SOLUTIONS TO THE THEORETICAL COMPETITION

  1 st ASI AN PH Y SI CS OLY M PI AD K ARAWACI , I N DON ESI A SOLU T I ON S T O T H E T H EORET I CAL COM PET I T I ON APRI L 2 5 , 2 0 0 0 T im e a va ila ble : 5 hours READ THIS FIRST : 1. Use only the pen provided.

  2. Use only the marked side of the paper

  3. Each problem should be answered on separate sheets

  4. In your answers please user primarily symbols, equations, numbers, graphs, tables and as little text as possible.

  5. Write at the top of every sheet in your report:

• Your candidate number (APhO identification number).
• The problem number and section identification, e.g.2/a.
• Number each sheet consecutively.

  6. Write on the front page the total number of sheets in your report T his se t of proble m s c onsist s of …. pa ge s

  Solut ion Proble m 1 Ec lipse s of t he J upit e r’s Sa t e llit e

  We know

  T v π π ω

  R T and v R

  ^E 2 ^{E E E} _{J J J J}

  2

  2 ,

  2

    =    

  a. ( Total Point : 1 ) Assume the orbits of the earth and Jupiter are circles, we can write the centripetal force = equal gravitational attraction of the Sun. _{2 2}

  E J R v R v

  V J = velocity of Jupiter Hence _{2 J E}

  V E = velocity of the Earth

  (0.5 point) where G = universal gravitational constant M S = mass of the Sun M E = mass of the Earth M J = mass of the Jupiter R E = radius of the orbit of the Earth

  = =

  2 ^{E E} ₂ ^{E s E E} _{J S J J J J} M M M V G R R M M M V G R R

  π π ω = = = = we get _{R v E 3 2} T E  R  _{E E}

  = _{R J =  } T R _{J v J J}

    ₆ R 779.8

  10 km _J = × ( 0.5 point )

  b. ( Total Point: 1 ) The relative angular velocity is

  1

  1

  

  2

  ω ω = E − ω J = π  −  

  365 11.9 365

  ×  

  0.0157 rad day /

  =

  ( 0.5 point ) and the relative velocity is ₆

  

v R 2.36 10 km day /

= ω E = ×

₃

  27.3 10

  km = ×

  ( 0.5 point )

  c. ( Total Point: 3 ) The distance of Jupiter to the Earth can be written as follows d t R R

  

= J − E

( )

  d t .d t R R . R R

  = J − E J − E ( ) ( ) ( ) ( ) ₁ (1.0 point)

  2 R R cos t

  

= J E − E J ω

^{2 2 2} ( )

• d t R R

  ( ) _{2 1}  R    ^E + R 1 2 cos t ...

≈ J − ω

 

      R _J

      R

   

^E

+ R 1 cos t ...

  ≈ J − ω   R

_J

    The relative error of the above expression is the order of ₂

   R  _E

  4 %

  

≈

  R _J

   

  The observer saw M begin to emerge from the shadow when his position was at d(t) and he saw the next emergence when his position was at d(t +T0)/ Light need time to travel the distance so the observer will get apparent period T instead of the

  ∆ d =d(t+T )-d(t) true period T . o

  ∆ = E ω − ω ( )

• d R (cos t cos t T )

  R T sin t ≈ ω ω _E

  (1.0 point)

  ω T ω t ω T ≈ ≈ −

• because 0.03,sin ..., cos 1 ...

  We can also get this approximation directly from the geometrical relationship from Figure 1.

  (1.0 point) or we can use another method.

• =

  d t T T c c d t

  km/s (1.0 point)

  5

  Hence C=2.78 x 10

  R T c ω =

  15 ^E

  we get

  ω = +

  max ^E R T T T c

  (1.0 point)

  ≈ + = +

  ω ω ∆ − ≈ = ∆

  R T t T T T c c

  ; velocity of light sin _E

  From the figure above we get ( )

  d. ( Total Point: 2) ( ) ( )

  (1.0 point)

    ≈ ≈

  ∆ ≈   ≈ + +  

  ω α ω

ω ω φ

ω φ

  d T R T T R t T

  0.19 ^{E E}

  2 0.03 and

  (1.0 point) cos sin

  β φ α ω π β θ = +

  2 T

  2

e. Total Point : 2 from

  (1.0 point)

  0.1 0.1 1.9 10 1.6 10 ^ion

  

× × ×

∆ = =

×

  (0.5 point)

  b. Electrons from the ions-pairs produced by

  α

  particles from a radioactive sources of activity A (=number of

  α

  particles emitted by the sources per second) which enter the detector with detection efficiency 0.1, will produce a collected current. _{α ir 5 19}

  Q

  0.68 4.5 10

  I AN e t A A

  ρ − −

  = = ×

= × × × × ×

  (1.0 point) With I min =10

  A, the _{12 1 1 min 15} 10 dis 330 dis

  1.6 1.9 10

  s

A s

− −

−

  − = = × ×

  V V mV − −

  1.9 10 1.6 10

  Solut ion Proble m 2 De t e c t ion of Alpha Pa rt ic le s

  1.9 10

  a. From the given range-energy relation and the data supplied we get _{2 2 3 3}

  5.50

  6.69 0.318 0.318

  R E MeV MeV α    

  

= = =

       

  (0.5 point) since W

  ion- ρ air

  = 35 eV, then _{6 5 α ir} 6.69 10

  35 ^ion

  (0.5 point) Hence _{5 19 11}

  N ρ −

  × = = ×

  (0.5 point) Size of voltage pulse: ₁₁ with

  45 4.5 10 ^{air air}

  N e Q

  V C C

C pF

ρ −

  − ∆ ∆ = =

  = = ×

• 12
• 1

  Since 1 Ci = 3.7 x 10

  ∆

  2 E dS rlE

  The surface integral .

  σ l.

  and length l , the charge contained within it is

  τ

  If we construct a Gaussian surface which is a cylinder of radius

  d. By symmetry , the electric field is directed radially and depends only on distance the axis and can be deducted by using Gauss’ theorem.

  (0.5 point)

  = =

  0.68 G

  250 368

  particles in problem (a), to achieve q 0.25 V= 250mV voltage signal, the necessary gain of the voltage pulse amplifier should be

  α

  V = 0.68 mV generated at the anode of the ionization chamber by 6.69 MeV

  (0.5 point) For the voltage signal with height

  10

  c. With time constant

  dis s

  then _{9 min 10} 330

  8.92 10 3.7 10

  

A Ci Ci

−

  = = × ×

  (1.0 point)

  ( ) ^{-12 3}

  = Ω = Ω    

  with C=45×10 F

  10 1000

  22.22

  45 RC s

  R M M τ

  − = =  

  π = ∫ Figure 1 : The Gaussian surface used to calculate the electric field E.

  (1.0 point) Since the field E is everywhere constant and normal to the curved surface. By Gauss’s theorem :

  

τ ,

  λ πε   − = −    

  r V r V d

  2

  2 ln

  ( )

  Thus

  − = − ∫

  τ λ πε

  dr V r V r

  2 ^d

  ( ) ₂

  if we call the potential of inner wire V , we have

  ) with respect to

  ( )

  τ

  and the potential V can be found by integrating E(

  = −

  dV E dr

  then

  τ ,

  Since E is radial and varies only with

  πε = =

  λ π ε λ

  l rlE r r

  2

  2 so E

  (1.0 point) We can use this expression to evaluate the voltage between the capacitor’s conductors

  D

  by setting , giving a potential difference of

  r =

  2 D

  λ  

  ln

  V =  

  2 d

  πε  

  since the charge Q in the capacitor is l , and the capacitance C is defined by Q=CV,

  σ

  the capacitance per unit length is

  2 L

  πε _D

  ln _d (1.0 point)

  d

  The maximum electric field occurs where r minimum, i.e. at r . if we set the

  =

  2

  d

  field at r equal to the breakdown field E b , our expression for E® shows that the

  =

  2 charges per unit length in the capacitor must be E b d. Substituting for the potential

  σ π

  difference V across the capacitor gives

  1 D

   

  ln

  V E d = b  

  2 d

   

6 Taking E b = 3 x 10 V , d= 1mm, and D= 1 cm , gives V = 3.453.45 kV.

  (1.0 point).

  Solut ion t o Proble m 3 St e w a rt -T olm a n Effe c t

  Consider a single ring first Let us take into account a small part of the ring and introduce a reference system in which this part is a rest. The ring is moving with certain angular acceleration Thus, our

  α.

  reference system is not an inertial one and there exists certain linear acceleration in it. The radial component of this acceleration may be neglected as the ring is very thin and no radial effects should be observed in it. The tangential component of the linear acceleration along the considered part of the ring is r . When we speak about the

  α

  reference system in which the positive ions forming the crystal lattice of the metal are at rest. In this system certain inertial force acts on the electrons. This inertial force has the value mr and its oriented in a opposite side to the acceleration mentioned above.

  α

  An interaction between the electrons and crystal lattice does not allow electrons to increase their velocity without any limitations. This interaction, according to the Ohm’s law, is increasing when the velocity of electrons with respect to the crystal lattice in increasing. At some moment equilibrium between the inertial force and the breaking force due the interaction with the lattice is reached. The net results is that the positive ions and the negative electrons are moving with different velocities; its means that in the system in which the ions are at rest an electric current will flow! The inertial force is constant and in each point is tangent to the ring. It’s acts onto the electrons in the same way as certain fictitious electric field tangent to the ring in each point. Now we shall find value of this fictitious electric field. Of course, the force due to it should be equal to the inertial force. Thus:

  eE = mr α

  Therefore:

  mr α E = e

  In the ring 9 at rest) with resistance R, the field of the above value would generate a current: 2 rE

  

π

  I =

  

R Thus, the current in the considered ring should be : ₂ 2 mr

  

π α

  I =

  

R

  It is true that the field E is a fictitious electric field. But it describes a real action of the inertial force onto electrons. The current flowing in the ring is real! . The above considerations allow us to treat the system described in the system described in the text of the problem as a very long solenoid consisting of n loops per unit of length (along the symmetry axis), in which the current I is flowing. It is well known that the magnitude of the field B inside such solenoid (far from its end) is homogenous and its value is equal:

  B

  I = µ η

  I Β = µµ η

  where denotes the permeability of vacuum. Thus, since the point at the axis is not

  µ

  rotating, it is at rest both in the non-inertial and in the laboratory frame, hence the magnetic field at the center of the axis in the laboratory frame is ₂ 2 nmr

  πµ B

  =

eR

  It seems that this problem is very instructive as in spite of the fact that the rings are electrically neutral, in the system – unexpectedly., due to a specific structure of matter – there occurs a magnetic field. Moreover, it seems that due to this problem it is easier to understand why the electrical term: electromotive force” contains a mechanical term “force” inside.

  M a rk ing Sc he m e

  1. Reference system in which the ions are at rest 2p

  2. Inertial force along the ring 2p

  3. Equilibrium of two forces acting onto the electrons 2p

  4. Electric field E 1p

  5. Current I 1p

  6. Equivalence of the systems considered in the text and solenoid 1p

  7. Magnetic field ( final formula) 1p