[User]
APHO II   2001    Theoretical Question 2                                                                      p. 1 / 2

                                           Theoretical Question 2
                  Motion of an Electric Dipole in a Magnetic Field
    In the presence of a constant and uniform magnetic field                         , the translational motion of a
system of electric charges is coupled to its rotational motion. As a result, the conservation laws
for the momentum and the component of the angular momentum along the direction of                                 are
modified from the usual form. This is illustrated in this problem by considering the motion of an
electric dipole made of two particles of equal mass m and carrying charges q and
respectively ( q > 0 ). The two particles are connected by a rigid insulating rod of length                     , the
mass of which can be neglected.        Let          be the position vector of the particle with charge q,
that of the other particle and         =     －      . Denote by             the angular velocity of the rotation
around the center of mass of the dipole. Denote by                      and            the position and the velocity
vectors of the center of mass respectively. Relativistic effects and effects of electromagnetic
radiation can be neglected.

      Note that the magnetic force acting on a particle of charge q and velocity                         is     × ,
where the cross product of two vectors              ×          is defined, in terms of the x, y, z, components of
the vectors, by
                              (   ×        )x ＝ (       )y (      )z － (      )z (     )y
                              (    ×       )y ＝ (       )z (      )x － (      )x (     )z
                              (   ×    )z ＝ (           )x (     )y － (       )y (    )x.

(1) Conservation Laws

   (a) Write down the equations of motion for the center of mass of the dipole and for the
       rotation around the center of mass by computing the total force and the total torque with
       respect to the center of mass acting on the dipole.

   (b) From the equation of motion for the center of mass, obtain the modified form of the
       conservation law for the total momentum. Denote the corresponding modified conserved
       quantity by        . Write down an expression in terms of                            and   for the conserved
       energy E.

   (c) The angular momentum consists of two parts. One part is due to the motion of the center
       of mass and the other is due to rotation around the center of mass. From the modified
       form of the conservation law for the total momentum and the equation of motion of the
       rotation around the center of mass, prove that the quantity J as defined by

                                             J=(          × ＋I         )．

       is conserved.

                                                           1

APHO II    2001   Theoretical Question 2                                                           p. 2 / 2

       Note that

                                                   ×       ＝    ×

                                           ‧(      ×   ) ＝ (    ×    )‧

       for any three vectors           ,     and       . Repeated application of the above first two
       formulas may be useful in deriving the conservation law in question.

In the following, let       be in the z-direction.

(2) Motion in a Plane Perpendicular to

    Suppose initially the center of mass of the dipole is at rest at the origin,         points in the
    x-direction and the initial angular velocity of the dipole is         (   is the unit vector in the
    z-direction).

   (a) If the magnitude of            is smaller than a critical value    , the dipole will not make a
          full turn with respect to its center of mass. Find     .

   (b) For a general           > 0, what is the maximum distance           in the x-direction that the
          center of mass can reach?

   (c) What is the tension on the rod? Express it as a function of the angular velocity        .

                                                       2

Solution:
APHO II   2001   Theoretical Question 2                                                     p. 1 / 4

[Solution]                         Theoretical Question 2

          Motion of an Electric Dipole in a Magnetic Field

(1) Conservation Laws

   (1a)          = (         ),           = (       ),             ,

   Total force        on the dipole is

   so that

                  M       = q ×              (M = 2m)                                 (1)

   Computing the torque for rotation around the center of mass, we obtain

                                                                                (2)

   where

                   I =                                                                (3)

   (1b) From eq.(1), we obtain the conservation law for the momentum:

                       = 0,            =M        q ×                                 (4)

   From eq.(1) and eq.(2), one obtains the conservation law for the energy.

                          = 0,         E =                I                          (5)

   (1c) Using eq.(4) and eq.(2),

             (     × )            =   (     × )         = q     ×( × )

                                   = q( × )×                  = q( × )     × )

                                   = q         ×(     × ))     = q     × )× )

                                   = q ×(        × )

                                   = I

                                                  1

APHO II   2001      Theoretical Question 2                                                     p. 2 / 4

[Solution] (continued)                  Theoretical Question 2
    Motion of an Electric Dipole in a Magnetic Field
    we obtain the conservation law

                       =0           = (        × I         )                           (6)

    for the component of the angular momentum along the direction of .

(2) Motion in a Plane Perpendicular to

   (2a) Write

            =       {cos (t) sin (t) },                    (0) = 0,        (0) =        (7)

   Note that

                       =                                                                 (8)

   From eq.(4), we have

                M        =      q B (sin            cos         )                      (9)

   At t = 0, we have            = 0,         = 0 so that

                      = q B                                                             (10)

   Hence from eqs.(9) and (10) we have

                           =           sin ,            =              (1cos       )   (11)

   From conservation of energy, i.e. Eq.(5), we have

                               I                  (1cos        )=     I

                                           (1cos     )=                              (12)

   where

                                   =                =                                   (13)

   In order to make a full turn,             can not become zero so that

                                                    2

APHO II   2001    Theoretical Question 2                                                                   p. 3 / 4

[Solution] (continued)                  Theoretical Question 2
          Motion of an Electric Dipole in a Magnetic Field

                                                           =                                      (14)

   (2b) From Eq.(6), we have

                               P         =                                                         (15)

   where P is the magnitude of             .

   At     t  0,     we have        =              so that

                                                                                (16)

   From eq.(12), one can see that           so that                            . Thus         reaches a
   maximum        when       takes its minimum value.

   When               , the minimum value of                     is       so that

                                                             ,                                      (17)

   When               , the minimum value of                     is                so that

                                                                               ,                    (18)

   When               ,

   When          is close to    ,   let                          then

   so that it will take             for                 , i.e. for      to reach      .   Hence

                                               ,                                                    (19)

   (2c) Tension on the rod comes from three sources:

                                                        3

APHO II    2001   Theoretical Question 2                                                      p. 4 / 4

[Solution] (continued)               Theoretical Question 2
           Motion of an Electric Dipole in a Magnetic Field

          (i) Coulomb force =                                                          (20)

             Positive value means compression on the rod.

          (ii) Centrifugal force due to rotation of the rod =                          (21)

      (iii) Magnetic force on the particles due to the motion of the center of the mass

             =

   Taking the square of both sides of eq.(4) and using the initial condition for the
   value of    , we obtain

                                                                                (22)

   Combining the three forces, we have

                  tension on the rod                                                   (23)

   A positive value means compression on the rod.

                                             4

APhO 2001 Theory T2


[User] A glass tube (radius $r_1$) is placed inside a thicker glass tube so that their axes coincide. Next they are both placed upright in water. Find how big should be the inner radius $r_2$ of the thicker tube so that the water level inside both of the glass tubes would be the same. Assume that the walls of the tubes are insignificantly thin.

Solution:
Vaatleme jõudude tasakaalu kummaski torus: reservuaari veetasemest
ülespoole jääva vee kaalu $\rho g h S$ tasakaalustab kapillaarjõud
$\sigma p \cos\alpha$, kus $p$ on vee ja klaasi kontaktjoone kogupikkus,
$h$ --- veetaseme kõrgus kapillaaris, $S$ --- toru ristlõikepindala ja
$\alpha$ --- nurk veepinna puutuja ja klaasi pinna vahel, mis sõltub
märgamise määrast, kuid on mõlema toru jaoks sama (õigeks loetakse ka lahendused,
kus eeldades täielikku märgamist jäetakse tegur $\cos\alpha$ ära). Niisiis $h=\sigma p
\cos\alpha/\rho g S$. Suure toru jaoks $p_2=2\pi (r_2+r_1)$ ja
$S_2=\pi(r_2^2-r_1^2)$; väikse toru jaoks $p_1=2\pi r_1$ ja $S_1=\pi
r_1^2$. Seosest $h_1=h_2$ saame $p_2/p_1=S_2/S_1$, millest ülaltoodud
avaldiste asendamise teel omakorda saame
$1+r_2/r_1=(r_2/r_1)^2-1$. Viimane seos kujutab endast suhte $x=r_2/r_1$
jaoks ruutvõrrandit $x^2-x-2=0 \Rightarrow x=2$ (negatiivne lahend ei
oma füüsikalist tähendust). Niisiis $r_1=2 r_2$.

Pindpinevus

Klaastoru (raadius $r_1$) asetatakse jämedama klaastoru sisse nii, et nende teljed ühtivad. Seejärel
pannakse mõlemad püsti vette. Leidke, kui suur peaks olema jämedama toru
siseraadius $r_2$, et veetase oleks mõlemas klaastorus sama. Eeldage, et torude
seinad on tühiselt õhukesed.

Reservuaari veetasemest ülespoole jääva vee kaalu peab tasakaalustama kapillaarjõud vee ja klaasi kontaktjoonel.


[User]
In this problem, you will explore yet another slick derivation of Kepler’s first law. As
    usual, suppose the orbit lies in the xy plane. This derivation resolves around writing r(t) = r(t)r̂(θ)
    where r̂ = cos θ x̂ + sin θ ŷ, and solving a differential equation for v(θ).

      (a) Show that dv/dθ is proportional to r̂, times constants and conserved quantities.

      (b) Integrate this result to find v(θ). What nice geometric property does it have?

        (c) Plug this result into L = r × p to show that r(θ) is a conic section.

    6      Fluids
[2] Problem 26 (BAUPC). Two cylindrical containers, A and B, have the same shape and contain
    equal volumes of water. In addition to the water, B contains an immersed balloon, attached to the
    bottom with a string. Consider the following reasoning.

           The total upward force exerted by the bottom of container A is equal to the weight of
           the water in A, likewise for B. Since the areas of the bottoms are the same, and the
           weights of the water are the same, the pressures at the bottoms are the same.

                     01T
    Is this reasoning correct or incorrect? Explain your answer.

Solution:
(a) Using the usual variables, we have

                                             dv    GM                     dθ
                                                = − 2 r̂,       L = mr2      .
                                             dt     r                     dt
           The desired quantity is
                                               dv   dv dt    GM m
                                                  =       =−      r̂.
                                               dθ   dt dθ     L
      (b) Performing a straightforward integration, we find

                                                  GM m
                                       v(θ) = −        (sin θ + c1 , − cos θ + c2 )
                                                   L
           where c1 and c2 are constants of integration. The nice result is that v(θ) is a circle! That’s
           the hidden reason why this derivation will be so quick.

        (c) Carrying out the cross product, the magnitude of the angular momentum is

                                     GM m2 r
                              L=−            ((cos θ)(− cos θ + c2 ) − (sin θ)(sin θ + c1 )).
                                       L
           We can easily solve this to get

                                          1   GM m2
                                            =       (1 + c1 sin θ − c2 cos θ).
                                          r    L2
           This can be put into the standard form for a conic section by just shifting θ to get rid of the
           sin θ term.

    6      Fluids
[2] Problem 26 (BAUPC). Two cylindrical containers, A and B, have the same shape and contain
    equal volumes of water. In addition to the water, B contains an immersed balloon, attached to the
    bottom with a string. Consider the following reasoning.

           The total upward force exerted by the bottom of container A is equal to the weight of
           the water in A, likewise for B. Since the areas of the bottoms are the same, and the
           weights of the water are the same, the pressures at the bottoms are the same.

    Is this reasoning correct or incorrect? Explain your answer.

    Solution. The reasoning is incorrect. As usual, the pressure depends only on the depth of the
    water, which is greater for container B. The net forces still balance: the greater upward force of the
    bottom on the water in container B is compensated for by the downward force of the balloon on

                    01T
    the water.

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


[User]
A hemispherical shell of radius R has uniform charge density σ and is centered at the
    origin. Find the electric field at the origin. (Hint: combine the previous two ideas.)

Solution:
Place a point charge q at the origin. To find the magnitude of the field, we will compute
    the force on the hemisphere divided by q. The force on the hemisphere is
                                        Z                         Z
                                             q                qσ
                                                  σ dS =            dS.
                                          4πϵ0 R2         4πϵ0 R2

    By idea 5, we can replace the surface of integration with a flat disk, so | dS| = πR2 . Thus, the
                                                                               R

    force is F = qσ/4ϵ0 , so the field is
                                                         σ
                                                   E=       .
                                                        4ϵ0


[User]
01@  IPhO 2023, problem 3, parts B and C. A nice problem on the shape of a
    meniscus, which also explains why pieces of cereal clump together in a bowl of milk.

      Example 21: IPhO 2022 3B

      Slightly wet sand is much stronger than either dry sand or very wet sand, which allows the
      construction of large structures like sand castles. Why is this, and how does the strength
      depend on the typical size r of the sand grains?

      Solution
      When a pile of sand is dry, the only force keeping it in place is friction, which is weak. When
      it’s very wet, it’s essentially just water, which will simply collapse. But when it’s slightly
      wet, adjacent sand grains have a small layer of water connecting them. Since sand grains are
      small, this implies a huge total surface area, and thus large surface tension effects.

      There are actually two conceptually distinct components to the effect. First, the bit of water
      connecting two sand grains will provide a surface tension force F ∼ γr. Second, as you
      saw in problem 29, the water has a pressure lower by ∆P ∼ γ/r, leading to an attractive
      pressure force (∆P )A ∼ γr. In either case, that means the force needed to displace a single
      grain of sand scales with r. The number of sand grains in a fixed cross-sectional area scales
      as 1/r2 , so the weight a sand castle can bear scales as 1/r. Thus, fine-grained sand is stronger.

      This is another example of the subtleties of granular media, first mentioned in M2. Neither
      sand nor water are strong on their own, but they’re strong together. Water provides the
      forces, while the sand provide the structure which lets those forces be effective.

Solution:
IPhO 2023, problem 3, parts B and C. A nice problem on the shape of a
    meniscus, which also explains why pieces of cereal clump together in a bowl of milk.

       Example 21: IPhO 2022 3B

       Slightly wet sand is much stronger than either dry sand or very wet sand, which allows the
       construction of large structures like sand castles. Why is this, and how does the strength
       depend on the typical size r of the sand grains?

       Solution
       When a pile of sand is dry, the only force keeping it in place is friction, which is weak. When
       it’s very wet, it’s essentially just water, which will simply collapse. But when it’s slightly
       wet, adjacent sand grains have a small layer of water connecting them. Since sand grains are
       small, this implies a huge total surface area, and thus large surface tension effects.

       There are actually two conceptually distinct components to the effect. First, the bit of water
       connecting two sand grains will provide a surface tension force F ∼ γr. Second, as you
       saw in problem 29, the water has a pressure lower by ∆P ∼ γ/r, leading to an attractive
       pressure force (∆P )A ∼ γr. In either case, that means the force needed to displace a single
       grain of sand scales with r. The number of sand grains in a fixed cross-sectional area scales
       as 1/r2 , so the weight a sand castle can bear scales as 1/r. Thus, fine-grained sand is stronger.

       This is another example of the subtleties of granular media, first mentioned in M2. Neither
       sand nor water are strong on their own, but they’re strong together. Water provides the
       forces, while the sand provide the structure which lets those forces be effective.


[User]
3. A chain of length l and linear density λ hangs from a horizontal support with both ends A and
   B fixed to a horizontal support as shown. The two fixed ends are close to each other. At time t
   = 0 the end A is released. All vertical distances (x) are measured with respect to the horizontal
   support with the downward direction taken as positive (A and B are initially at x = 0).

                                                             9
   (a) Obtain the momentum P of the center of mass of the system when the end A has fallen by
       a distance x, in terms of x and speed ẋ.

         P =
                                           C1
                                                                                [2]

   (b) Assume that the end A is falling freely under gravity, i.e., ẍ = g. Obtain the tension T at      [1]
       the fixed end B just before the chain completes the fall and becomes entirely vertical.

         T =

                      HB
   Experimentally, the value of tension is found to be different from the above result. We adopt an
   alternative approach assuming conservation of mechanical energy.
    (c) Obtain the potential energy U (x) of the chain and plot it versus x. Take the potential
        energy of a point mass placed at the horizontal support to be zero.
                                                                                [21/2]

INPhO 2019 Page 4          Questions & Summary Answers                    Last four digits of Roll No.:
   (d) Obtain the speed ẋ when the end A has fallen by a distance x. Assume that all sections of         [21/2]
       the falling (right side) part of the chain have the same speed.

         ẋ =

    (e) Hence obtain T (x) at B as a function of x and related quantities. You are advised to simplify      [3]
        your expression as far as possible.

         T (x) =

    (f) Qualitatively sketch T (x) versus x.                                                                [1]

                   T (x)

                                     C1                   9
                                                                                x

                HB               Detailed answers can be found on page numbers:

Solution:
3. A chain of length l and linear density λ hangs from a horizontal support with both ends A and
   B fixed to a horizontal support as shown. The two fixed ends are close to each other. At time t
   = 0 the end A is released. All vertical distances (x) are measured with respect to the horizontal
   support with the downward direction taken as positive (A and B are initially at x = 0).

INPhO 2019 Page 6               Questions & Answers                                 Last four digits of Roll No.:

   (a) Obtain the momentum P of the center of mass of the system when the end A has fallen by                       [2]
       a distance x, in terms of x and speed ẋ.

        P =

                                   C1
         Solution: Let total mass of the chain to be
         The mass of the left side of the chain = λ

                                              l+x
                                                           9
                                                     M =λl.
                                                     l+x
                                                    2
         CM of the left side of the chain =
                                                4
                                                       l−x

         The mass of the right side of the chain = λ
                                                        2

              HB

         CM of the right side of the chain = x +

                              xCM =

                           λlxCM =
                                      λ

                                    λl2 λlx λx2
                                        +
                                              l+x

                                              −
                                               2
                                                    l−x
                                                     4

                                                         l+x
                                                          4

                                                                   +λ
                                                                    λl

                                                                             l−x
                                                                              2

                                                                                x+
                                                                                l−x
                                                                                4

                                                                                (3.1)

                                                                                (3.2)
                                     4    2    4
                                        l−x
                       P = λlẋCM = λ         ẋ                                                          (3.3)
                                         2

   (b) Assume that the end A is falling freely under gravity, i.e., ẍ = g. Obtain the tension T at                 [1]
       the fixed end B just before the chain completes the fall and becomes entirely vertical.

        T =

         Solution:

                                                ẍ = g                                                    (3.4)
                                                               p
                                                ẋ = gt =          2gx                                    (3.5)
                                                Ṗ = M g − T                                              (3.6)

INPhO 2019 Page 7                Questions & Answers

         From Eq. (3.3)

                                                  λ
                                           Ṗ =     [ẍ(l − x) − ẋ2 ]                      (3.7)
                                                  2
                              λ
                                [gl − 3gx] = M g − T                                        (3.8)
                              2
                                             Mg        3x     λlg     3x

                                     T (x) =        1+      =      1+                       (3.9)
                                              2         l      2       l
                                 T (x = l) = 2λlg                                          (3.10)

   Experimentally, the value of tension is found to be different from the above result. We adopt an
   alternative approach assuming conservation of mechanical energy.
    (c) Obtain the potential energy U (x) of the chain and plot it versus x. Take the potential       [21/2]
        energy of a point mass placed at the horizontal support to be zero.

                                              C1                         9
                      HB
         Solution: In general, U = −λlgxCM . Using Eq. (3.2)

                                                   −λg 2
                                            U=        (l + 2lx − x2 )                      (3.11)
                                                    4

                                   U   0                                 L   x

                            −λgl2 /4

                            −λgl2 /2

   (d) Obtain the speed ẋ when the end A has fallen by a distance x. Assume that all sections of     [21/2]
       the falling (right side) part of the chain have the same speed.

INPhO 2019 Page 8                 Questions & Answers                                Last four digits of Roll No.:

         ẋ =

          Solution: From Eq. (3.3), kinetic energy of the chain

                                                 1  l−x

                                           K(x) = λ     ẋ2                                              (3.12)
                                                 2   2
          As the total energy is conserved,

                                      U (x) + K(x) = U (t = 0)                                           (3.13)
                                                           "                  #1/2
                                                               g(2lx − x2 )
                                                   ẋ2 =                                                 (3.14)
                                                                 (l − x)
                                                           "                  #1/2
                                                         g(2lx − x2 )
                                                    ẋ =                                                 (3.15)
                                                           (l − x)

    (e) Hence obtain T (x) at B as a function of x and related quantities. You are advised to simplify               [3]
        your expression as far as possible.

         T (x) =

          Solution:                  C1                     9
                                                d 2
                                                   ẋ = 2ẋẍ
                                                dt
          Substituing Eq. (3.15) in the above equation yields

                HB                             ẍ = g +

          Combining Eqs. (3.6), (3.7) and (3.16)

                                       T = λlg −
                                                     λ
                                                     2
                                                          g(2lx − x2 )
                                                           2(l − x)2

                                                       [ẍ(l − x) − ẋ2 ]
                                                                                (3.16)

                                                                                (3.17)
                                                  λg
                                          =             [2l2 + 2lx − 3x2 ]                               (3.18)
                                               4(l − x)

    (f) Qualitatively sketch T (x) versus x.                                                                         [1]

INPhO 2019 Page 9                 Questions & Answers

                  T (x)

                                                                                x

          Solution:

                                                  C1            9
                                   T

                       HB      λgl/2
                                       0
                                                           x
                                                                         L

                                 Detailed answers can be found on page numbers:

INPhO 2019 Problem 3


[User]
8. AIR IN A SUBMARINE (6 points)

— Johan
Runeson, Jaan Kalda. A submarine of unknown
nationality is travelling near the bottom of the
Balticsea,atthedepthof h=300m. Itsinterior
is one big room of volume V =10m3
filled with
air(M =29g/mol)atpressure p0 =100kPa and
temperature t0 = 20◦
C. Suddenly it hits a rock
and a large hole of area A =20cm2
is formed at
the bottom of the submarine. As a result, the
submarine sinks to the bottom and most of it
is filled fast with water, leaving a bubble of air
at increased pressure (no air escapes the submar-
ine). The density of water ρ = 1000kg/m3
and
free fall acceleration g =9.81m/s2
. Molar heat
capacitance of air by constant volume cV = 5
2R,
where R =8.31J/Kmol is the gas constant.
i) (2 points) What is the volume rate (m3
/s) at
which the water flows into the submarine im-
mediately after the formation of the hole?
ii)(2points)Theflowrateissolargethatthesub-
marine is filled with water so fast that the heat
exchange between the gas and the water can be
neglected (this applies also to the next question).
What is the volume of the air bubble once water
flow has stopped?
iii) (2 points) The water stream rushing into
the submarine creates turbulent water inside the
submarine; what is the total kinetic energy of
this water turbulence (which later ends up being
dissipated as heat), once the inflow has stopped
due to equalized pressures?
9. BLACK BOX (?? 1point) — Jaan Kalda, Mi-
hkelHeidelberg. Theblackboxhasthreeterminal
wires: “blue”, “black” and “white”, and contains
in a star configuration: a battery, a capacitor, an
inductor in series with a diode. You may con-
sider the diode to be ”ideal” — it conducts cur-
rent perfectly one way and not at all the other
way. You may neglect internal resistance of the
battery and capacitor, but the inductor has con-
siderable internal resistance. The multimeter’s
internal resistance when measuring voltages is
Rm =10MΩ anditdisplaysanewreadingevery
t=0.4s.
i) (3 points)Draw the electrical circuit that is in-
side the black box. Motivate your solution with
measurements.
ii) (1 point) Determine the electromotive force
of the battery.
iii) (1point)Determine the internal resistanceof
the inductor.
iv) (3 points) Estimate the value of the capacit-
ance C.
v)(3points)Estimatethevalueoftheinductance
L.
Equipment: Black box, multimeter, stop-
watch.

Solution:
8. AIR IN A SUBMARINE

i) We are supposed to calculate the volume
rate (in m3
s ) at which the water flows in. We
know A = 10cm2
. We apply Bernoulli’s equa-
tion, where the initial point is in the sea and the
final point is in the hole:
Pi +
1
2
ρv2
i
| {z }
=0
= Pf +
1
2
ρv2
f (5)
from which we get:
vf =
√
2
∆P
ρ
≈
√
2gh=76.72m/s. (6)
Here vf is the speed at which the water flows
in. This we can insert into the equation for the
volume rate:
Q = Avf =0.153
m3
s
≈150
litres
s
. (7)
ii) Atmospheric air consists mainly of diatomic
nitrogen and oxygen gas. At the temperatures
involved these molecules have f = 5 degrees of
freedom: 3 translational and 2 rotational. The
adiabatic constant γ is γ =(f +2)/f =7/5. One
can also obtain this result from γ=(cV +R)/cV .
For adiabatic compression we have
piV
γ
i
= pf V
γ
f
. (8)
The final pressure is the pressure from the sea,
which is approximately p0+ρgh=(105
+1000·
9.8·300)Pa≈3MPa. This gives
Vf =Vi
(
pi
pf
)5
7
≈0.9m3
(9)
Note: The final temperature is only about 2.6
times the initial temperature, such that the vi-
brational degrees of freedom of the molecules
does not have to be considered.
iii) The work W done on the system (consist-
ing of the whole submarine) by the surrounding
water is W = Pc∆V, where Pc is the constant
pressure of the surrounding water. There is no
heat exchange, so this work must be equal to the
change in internal energy of the system:
W =∆Ugas +∆Uwater = cV n∆T +Kturb (10)
where Kturb is the quantity that we are after and
get:
Kturb = Pc∆V −cV n∆T. (11)
We need the value of n (NB. you can also figure
it out from the ideal gas law):
n=
m
M
(12)
where m = 1.23
kg
m3 ·10m3
= 12.3kg and M =
0.02897
kg
mol. Plugging in the values we get n ≈
424mol. The final temperature can be calcu-
lated from the initial temperature by using that
p1−γ
Tγ
is conserved.
By plugging in all the other values we get:
Kturb ≈2.2×107
J. (13)
Alternative solution: Alternatively, one can
look at the gas and the water (inside the submar-
ine) as separate subsystems. The work done on
the gas is equal to the change in internal energy
of the gas:
∫
Pg dV = cV n∆T. (14)
The work done on the water inside the submar-
ine by the water outside the submarine is Pc∆V.
The water inside the submarine also does work
on the gas given by
∫
Pg dV. The change in in-
ternal energy of the water in the submarine is
then
Kturb = Pc∆V −
∫
Pg dV = Pc∆V −cV n∆T
(15)
where the last equality follows from eqn (??).
From here one proceeds as already written
above.

AIR IN A SUBMARINE


[User]
4. Questions consist of sub-questions. Write your detailed answer in the blank
   space provided below the sub-question and final answer to the sub-question in
   the smaller box which follows the blank space.

Solution:
4. (a) k =                  −1
             2 (R2 − ω 2 )1/2
     (b)
                            k

                                                             R/2   w

                                                                       1

      (c)
            +1 cm

                          Pn
            0.5 cm

                     P2
                     P1
                     O
                                   F       C

            − 0.5 cm

            −1 cm

INPhO 2013 Problem 4


[User]
Physics Cup 2013, problem 9.

    5      Heat Conduction

  Idea 10: Fourier’s Law
  In T1, you investigated heat conduction from the standpoint of kinetic theory. Now we
  revisit the subject from the standpoint of hydrodynamics. The flux of heat (i.e. the power
  per unit area) due to thermal conduction is proportional to the temperature gradient,

                                                     dT
                                           J = −κ       .
                                                     dx
  By considering the net heat flowing in and out of a slab of width dx, we have

                                       du    dJ   d2 T
                                          =−    =κ 2.
                                       dt    dx   dx
  where u is the energy density. Intuitively, this shows how heat conduction works to smooth
  out temperature gradients. For example, if the temperature had a local minimum, then du/dt
  would be positive at that point, as heat flows in from all directions.

  Example 12

  Find the steady state heat flow through a rectangular slab of area A and thickness d, whose
  left and right ends are held at temperature T and T + ∆T .

  Solution
  By the argument in the above idea, in the steady state the energy density within the slab
  is constant, so d2 T /dx2 = 0 and the temperature gradient is uniform. In this steady state,
  heat simply flows uniformly through the slab, without accumulating anywhere in it, just
  like current flows uniformly through a resistor. (Since this “internal” steady state is often
  reached quickly, in many problems it is just assumed to hold, without being explicitly stated.)

  Therefore, the temperature gradient is just ∆T /d everywhere in the slab, so

                                                     κA∆T
                                       P = JA =           .
                                                       d
  This is single most important equation for heat conduction problems.

  Remark: Continuity Equations

  We can generalize the results above to three dimensions. In this case, the flux of heat is
  described by a vector,
                                        J = −κ ∇T.
  The conservation of energy equation becomes
                                     du
                                        = −∇ · J = κ ∇2 T.
                                     dt
  The first equality here is known as a continuity equation, and it is the general way that the
  density of a conserved quantity is related to a flow. For example, the continuity equation for

       fluids, which we saw in one-dimensional form in M7, is written in three dimensions as
                                              dρ
                                                 = −∇ · (ρv)
                                              dt
       and expresses conservation of mass. The continuity equation for charge, which we’ll see in
       E4, is very similar,
                                               dρ
                                                  = −∇ · J
                                               dt
       where ρ is a charge density and J is a current density.

Solution:
See the official solutions here.

   5    Heat Conduction
       Idea 10: Fourier’s Law
       In T1, you investigated heat conduction from the standpoint of kinetic theory. Now we
       revisit the subject from the standpoint of hydrodynamics. The flux of heat (i.e. the power
       per unit area) due to thermal conduction is proportional to the temperature gradient,

                                                          dT
                                                J = −κ       .
                                                          dx
       By considering the net heat flowing in and out of a slab of width dx, we have

                                            du    dJ   d2 T
                                               =−    =κ 2.
                                            dt    dx   dx
       where u is the energy density. Intuitively, this shows how heat conduction works to smooth
       out temperature gradients. For example, if the temperature had a local minimum, then du/dt
       would be positive at that point, as heat flows in from all directions.

       Example 12

       Find the steady state heat flow through a rectangular slab of area A and thickness d, whose
       left and right ends are held at temperature T and T + ∆T .

       Solution
       By the argument in the above idea, in the steady state the energy density within the slab
       is constant, so d2 T /dx2 = 0 and the temperature gradient is uniform. In this steady state,
       heat simply flows uniformly through the slab, without accumulating anywhere in it, just
       like current flows uniformly through a resistor. (Since this “internal” steady state is often
       reached quickly, in many problems it is just assumed to hold, without being explicitly stated.)

       Therefore, the temperature gradient is just ∆T /d everywhere in the slab, so

                                                          κA∆T
                                            P = JA =           .
                                                            d
       This is single most important equation for heat conduction problems.

       Remark: Continuity Equations

       We can generalize the results above to three dimensions. In this case, the flux of heat is
       described by a vector,
                                             J = −κ ∇T.
       The conservation of energy equation becomes
                                          du
                                             = −∇ · J = κ ∇2 T.
                                          dt
       The first equality here is known as a continuity equation, and it is the general way that the
       density of a conserved quantity is related to a flow. For example, the continuity equation for
       fluids, which we saw in one-dimensional form in M7, is written in three dimensions as
                                              dρ
                                                 = −∇ · (ρv)
                                              dt
       and expresses conservation of mass. The continuity equation for charge, which we’ll see in
       E4, is very similar,
                                               dρ
                                                  = −∇ · J
                                               dt
       where ρ is a charge density and J is a current density.


[User]
Keevasse vette kallatakse külmutatud jõhvikaid. Vee temperatuur langes väärtuseni $t=\SI{89}{\degreeCelsius}$.
Mitu korda oli vee mass suurem jõhvikate massist? Kuna jõhvikad olid väikesed ja sulasid väga kiiresti, siis
võib vee soojusvahetuse ümbritseva keskkonnaga arvestamata jätta. Jõhvikate algtemperatuur oli
$t_2=\SI{-18}{\degreeCelsius}$. Jää erisoojus $c_j=\SI{2100}{J/(kg.C)}$, vee erisoojus $c_v=\SI{4200}{J/(kg.\degreeCelsius)}$, jää sulamissoojus $L=\SI{330}{kJ/kg}$. Jõhvikate suure veesisalduse tõttu võib neid käsitleda jääna.

Solution:
Vee algtemperatuur oli $t_1=\SI{100}{\degreeCelsius}$. Olgu vee ja jõhvikate massid vastavalt $M$ ja $m$. Jõhvikate soojendamiseks kuluv soojushulk tuleb vee jahtumise arvelt. Vee jahtumisel eralduv soojushulk oli
\[
Q_j=Mc_V(t_1-t).
\]
Jõhvikate soojendamise käigus tuli 1) soojendada külmunud jõhvikad sulamistemperatuurini, 2) sulatada külmunud jõhvikad ja 3) soojendada sulanud jõhvikad vee temperatuurini.
Leiame igas etapis kulunud soojushulga:
\[
\begin{aligned}
Q_{s1}&=mc_j(0-t_2)=-mc_jt_2,\\
Q_{s2}&=mL,\\
Q_{s3}&=mc_v(t-0)=mc_vt.
\end{aligned}
\]
Liidame jõhvikate soojendamiseks kulunud soojushulgad ja võrdsustame saadud summa vee jahtumisel eraldunud soojushulgaga. Saadud võrrandist avaldame vee ja jõhvikate masside suhte:
\[
-mc_jt_2+mL+mc_vt=Mc_V(t_1-t),
\]
kust
\[
\frac{M}{m}=\frac{-c_jt_2+L+c_vt}{c_V(t_1-t)}.
\]
Arvuliseks vastuseks saame \num{16}.

Jõhvikad

Keevasse vette kallatakse külmutatud jõhvikaid. Vee temperatuur langes väärtuseni $t=\SI{89}{\degreeCelsius}$.
Mitu korda oli vee mass suurem jõhvikate massist? Kuna jõhvikad olid väikesed ja sulasid väga kiiresti, siis
võib vee soojusvahetuse ümbritseva keskkonnaga arvestamata jätta. Jõhvikate algtemperatuur oli
$t_2=\SI{-18}{\degreeCelsius}$. Jää erisoojus $c_j=\SI{2100}{J/(kg.C)}$, vee erisoojus $c_v=\SI{4200}{J/(kg.\degreeCelsius)}$, jää sulamissoojus $L=\SI{330}{kJ/kg}$. Jõhvikate suure veesisalduse tõttu võib neid käsitleda jääna.

Energia jäävuse seaduse kohaselt peab jõhvikate soojendamiseks kuluv soojushulk tulema vee jahtumise arvelt.


[User]
Optiline süsteem koosneb kahest nõguspeeglist ja kumerläätsest (vt joonist), mille optilised peateljed ühtivad. Ringikujulise ristlõikega valgusimpulss siseneb süsteemi optilise peatelje sihis ning valgusvihu telg ühtib sellega. Peeglite kõverusraadiused on $R_1 = \SI{8}{m}$ ja $R_2 = \SI{4}{m}$ ning peeglite vahekaugus $L = \SI{6}{m}$. Peeglite läbimõõdud on $d_1 = \SI{160}{mm}$ ja $d_2 = \SI{96}{mm}$. Kiire läbimõõt on $D = \SI{192}{mm}$. Läätse läbimõõt on suurem valgusvihu omast. Suurema peegli keskel on ava läbimõõduga $d_0 = \SI{1}{mm}$. Joonistage valguse intensiivsuse ajaline kulg kumerläätse fookuses $f$. Eeldage, et süsteemi saabuva impulsi kestvus $\tau \ll L/c$. Valguse kiirus $c = \SI{3e8}{m/s}$.

\begin{center}
	\includegraphics[width=0.95\linewidth]{2006-v3g-09-yl}
\end{center}

Solution:
Alustuseks paneme tähele suurepärast seost arvandmetes. Nimelt ühtivad peeglite fookused, sest peeglite vahemaa on võrdne peeglite fookuskauguste summaga. Nõguspeegli fookuskaugus on teatavasti pool raadiusest. Paralleelne kiirtekimp koondub peegli fookuses. Seepärast jääb paralleelne kiirtekimp antud süsteemis pärast kahekordset peegeldumist ikkagi paralleelseks. Kuid paneme tähele, et kiire laius väheneb kaks korda, sest ühe läätse fookuskaugus on kaks korda suurem teise omast.

Valguskiir jääb niisiis peeglite vahele pendeldama seniks, kuni tema kaugus teljest on väiksem kui $d_0/2 = \SI{0.5}{mm}$, kusjuures esialgne kauguste vahemik, mis jõuab peegliteni, on $d_2/2 = \SI{48}{mm}$ kuni $d_1/2 = \SI{80}{mm}$, ehk pilu läbimiseks peab valgusvihk koonduma $d_2/d_0 = 96$ kuni $d_1/d_0 = 160$ korda. Pärast $n$-kordset edasi-tagasi peegeldumist väheneb kiire kaugus teljest $2^n$ korda. Paneme tähele, et 128 on kahe aste ($2^7 = 128$). Seepärast väljub peeglile langenud kiir pilust kahes järgus: esimene osa pärast seitsmendat edasi-tagasi liikumist ning teine osa pärast kaheksandat. Üks edasi-tagasi liikumine peeglite vahel tekitab ajalise viivise $2L/c$.

Kumerläätsele langev paralleelne kiirtekimp koondub fookuses, kusjuures fookuseni jõudmise aeg ei sõltu kiire asukohast (teljelähedased kiired läbivad paksema klaasikihi kui kaugemad kiired, klaasi läbib valgus aga aeglasemalt). Niisiis on fookuses oodata kolme impulssi: peegli ümbert tulnud osa, pärast seitset edasi-tagasi peegeldumist tulnud osa ning pärast kaheksat. Nende impulsside ajaline vahe on vastavalt $14L/c$ ja $2L/c$.

Nüüd tuleb veel leida impulsside suhtelised intensiivsused, mis on võrdsed vastavate rõngaste pindaladega esialgse kiire ristlõikes. Esimese peegli läbimõõdust väiksem osa ei läbigi süsteemi. Kõige välimise rõnga pindala on võrdeline arvuga $192^2 - 160^2$, järgmisele impulsile vastav rõnga pindala $128^2 - 96^2$ ning viimasele impulsile vastava rõnga korral $160^2 - 128^2$. Need arvud suhtuvad kui: $32^2(6^2 - 5^2) : 32^2 (4^2 - 3^2) : 32^2 (5^2 - 4^2) = 11 : 7 : 9$.

\begin{center}
	\includegraphics[width=0.6\linewidth]{2006-v3g-09-lah}
\end{center}

Sähvatus

Optiline süsteem koosneb kahest nõguspeeglist ja kumerläätsest (vt joonist), mille optilised peateljed ühtivad. Ringikujulise ristlõikega valgusimpulss siseneb süsteemi optilise peatelje sihis ning valgusvihu telg ühtib sellega. Peeglite kõverusraadiused on $R_1 = \SI{8}{m}$ ja $R_2 = \SI{4}{m}$ ning peeglite vahekaugus $L = \SI{6}{m}$. Peeglite läbimõõdud on $d_1 = \SI{160}{mm}$ ja $d_2 = \SI{96}{mm}$. Kiire läbimõõt on $D = \SI{192}{mm}$. Läätse läbimõõt on suurem valgusvihu omast. Suurema peegli keskel on ava läbimõõduga $d_0 = \SI{1}{mm}$. Joonistage valguse intensiivsuse ajaline kulg kumerläätse fookuses $f$. Eeldage, et süsteemi saabuva impulsi kestvus $\tau \ll L/c$. Valguse kiirus $c = \SI{3e8}{m/s}$.

\begin{center}
	\includegraphics[width=0.95\linewidth]{2006-v3g-09-yl}
\end{center}

Ülesandes on arvandmed valitud nõnda, et peeglite fookused ühtiksid. Tõepoolest, nõguspeeglite fookuskaugus on pool raadiusest ja peeglite vahemaa on võrdne peeglite fookuskauguste summaga. See tähendab, et peale igat edasi-tagasi peegeldumist püsivad valguskimbu kiired paralleelselt, kusjuures kimp muutub 2 korda kitsamaks (sest fookuskauguste suhe on 2).


[User] Rong, mis sõidab kiirusega $v = \SI{50}{km/h}$, sisenes hästi pikka tunnelisse. Nii rongi kui tunneli ristlõiget lugeda ruuduks küljepikkusega vastavalt $a = \SI{4}{m}$ ja $b = \SI{6}{m}$. Hinnake, milline on tuule kiirus rongi aknast mõõdetuna.

Solution:
Loeme õhu kokkusurumatuks. Sellisel juhul rongi ja tunneli seina vahele juurde
voolanud õhu ruumala võrdub rongi eest välja surutud õhu ruumalaga:
\[
\left(b^{2}-a^{2}\right) u^{\prime}=a^{2} v,
\]
kus $u'$ on õhu kiirus tunneli suhtes. Seega otsitav kiirus on
\[
u=v+u^{\prime}=v\left(1+\frac{a^{2}}{b^{2}-a^{2}}\right)=\frac{v b^{2}}{b^{2}-a^{2}}=\SI{90}{km/h}.
\]
Tegelikult on kiirus veidi väiksem, sest rongi ees surutakse õhk mingil määral kokku ja rongi ees ning taga tekib teatud õhuvool.

Tunnel

Rong, mis sõidab kiirusega $v = \SI{50}{km/h}$, sisenes hästi pikka tunnelisse. Nii rongi kui tunneli ristlõiget lugeda ruuduks küljepikkusega vastavalt $a = \SI{4}{m}$ ja $b = \SI{6}{m}$. Hinnake, milline on tuule kiirus rongi aknast mõõdetuna.

Rongi taustsüsteemis on rong paigal ning õhk uhab mööda kiirusega $v$, kusjuures õhu jaoks kehtib massi jäävuse seadus.


[User]
6. Adequate space has been provided in the answersheet for you to write/calculate your answers. In case
     you need extra space to write, you may request additional blank sheets from the invigilator. Write
     your roll number on the extra sheets and get them attached to your answersheet and indicate number
     of extra sheets attached at the top of this page.
                                                       9

Solution:
6. Consider n moles of a monoatomic non-ideal (realistic) gas. Its equation of state may be described
   by the van der Waal’s equation
                                                !
                                         an2          V

                                       P+ 2             − b = RT
                                         V            n

   where a and b are positive constants and other symbols have their usual meanings. The internal
   energy change of a realistic gas can be given by
                                                       dP

                                 dU = CV dT + T                     − P dV
                                                       dT       V

                                                                    9
   As indicated in the above expression, the derivative of pressure is taken at constant volume.
   We take one mole of the gas (n = 1) through a Diesel cycle (ABCDA) as shown in the following

                                              C1
   P -V diagram (diagram is not to scale). During the whole cycle assume that the molar heat
   capacity at constant volume (CV ) remains constant at 3R/2. Path AB and CD are reversible
   adiabats.

                                  P
                                          B
                            PB = PC

                      HB                          C

                                                                             D

                                                                             A
                                                                                V
                                         VB     VC                     VA = VD

   (a) Obtain the temperature at B (TB ) in terms of temperature at A (TA ), VA , VB and constants      [21/2]
       only.

         TB =

         Solution:
                                                           RT

                                      dU = CV dT +             − P dV
                                                          V −b
         AB and CD are reversible adiabats, hence entropy change during these processes are
         zero.

INPhO 2019 Page 14                Questions & Answers                   Last four digits of Roll No.:

                                            Z B        D dQ
                                              dQ
                                                        Z
                         ∆SAB = ∆SCD =            =          =0                               (6.1)
                                           A T        C   T
                                Z B
                                     dQ      CV            R
                                          Z           Z
                                        =       dT +          dV = 0                          (6.2)
                                 A    T      T           V −b
                                     TB           VB − b

                               CV ln    = −R ln                                               (6.3)
                                     TA           VA − b
                                               VB − b −R/CV          VB − b −2/3

                                     TB = TA                  = TA                            (6.4)
                                               VA − b                VA − b

   (b) Let temperature at A to be TA =100.00 K, VA = 8.00 l, VB = 1.00 l, VC = 2.00 l, a = 1.355        [11/2]
       l2 ·atm/mol2 , and b = 0.0313 l/mol. Calculate the highest temperature reached during the
       whole cycle.

         Highest temperature =

                                                         9
          Solution: Highest temperature during cycle is at TC .

                                     C1
          which gives, from van der Waal equation
                                                   TB = 407.5 K                               (6.5)

                                            PB = 33.18 atm = PC                               (6.6)
                                         ⇒ TC = 803.76 K ≈ 804 K                              (6.7)

               HB
          Different value of TC (within a range) obtained due to reasonable roundoff in previous
          step(s) will be credited.

    (c) Calculate the efficiency η of the cycle.

         Value of η =
                                                                                [3]

          Solution:

                           dQin = dQBC = dU + PB dV                                           (6.8)
                                                   RT
                                       = CV dT +       dV                                     (6.9)
                                                  V −b
                                                        a

                                       = CV dT + P + 2 dV                                   (6.10)
                                                       V
                                                                              a VC
                                      Qin = CV (TC − TB ) + PB (VC − VB ) −                 (6.11)
                                                                              V VB
                                     Qout = QDA = CV (TA − TD )                             (6.12)
                                               Qout
                                        η =1−       = 67.7% ≈ 68%                           (6.13)
                                               Qin

          Different value of η (within a range) obtained due to reasonable roundoff in previous
          step(s) will be credited.

INPhO 2019 Page 15            Questions & Answers
   (d) Draw the corresponding T -S (entropy) and V -T diagram for the Diesel cycle. Wherever   [7]
       possible, mention the numerical values of T, V, and S on the diagrams.
               T

                                                                                S

               V

                                          C1            9
                     HB
                                                                                T

        Solution:

                                         dQ      CV dT         R
                                      Z       Z           Z
                             ∆SBC =         =           +         dV                (6.14)
                                         dT        T         V −b
                                            TC        VC − b
                                    = CV ln    + R ln                               (6.15)
                                            TB        VB − b
                                    = 1.73R = ∆SDA                                  (6.16)

INPhO 2019 Page 16              Questions & Answers                     Last four digits of Roll No.:

                               T (K)
                                                            C
                                 803

                                        B
                                 408
                                 316                            D

                                 100
                                        A
                                                                    S (J/K)

                                                    1.73R

         Solution:
                                V (l)
                                        A      D
                                   8

                                   C1
                                   2
                                   1
                                                     B
                                                           9    C

                                                                     T (K)
                                        100   316    408     803

                              Detailed answers can be found on page numbers:

               HB

INPhO 2019 Problem 6


[User]
Consider a potential of the form V (r) ∝ −1/rn . It turns out that for some value of
    n, it is possible for a particle to orbit in a circle passing through the origin. What is n?

Solution:
This is a classic problem, which was common in mechanics books in the 1800s. Technically,
    it’s not really physical since the potential blows up near the origin, so the particle has to be aimed
    perfectly to pass straight through it rather than get deflected through some angle, but it’s still cute.
    Letting V (r) = −a/rn and using the effective potential results from M6, we have
                                             2
                                        1     dr            a      L2
                                          m         =E− n −            .
                                        2     dt           r     2mr2
    We also know that
                                                 dθ      L
                                                     =
                                                  dt   mr2
    and since we’re interested in the trajectory’s shape, we multiply by (dθ/dt)−2 to get
                                      2
                                                m2                 L2 r2

                                 1     dr               4     a
                                   m         = 2 Er − n−4 −                .
                                 2     dθ       L           r       2m
    The equation of a circle through the origin in polar coordinates is r = b sin θ, so
                                        2
                                         dr
                                               = b2 cos2 θ = b2 − r2 .
                                        dθ
    We therefore must have, for appropriate constants E, L, and b, that
                                                               L2 r2

                                  2   2    2m      4     a
                                 b − r = 2 Er − n−4 −                  .
                                            L          r        2m

    The final terms on each side cancel, so the first two terms on the right-hand side have to sum to a
    constant. This is only possible if E = 0 and n = 4. (The orbit can have finite L because v diverges
    when r goes to zero.)


[User]
2. GLASS PLATE (10 points)

— Rasmus Kisel
and Mihkel Heidelberg. Light from a laser with
a wavelength of λ and power of P falls perpen-
dicularly on to a thin glass plate of thickness
a = (100.25λ)/n where n is the index of re-
fraction of glass. The reflection coefficient of the
glass surfaces is r, which can be interpreted as
the probability for a photon to be reflected from
a single glass surface.
i) (2 points)Calculate the force Fa applied to the
glass by the laser light when the incident side of
the glass is painted totally black.
ii) (3 points)Calculate the force Fb applied to the
glass by the laser light when the opposite side of
the glass is painted totally black.
iii) (5points)Calculate the force Fc applied to the
glass by the laser light when neither of the sides
is painted.

Solution:
2. Glass plate

i) The energy and impulse of a photon are re-
lated by E = pc. Because the incident side is
black,itwillabsorbthephotonsandfeeltheforce
due to their momentum: Fa = dp
dt , where dp
dt =
1
c
dE
dt = P
c is the impulse of the photons gener-
ated per unit time. Fa = P
c .
ii) A fraction r of the photons will be reflected,
sotheywillexerttwicetheforce,becausethemo-
mentum change is twice as much as compared to
absorbingthem. Afraction1−rwillbeabsorbed
andwillgiveaforcesimilartothepreviousexpres-
sion.
Fb = 2r
P
c
+ (1 − r)
P
c
= (1 + r)
P
c
.
iii) The light will be reflected many times and
we will have to consider interference. Since en-
ergy is proportional to the square of the amp-
litude of the electric field, if a fraction r of the
photons and therefore of the energy is reflected,
theamplitudeofthereflectedwave’selectricfield
is
√
r of the incident electric field. Let’s calculate
the total electric field amplitude of the wave that
goes through. Denote by Ei the electric field of
the incident wave and by Et the electric field of
the wave that has gone through the other side of
the glass. On the incident side the proportion of
the electric field amplitude that goes through is √
1 − r, of which a part
√
1 − r will go through
and part
√
r will bounce pack. So a wave with
amplitudeE0 = (1−r)Ei wentstraightthrough
andawavewithamplitude
√
(1 − r)rEi isboun-
cingbackthroughtheglass. Nowitwillgobackto
thefirstside,reflectbackfromthereandofthata
proportion
√
1 − rwillgothroughandaddtoEt.
However the wave has traveled an optical path
2nd = 200.5λ, so it will be 180 degrees out of
phase and we get E1 = −(1 − r)rEi. Repeat-
ing the procedure E2 = (1 − r)r2
Ei and so on
En = (1−r)(−r)n
Ei. Summingalltheelectric
fields that go through the other side of the glass
we get:
Et =
∞ ∑
n=0
En = (1−r)Ei
∞ ∑
n=0
(−r)n
=
1 − r
1 + r
Ei.
Powerisproportionaltothesquareoftheelectric
field:
Pt
P
=
(
Et
Ei
)2
=
(
1 − r
1 + r
)2
.
Sincepowerisalsoproportionaltothenumberof
photons, this fraction of the photons have gone
through and haven’t given their momentum to
the glass. By energy conservation the rest have
been reflected back toward the laser and since
their momentum has been reversed:
Fc = 2
(
P − Pt
c
)
=
2P
c
(
1 −
(
1 − r
1 + r
)2
)
=
8Pr
c(1 + r)2
.
Note: if we would calculate directly the total
electricfieldof thereflected light,wewouldneed
to take into account that on the first reflection
of the laser light from the glass there is a phase
shift of 180 degrees due to reflection from a sur-
facewithahigherrefractiveindex(airtoglasssur-
face).
Alternative solution: The light is bouncing
back and forth. Inside the glass there is some
sum of waves with the same frequency, which we
can represent as a single wave moving forward
and a single wave moving backwards. Denote
by E the wave amplitude of the forward moving
wave right under the surface in the glass. The
wave amplitude is constant at that point and it
is made of the transmitted part of the incoming
light
√
1 − rEi and from the 180 degrees out of
phasepartofEitselfasitbouncestwiceinsidethe
glasstoarriveatthesamepointbackagain: −rE.
We get the relation:
E =
√
1 − rEi −rE → E =
√
1 − r
1 + r
Ei.
The wave that has gone all the way through the
plate is
Et =
√
1 − rE =
1 − r
1 + r
Ei.
The rest follows as in the previous solution.

GLASS PLATE


[User]
INPhO 2020, problem 1. A very contrived application of partial derivatives.

    2      The Second Law
         Idea 5: Entropy

         The increase in entropy of a body at temperature T that receives heat d̄Q is
                                                             d̄Q
                                                    dS =
                                                              T

       if the process is reversible; moreover, entropy is a state function.

       Idea 6: Second Law
      The second law of thermodynamics takes several equivalent forms, which are all useful.

        • For an isolated system, dS > 0 for any spontaneous process.

        • Clausius: no process is possible whose sole result is the transfer of heat from a colder to
          a hotter body.

        • Kelvin: no process is possible whose sole result is the complete conversion of heat into
          work.

        • Carnot: of all heat engines working between two given temperatures, none is more
          efficient than a Carnot engine.

       Example 4: Mixing Entropy

      A container of volume V is divided in half by a partition. The two halves contain pure
      nitrogen and oxygen gas, respectively, both at pressure P and temperature T . The partition
      is removed and the gases are allowed to mix. How much does the entropy increase?

       Solution
      We cannot apply the formula dS = d̄Q/T , because the gas is not in equilibrium during
      the process. However, because entropy is a state function, we can calculate its change by
      considering a path that only goes through equilibrium states.

       Suppose we introduce a piston at one end of the container, which is magically transparent
       to oxygen atoms, and slowly and isothermally move it to the center of the container. This
       puts the nitrogen back where it started. The change in entropy of the nitrogen is
                                 Q    W          Vi
                        ∆SN =      =−   = nR log    = −nR log 2 = −N kB log 2
                                 T    T          Vf

      where N is the number of nitrogen molecules, which is also the number of oxygen molecules.
      If we do the same for the oxygen with another piston, introduced from the other side, we
      get another, equal decrease in the entropy. This takes us back to the original state, so since
      entropy is a state function, the increase in entropy when we allowed the gases to mix was

                                        ∆S = −2∆SN = 2N kB log 2.

Solution:
See the official solutions here.

    2      The Second Law
         Idea 5: Entropy

         The increase in entropy of a body at temperature T that receives heat d̄Q is
                                                            d̄Q
                                                    dS =
                                                             T
         if the process is reversible; moreover, entropy is a state function.

         Idea 6: Second Law
         The second law of thermodynamics takes several equivalent forms, which are all useful.

           • For an isolated system, dS > 0 for any spontaneous process.

           • Clausius: no process is possible whose sole result is the transfer of heat from a colder to
             a hotter body.

           • Kelvin: no process is possible whose sole result is the complete conversion of heat into
             work.

           • Carnot: of all heat engines working between two given temperatures, none is more
             efficient than a Carnot engine.

         Example 4: Mixing Entropy

         A container of volume V is divided in half by a partition. The two halves contain pure
         nitrogen and oxygen gas, respectively, both at pressure P and temperature T . The partition
         is removed and the gases are allowed to mix. How much does the entropy increase?

       Solution
      We cannot apply the formula dS = d̄Q/T , because the gas is not in equilibrium during
      the process. However, because entropy is a state function, we can calculate its change by
      considering a path that only goes through equilibrium states.

       Suppose we introduce a piston at one end of the container, which is magically transparent
       to oxygen atoms, and slowly and isothermally move it to the center of the container. This
       puts the nitrogen back where it started. The change in entropy of the nitrogen is
                                  Q    W          Vi
                         ∆SN =      =−   = nR log    = −nR log 2 = −N kB log 2
                                  T    T          Vf

      where N is the number of nitrogen molecules, which is also the number of oxygen molecules.
      If we do the same for the oxygen with another piston, introduced from the other side, we
      get another, equal decrease in the entropy. This takes us back to the original state, so since
      entropy is a state function, the increase in entropy when we allowed the gases to mix was

                                        ∆S = −2∆SN = 2N kB log 2.


[User] An optical fiber consists of a cylindrical glass core with a refractive index $n_1=\SI{1,46}{}$ and a coating with a refractive index $n_2=\SI{1,44}{}$ that surrounds the core as a tube. Using classical optics, find the apex angle of a cone of light leaving a long optical fiber.

Solution:
Pikas optilises kius jäävad levima vaid sellised kiired, mille jaoks toimub südamiku ja katte lahutuspinnal täielik sisepeegeldumine. Kui valgus langeb lahutuspinnale täieliku sisepeegeldumise piirnurgast väiksema nurga all, siis toimub korraga nii peegeldumine kui ka murdumine. Pärast mitmeid peegeldusi väheneb nende kiirte intensiivsus praktiliselt nullini, sest peaaegu kogu valgus on kiu külgedelt välja murdunud. Täieliku sisepeegeldumise piirnurk
\[
\alpha=\arcsin(n_2/n_1)=\ang{80,5}.
\]
Piirnurgale vastavad kiired levivad kiu telje suhtes nurga $\ang{90}-\alpha$ all. Pärast kiu otsast väljamurdumist on nende kiirte nurk kiu telje suhtes
\[
\beta=\arcsin[n_1\sin(\ang{90}-\alpha)]=\arcsin[n_1\cos(\alpha)].
\]
Valguskoonuse tipunurk $\theta$ on sellest kaks korda suurem:
\[
\theta=2\beta=2\arcsin[n_1\cos(\alpha)]=\ang{28}.
\]
Kuna $\cos(\arcsin(a))=\sqrt{1-a^2}$, siis on võimalik vastus esitada kujul
\[
\theta=2\arcsin(\sqrt{n_1^2-n_2^2}).
\]

\begin{center}
 \includegraphics[width=0.35\textwidth]{2016-lahg-04-kiud}
\end{center}

Optiline kiud

Optiline kiud koosneb silindrikujulisest klaassüdamikust murdumisnäitajaga $n_1=\SI{1,46}{}$ ja seda toruna ümbritsevast kattest murdumisnäitajaga $n_2=\SI{1,44}{}$. Leidke pikast optilisest kiust väljuva valguskoonuse tipunurk, lähtudes klassikalisest optikast.


[User]
Question B1
Virial Reality
The ideal gas law states that P Vm = RT , where Vm = V /n is the volume per mole of gas. However,
any real gas will exhibit deviations from the ideal gas law, described by the virial expansion,

                                                B(T ) C(T )
                              P Vm = RT 1 +           +        + ... .
                                                 Vm       Vm2

For gases with low density, the higher-order terms are negligible, so in this problem we will neglect
all of the temperature-dependent terms in parentheses except for B(T )/Vm . The table below shows
measurements of B for nitrogen gas (N2 ) at atmospheric pressure, P = 1.01 × 105 Pa.

                                       T (K)    B (cm3 /mol)
                                        100         −160
                                        200         −35
                                        300         −4.2
                                        400          9.0
                                        500         16.9
                                        600         21.3

a. According to the ideal gas law, what is the value of Vm at temperatures 100 K, 300 K, and 600
   K? Give your answers in SI units.

b. What is the percentage change in Vm at these temperatures if one accounts for B(T )?

c. In 1910, van der Waals was awarded the Nobel Prize for formulating the equation

                                          a
                                    P + 2 (Vm − b) = RT
                                         Vm

   which accurately describes many real gases. According to this equation, what is the form of
   B(T )? You may assume that b ≪ Vm .

d. Using the data above, extract the values of a and b. Give your answers in SI units.

e. In this problem, we have neglected terms in the virial expansion beyond B(T ), which is a good
   approximation as long as the volume correction due to B(T ) itself is small. Assuming the van
   der Waals equation holds, numerically estimate the temperature range within which the volume
   correction due to B(T ) is at most 10%, for nitrogen gas at atmospheric pressure.

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                       Part B                                              9

Solution:
Question B1
Virial Reality
The ideal gas law states that P Vm = RT , where Vm = V /n is the volume per mole of gas. However,
any real gas will exhibit deviations from the ideal gas law, described by the virial expansion,

                                                B(T ) C(T )
                              P Vm = RT 1 +           +        + . . .   .
                                                 Vm       Vm2
For gases with low density, the higher-order terms are negligible, so in this problem we will neglect
all of the temperature-dependent terms in parentheses except for B(T )/Vm . The table below shows
measurements of B for nitrogen gas (N2 ) at atmospheric pressure, P = 1.01 × 105 Pa.
                                       T (K)    B (cm3 /mol)
                                        100         −160
                                        200         −35
                                        300         −4.2
                                        400          9.0
                                        500         16.9
                                        600         21.3
a. According to the ideal gas law, what is the value of Vm at temperatures 100 K, 300 K, and 600
   K? Give your answers in SI units.

                                               Solution
     Plugging in the values and converting to m3 /mol gives
                                       
                                                  −3 3
                                       8.23 × 10 m /mol          T = 100 K
                                       
                                       
                                RT
                          Vm =      = 2.47 × 10−2 m3 /mol         T = 300 K
                                 P     
                                       4.94 × 10−2 m3 /mol
                                       
                                                                  T = 600 K

b. What is the percentage change in Vm at these temperatures if one accounts for B(T )?

                                               Solution
     The fractional change in volume is
                                           
                                           −1.9%
                                                              T = 100 K
                               ∆Vm   B(T ) 
                                   ≈      ≈ −0.02%             T = 300 K
                                Vm    Vm   
                                           0.04%
                                           
                                                               T = 600 K

c. In 1910, van der Waals was awarded the Nobel Prize for formulating the equation

                                          a
                                    P + 2 (Vm − b) = RT
                                         Vm
   which accurately describes many real gases. According to this equation, what is the form of
   B(T )? You may assume that b ≪ Vm .

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                    Part B                                         15

                                             Solution
     We have
                                      a     RT      RT                 b
                                P+       =        ≈              1+            .
                                     Vm2   Vm − b   Vm                Vm
     Rearranging yields
                                                  b   a
                                   P Vm ≈ RT 1 +    −
                                                 Vm Vm RT
     from which we conclude
                                                          a
                                           B(T ) = b −      .
                                                         RT

d. Using the data above, extract the values of a and b. Give your answers in SI units.

                                             Solution
     First, we convert the data points to SI units, giving

                                         T (K)   B(m3 /mol)
                                          100    −1.6 × 10−4
                                          200    −3.5 × 10−5
                                          300    −4.2 × 10−6
                                          400    9.0 × 10−6
                                          500    1.69 × 10−5
                                          600    2.13 × 10−5

     Now, plotting B versus 1/T , the slope is −a/R and the y-intercept is b.

     The final result is
                             a = 0.18 J m3 /mol2 ,   b = 6 × 10−5 m3 /mol.
     Any answer within 20% of this is acceptable.

e. In this problem, we have neglected terms in the virial expansion beyond B(T ), which is a good
   approximation as long as the volume correction due to B(T ) itself is small. Assuming the van
   der Waals equation holds, numerically estimate the temperature range within which the volume
   correction due to B(T ) is at most 10%, for nitrogen gas at atmospheric pressure.

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                               Part B                                          16

                                         Solution
   We are looking for the temperature range where
                                                Vm   RT
                                    |B(T )| ≲      ≈     .
                                                10   10P
   At high temperatures, B(T ) always yields a small correction for nitrogen at atmospheric
   pressure. At low temperatures, B(T ) is dominated by the −a/RT term, so we require
                                           √
                                             10P a
                                    T ≳            = 50 K.
                                              R
   Any answer within 25% of this is acceptable.

                Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                        Part B                                            17

USAPhO 2022 Question B1


[User]
5. Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the
     Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary

                                 C2
     Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer
     calculations. Strike out any rough work that you do not want to be considered for evaluation.

Solution:
5. The following is the top view of an assembly kept on a smooth horizontal table.                         [20]

                               D
                        θ                   L
                                                                        y
                                                       vo
                                                            0                   x

                                       C2
   A massless inextensible string of length D lies with one end fixed, while the other is attached
   to one end of a uniform rod of length L. The system is initially at rest with the rod aligned
   along the x-axis and the string stretched to its natural length at an angle with the negative y-axis
   θ (cos θ = 1/3). At a certain instant, a bullet of the same mass m as the rod and negligible

                HB
   dimensions is fired horizontally along the positive y-direction. The bullet hits the rod at its right
   end with velocity vo and gets lodged in it, the impact being nearly instantaneous. What is the
   tension (T ) in the string immediately after the impact? Assume the string doesn’t break.

    T =

     Solution: Geometry of the problem dictates that the string becomes taut during the impact
     and hence exerts an impulse. We conclude:

INPhO 2020 Problem 5


[User]
Three balls
Three small identical balls (denoted as A, B, and C) of mass m each are connected with two
massless rods of length l so that one of the rods connects the balls A and B, and the other rod
connects the balls B and C. The connection at the ball B is hinged, and the angle between the rods
can change effortlessly. The system rests in weightlessness so that all the balls lie on one line. The
ball A is given instantaneously a velocity perpendicular to the rods. Find the minimal distance
d between the balls A and C during the subsequent motion of the system. Any friction is to be
neglected.

Solution:
Solution - Three balls
  Let the ball A be given an initial velocity v along the axis Y , which is perpendicular to the rod.
The total momentum of the system conserves, therefore the center-of-mass (CM) of the system
moves with a constant velocity:
                                                    mv   v
                                            vCM =      =
                                                    3m   3

along Y . In what follows, we will work in the CM frame of reference, which is an inertial system
of reference. Therefore, in the CM frame the laws of conservation of energy, momentum, and the
angular momentum hold true. The initial velocities of the three balls along Y are:
                                            2v              v
                                     vA =      ; vB = vC = − .
                                             3              3

Correspondingly, the total kinetic energy of the balls is:

                                      mv 2 4 1 1             mv 2

                                 E=            + +         =
                                        2     9 9 9           3

and the total angular momentum with respect to the CM equals:
                                            2v      v
                                   L=m         ` − m (−`) = mv`
                                             3      3

    In any moment the three balls form an isosceles triangle with an angle 2φ at the top vertex.
The distance between A and C is minimal when either φ = 0, or φ̇ = 0. For φ = 0, however, the
laws of conservation are not consistent with the rigidity of the rods. Therefore, at the minimal
distance φ̇ = 0, and in this particular instance the system behaves as a rigid body whose moment
of inertia with respect to the CM can be obtained through:

                                               L2  3
                                         I=       = m`2 .
                                               2E  2

On the other hand, the moment of inertia I could be found independently from geometric conside-
rations. Although, I could be found using the distances from the balls to the CM (medicentre of
a triangle), it is more convenient to use this relatively unknown formula for the moment of inertia
of a collection of point masses with respect to CM:
                                                                   2
                                           P
                                             i,j mi mj (r i − r j )
                                      I=           P                 .
                                                     i mi
In our case:

                              m2                        m
                           I=    (AB 2 + BC 2 + AC 2 ) = (2`2 + d2 ).
                              3m                        3
From (1) and (2), we obtain the minimal distance:
                                            r
                                              5
                                       d=`      ≈ 1.58`
                                              2

Three balls


[User]
The function cos−1 (1 − x) does not have a Taylor series about x = 0. However, it
    does have a series expansion about x = 0 in a different variable.

      (a) What is this variable, and what’s the first term in the series?

     (b) ⋆ What’s the next nontrivial term in the series?

       Idea 5: Binomial Theorem
       When the quantity xn is small, it is useful to use the binomial theorem,

                                       (1 + x)n = 1 + xn + O(x2 n2 ).

       It applies even when n is not an integer. In particular, n can be very large, very small, or
       even negative. The extra terms will be small as long as xn is small. If desired, one can find
       higher terms using binomial coefficients,
                                                    ∞
                                                 n
                                                    X  n m
                                          (1 + x) =      x
                                                       m
                                                      m=0

       where the definition of the binomial coefficient is formally extended to arbitrary real n.

    The binomial theorem is one of the most common approximations in physics. It’s really just taking
    the first two terms in the Taylor series of (1 + x)n , but we give it a name because it’s so useful.

Solution:
(a) We have
                                         d                       1
                                           arccos(1 − x) = p
                                        dx                   1 − (1 − x)2
         which is undefined at x = 0, so there is no Taylor series. But note that if we let y = cos−1 (1−x)
         and take the cosine of both sides, we have

                                                   cos y = 1 − x.

          Now y does have a good Taylor series near y = 0, which corresponds to where x = 0. At
          lowest order, we have
                                             1 − y 2 /2 ≈ 1 − x
         which implies that                            √
                                                    y ≈ 2x.
                                                   √
          More generally, the answer is a series in x. Since cosine is even, the next term is O(x3/2 ).

     (b) In order to get higher order terms, we can write

                                           cos−1 (1 − x) = cos−1 (1 − u2 )
                    √
         where u = x, and directly compute a Taylor series in u, using the usual rule for a derivative
         of an inverse function.

         That appraoch is straightforward, but for variety we’ll show a slightly trickier but much faster
         method. Write the answer as
                                            √
                                        y = 2x1/2 + Ax3/2 + O(x5/2 )

         where A is to be determined. By the Taylor series for cosine, we know that

                                                              y2   y4
                                      1 − x = cos y = 1 −        +    + O(y 6 )
                                                              2    24
         which is equivalent to
                                                   y2    y4
                                              x=      −     + O(x3 )
                                                    2    24
         since y 6 = O(x3 ). We can now plug in our expression for y. Equating the term linear in x
         just recovers the result of part (a), while equating the term quadratic in x (which vanishes
         on the left-hand side) gives the coefficient A. Concretely, we have

                                           y2      √
                                              = x + 2 A x2 + O(x3 )

         and
                                               y4  1
                                                  = x2 + O(x3 )
                                               24  6
                                        √
         from which we conclude A = 1/(6 2), so that
                                                       √       x3/2
                                     cos−1 (1 − x) =       2x + √ + O(x5/2 ).
                                                               6 2
         This technique is called “reversion of series”.

       Idea 5: Binomial Theorem
       When the quantity xn is small, it is useful to use the binomial theorem,

                                       (1 + x)n = 1 + xn + O(x2 n2 ).

       It applies even when n is not an integer. In particular, n can be very large, very small, or
       even negative. The extra terms will be small as long as xn is small. If desired, one can find
       higher terms using binomial coefficients,
                                                    ∞
                                                 n
                                                    X  n m
                                          (1 + x) =      x
                                                       m
                                                       m=0

       where the definition of the binomial coefficient is formally extended to arbitrary real n.

   The binomial theorem is one of the most common approximations in physics. It’s really just taking
   the first two terms in the Taylor series of (1 + x)n , but we give it a name because it’s so useful.


[User] A hole with an area of $S$ has been made in the bottom of a half-liter bottle. The bottle has $m$ grams of water in it. The lid of the bottle is removed and the bottle is thrown in the air with a starting speed of $v$. How fast is the water flowing out of the hole when the bottle is still moving up? How fast is it flowing out when the bottle is falling down? Explain.

Solution: Mõlemat juhtu, millal pudel liigub üles ning pudel liigub alla, võib vaadelda kui vabalangemist. Kuna pudelile ja veele mõjuvad jõud on vabalangemise korral samasugused, siis vesi ei voola pudelist välja kummalgi juhul. Seega vee väljavoolu kiirus on \SI{0}{m/s}.

Lendav pudel

Pooleliitrises pudelis, mille põhja on tehtud \SI{0.4}{V} väike auk pindalaga $S$, on $m$ grammi vett. Pudelil keeratakse kork pealt ning pudel visatakse õhku algkiirusega $v$. Kui kiiresti voolab vesi pudeli põhjas olevast august välja siis, kui pudel veel üles liigub? Kui kiiresti voolab vesi august välja sel hetkel, kui pudel alla kukub? Põhjendage.

Nii pudelile kui veele mõjuvad täpselt samad jõud ning neid visatakse sama algkiirusega.


[User]
7. Lifting by current (7 points)

Consider a loop of freely deformable conduct-
ing wire with insulation of length 2l, the two ends
of which are fixed (permanently) to the ceiling. A
load of mass m is fixed to the middle of the wire
(the mass of the wire is negligible). There is also a
horizontal magnetic field of induction B; free fall
acceleration is g. A current I is lead through the
wire. Neglect the field induced by the wires.
i. (2 pts) Sketch the shape of the wire.
ii. (2 pts) What is the maximal height by which the load can
be lifted in such a way (increasing the current, if necessary)?
iii. (2 pts) Write an equation from which it is possible to
determine the lifting height ∆h.
iv. (1 pt) How large current I0 is needed to lift the load by
∆h0 = l(1 − 3
π )?
Problem

Solution:
7. Lifting by current

i. (2 pts) The Ampère force pulls the wires to the side so that
the wires will take a curved shape. Since the Ampère force is
perpendicular to the wire, the mechanical tension is constant
along the wires. Let the tension be T and the curvature of the
wire at a certain point — R. Let us consider a short piece of
the wire, of length a ≪ R. Then the angle by which the tan-
gent of the wire rotates while the tangent point moves over an
arc of length a is given by α = a/R. Let us study the force bal-
ance in the perpendicular direction for that piece of wire: the
Ampère’s force IaB is balanced by the tension Tα = Ta/R.
So, R = T/IB which means that R is constant, and the wire
will take the form of a circle segment. To conclude, both halves
of the wire will take the form of a circle segment, the convex
sides of which are turned outside.
ii. (2 pts) The maximal height is achieved when the circle
segments form a perfect circle, in which case the lifting height
is ∆h = l(1 − 2
π ).
iii. (2 pts) If the central angle of the circle segments is 2α,
the tangents to the wires at the point where the load is fixed
forms angle α with the vertical direction. So, the lifting force
is mg = 2T cosα. From the other hand, R = l/2α = T/IB, ie.
α
mg
lIB
= cosα,
which is the equation from where one can determine the angle
α. Then, the lifting height
∆h = l − 2Rsinα = l −
2T
lIB
sinα = l

1 −
sinα
α

.
iv. (1 pt) From the previous result it can be seen that we need
to have sinα
α = π
3 , hence α = π
6 and
I =
mgα
lB cosα
=
mgπ
3
√
3lB
.
Problem

Lifting by current