Page 3 of 4

                                36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

                                              Th 2      ANSWER SHEET

Question    Basic formulas and ideas used                                   Analytical results                 Marking
                                                                                guideline
                       r        r
           Φ = N B0 ⋅ S
                dΦ
                                                           ε = N π a 2 B0 ω sin ω t                               0.5
           ε =−
                                                                (N π a 2 B0ω )
   1                                                                          2
                      dt
                                                            P =                                                   1.0
                  ε2                                                        2R
           P=
                  R
           r r       r
           B = B 0 + Bi
                   μ0 N
   2       Bi =             I                                μ N π aω   2
                      2a                                   R= 0                                                   2.0
                                                               4tan θ
                           By
           tanθ =
                           Bx

            r r r
           E = v×B
           v =ωr                                                   μ 0 b 2ω I
                 μ I
                                                           ε=N                                                    2.0
   3                                                                    2a
           B=N 0
                 2a
                br r
           ε = ∫ E dr
                  0

   4       ε = RI                                              μ b ω
                                                            R=N 0
                                                                            2
                                                                                0,5
                                                                 2a

                  μ0 I I ′                                        μ0 a
   5        f =                                             F=              I2                                    1.0
                  2π h                                              h

                                                                                1/ 2
                                                                  ⎛ mghx ⎞
   6       mg x = 4F d                                       I = ⎜⎜          ⎟
                                                                             ⎟
                                                                  ⎝ 4μ 0 a d ⎠
                                                                                1.0

           Γ grav > Γ mag                                                M l h2
   7                                                        δ z max =                                             2.0
                                                                         m xd

                                                                                Solution Th 2 Page 4 of 4

R.S.E.F.


[User]
Consider a cube of initial mass m and side length L in free space. In the lab frame,
    the cube has an initial velocity v0 ≪ c to the right, and plane electromagnetic waves of intensity I
    (in units of W/m2 ) approach the cube from the left and right, striking two faces of it head on. Find
    the displacement of the cube after a long time, for three cases:

      (a) The left and right faces of the cube are perfectly black, and emit negligible thermal radiation.
          (This is the easiest case, but it’s actually extremely unrealistic; can you see why?)

      (b) The left and right faces of the cube are perfectly black. In addition, they are kept in thermal
          equilibrium with each other, and emit thermal radiation so that the mass-energy of the cube
          stays constant in the cube’s frame.

      (c) The cube is perfectly reflective.

    For simplicity, you may always work to lowest order in v/c.

Solution:
We’ll set c = 1 for convenience, and expand everything to lowest order in v. There are
many ways to do this problem, though each one requires some careful bookkeeping. For instance,
you can do it like example 6, by transforming between the lab and cube frames. For variety, I’ll
present a slightly different method here.

 (a) This can be done without leaving the lab frame. Since the cube is running into one of the
     beams and directly away from the other, the rate of momentum transfer from each beam is
     multiplied by 1 + v and 1 − v, respectively. Then we have
                                               dp
                                                  = −2IL2 v.
                                               dt
     Integrating both sides with respect to time, using p0 ≈ mv0 , we get mv0 = 2IL2 ∆x, so that
                                                       mv0
                                                ∆x =        .
                                                       2IL2
     The reason this is unrealistic is that during this process, the cube will absorb an incredible
     amount of energy. The velocity decays on the characteristic time m/(IL2 ), which means that
     during this time, the cube absorbs a total energy of order m, which is enough to change its
     rest mass by a significant amount! The above result is still correct, because it only uses the
     fact that the initial momentum is mv0 and the final momentum is zero, but any real object
     would either get extremely hot and start emitting energy, or reflect away the energy. Those
     are the cases we consider in the next two parts.

 (b) We start by working in the cube frame. In this frame, the light beam coming in from the right
     has its intensity enhanced by two powers of 1 + v. To see this, I find it helpful to imagine the
     light beam as made of discrete photons.
     Suppose that in the lab frame, each photon had frequency f , and they happened to be spaced
     a wavelength λp    = 1/f apart. In the cube frame, each photon incoming from the right has
                    ′
     frequency f = (1 + v)/(1 − v)f ≈ (1 + v)f . In addition, the spacing between them is now
     1/f ′ , so the rate at which they hit the cube is enhanced by another factor of 1 + v. Therefore,
     the cube sees an incoming intensity I ′ ≈ (1 + 2v)I.
     Of course, this isn’t exactly how photons work, but the transformation of intensity doesn’t
     depend on exactly what the beam is made of, so this has to be the right answer in general.
     Similarly, the cube sees an incoming intensity of (1 − 2v)I from the left.
     So, if the cube didn’t emit any radiation, then in its own frame, its energy E ′ and momentum
     p′ satisfy
                                    dE ′               dp′
                                         = 2IL2 ,           = −4IL2 v.
                                     dt                dt
     Transforming back to the lab frame using the Lorentz transformations at first order in v,
                                         dp   dp′    dE ′
                                            ≈     +v      = −2IL2 v
                                         dt   dt      dt

         where we neglected time dilation since it’s second order in v. This is as we found in part (a).
         Now let’s add on the radiation emission. In the cube frame, an equal intensity I is emitted
         from both sides, so that
                                        dE ′          dp′
                                             = 0,         = −4IL2 v.
                                         dt           dt
         Transforming back to the lab frame, we have
                                                  dp
                                                     = −4IL2 v
                                                  dt
         from which we conclude
                                                         mv0
                                                   ∆x =       .
                                                         4IL2
         This is smaller than in part (a), which makes sense. Thermal radiation by itself can’t change
         the cube’s velocity in any frame. However, by removing energy, it reduces the cube’s inertia
         (or rather, prevents the inertia from increasing), making it easier to slow down.

     (c) In this case, a similar argument to the above gives

                                          dE ′             dp′
                                               = 0,            = −8IL2 v.
                                           dt              dt
         We now get twice the force as before, since the photon momenta get flipped upon reflection.
         Going back to the lab frame,
                                         dp                          mv0
                                            = −8IL2 v,        ∆x =        .
                                         dt                          8IL2
         Of course, it is also possible to get this answer by using the result of example 6 twice.


[User]
10. Ice-rally (7 points)

The car accelerates on a slippery ground
sothatthewheelsarealwaysatthelimit ofslipping (e.g. viausing
an electronic traction control). Suchanacceleration would result
in thevelocity vstimegraphasgiven in theFigure.
i)(2pt)Whatisthecoefficientoffriction,assumingafour-wheel
drive?
Because of a manual gear change, there is time period of
τ1 = 0.5s, during which there is no driving force (so that the
car decelerates due to air friction). Except for that period, the
acceleration follows the law given by the graph. As a result, the
terminalvelocity vt = 40m/sisreachedτ2 = 1.0slater thanit
wouldhavebeenreached,iftherewerenodelaycausedbythegear
change. Upon reaching the terminal velocity, the car continues
moving at constant speed. In your calculations, you can assume
that the air frictionwasconstant during the gearchange period.
ii)(2,5pt)At which speedthe gearwaschanged?
iii) (2,5 pt) How many meters shorter distance will be covered
duringthefirst100seconds,ascomparedtoidealacceleration(i.e.
without the delay due tothe gearchange)?

Solution:
10. Ice-rally

i. (2 pts) Since at the very beginning, the effect of the air
friction is negligible, the acceleration (i.e. the tangent of the
graph) gives us the ratio of the friction force Ff and the mass
m, i.e. μg. From the graph, this tangent is μg = 1.0m/s2
,
hence μ = 0.1.
ii. (2.5 pts) When the driving force stops, the acceleration
is reduced by Ff/m = μg, i.e. from the slope at the current
point of the graph we need to subtract the slope of it at the
origin. A close-up sketch of the graph around the period of
gear change is given in Fig. After the gear change, the new
graph follows the ideal graph, but is shifted rightwards by τ2,
this shift is marked also in Fig. Since 2τ1 = τ2, the ascend-
ing and descending slopes in that close-up sketch must be of
equal steepness, i.e. a = μg/2. So, the gear change takes place
at that speed, when the acceleration is twice smaller than at
zero-speed. From the graph we can find that v0 ≈ 25m/s.
iii. (2.5 pts) The distance difference is the surface area S
between the actual v(t) graph, and the ideal one. These graphs
coincide for v < v0 and upon achieving the value v = vt. So,
the area S is enclosed into the range vt > v > v0, where the
actual graph is, in fact, just the ideal graph, but shifted right-
wards by τ2. This area has a shape of a narrow curved stripe,
the horizontal width of which is at every value of v equal to τ2.
One can divide this stripe into tiny horizontal layers of height
δ and width τ2. If we sum up the surface areas of these layers,
we can bring τ2 before the braces; then, the sum of the layer
widths goes into the braces and yields vt − v0. So, the surface
area S = τ2(vt − v0) ≈ 15m.
Problem

Ice-rally