· theoretical competition 25 april 2010 page 1 of 5...

Theoretical Competition 25 April 2010 of 5 __________________________________________________________________________________________

Question Number 1

Theoretical Question 1 Particles and Waves

This question includes the following three parts dealing with motions of particles and waves: Part A. Inelastic scattering of particles Part B. Waves on a string Part C. Waves in an expanding universe

Part A. Inelastic Scattering and Compositeness of Particles A particle is considered elementary if it has no excitable internal degrees of freedom such

as, for example, rotations and vibrations about its center of mass. Otherwise, it is composite. To determine if a particle is composite, one may set up a scattering experiment with the

particle being the target and allow an elementary particle to scatter off it. In case that the target particle is composite, the scattering experiment may reveal important features such as scaling, i.e. as the forward momentum of the scattered particle increases, the scattering cross section becomes independent of the momentum.

For a scattering system consisting of an elementary particle incident on a target particle, we shall denote by 𝑄𝑄 the total translational kinetic energy loss of the system. Here the translational kinetic energy of a particle, whether elementary or composite, is defined as the kinetic energy associated with the translational motion of its center of mass. Thus we may write

𝑄𝑄 = 𝐾𝐾i − 𝐾𝐾f , where 𝐾𝐾i and 𝐾𝐾f are the total translational kinetic energies of the scattering pair before and after scattering, respectively.

In Part A, use non-relativistic classical mechanics to solve all problems. All effects due to gravity are to be neglected. (a) As shown in Fig. 1, an elementary particle of mass 𝑚𝑚 moves along the 𝑥𝑥 axis with 𝑥𝑥

-component of momentum 𝑝𝑝1 > 0. After being scattered by a stationary target of mass 𝑀𝑀, its momentum becomes 𝑝2.

𝑝𝑝1 𝑀𝑀

Fig. 1

𝑚𝑚

𝑝2 𝑦𝑦

𝑥𝑥

𝑚𝑚


Question Number 1

From data on 𝑝2, one can determine if the target particle is elementary or composite. We shall assume that 𝑝2 lies in the 𝑥𝑥-𝑦𝑦 plane and that the 𝑥𝑥- and 𝑦𝑦-components of 𝑝2 are given, respectively, by 𝑝𝑝2𝑥𝑥 and 𝑝𝑝2𝑦𝑦 .

(i) Find an expression for 𝑄𝑄 in terms of 𝑚𝑚, 𝑀𝑀, 𝑝𝑝1, 𝑝𝑝2𝑥𝑥 , and 𝑝𝑝2𝑦𝑦 . [0.2 point] (ii) If the target particle is elementary, the momenta 𝑝𝑝1, 𝑝𝑝2𝑥𝑥 , and 𝑝𝑝2𝑦𝑦 are related in a

particular way by a condition. For given 𝑝𝑝1, plot the condition as a curve in the 𝑝𝑝2𝑥𝑥 - 𝑝𝑝2𝑦𝑦 plane. Specify the value of 𝑝𝑝2𝑥𝑥 for each intercept of the curve with the 𝑝𝑝2𝑥𝑥-axis. In the same plot, locate regions of points of 𝑝2 corresponding to 𝑄𝑄 < 0, 𝑄𝑄 = 0, 𝑄𝑄 > 0, and label each of them as such.

[0.7 point] For a stationary composite target in its ground state before scattering, which region(s) of 𝑄𝑄 contains those points of 𝑝2 allowed? [0.2 point]

(b) Now, consider a composite target consisting of two elementary particles each with mass 1

2𝑀𝑀. They are connected by a spring of negligible mass. See Fig. 2. The spring has a force constant 𝑘𝑘 and does not bend sideways. Initially, the target is stationary with its center of mass at the origin O, and the spring, inclined at an angle 𝜃𝜃 to the 𝑥𝑥-axis, is at its natural length 𝑑𝑑0. For simplicity, we assume that only vibrational and rotational motions can be excited in the target as a result of scattering.

The incident elementary particle of mass 𝑚𝑚 moves in the 𝑥𝑥-direction both before and after scattering with its momenta given, respectively, by 𝑝𝑝1 and 𝑝𝑝2. Note that 𝑝𝑝2 is negative if the particle recoils and moves backward. A scattering occurs only if the incident particle hits one of the target particles and 𝑝𝑝2 ≠ 𝑝𝑝1. We assume all three particles move in the same plane before and after scattering.

(i) If the maximum length of the spring after scattering is 𝑑𝑑m , find an equation which

relates the ratio 𝑥𝑥 = (𝑑𝑑m − 𝑑𝑑0)/𝑑𝑑0 to the quantities 𝑄𝑄, 𝜃𝜃, 𝑑𝑑0, 𝑚𝑚, 𝑘𝑘, 𝑀𝑀, 𝑝𝑝1 and 𝑝𝑝2. [0.7 point]

𝜃𝜃

Fig. 2

𝑝𝑝1

𝑝𝑝2 𝑥𝑥

𝑚𝑚

𝑂𝑂

𝑑𝑑0

12𝑀𝑀

12𝑀𝑀


Question Number 1

(ii) Let 𝛼𝛼 ≡ sin2 𝜃𝜃. When the angle of orientation 𝜃𝜃 of the target is allowed to vary, the scattering cross section 𝜎𝜎 gives the effective target area, in a plane normal to the direction of incidence, which allows certain outcomes to occur as a result of scattering. It is known that for all outcomes which lead to the same value of 𝑝𝑝2, the value of 𝛼𝛼 must span an interval (𝛼𝛼min ,𝛼𝛼max ) and we may choose the unit of cross section so that 𝜎𝜎 is simply given by the numerical range (𝛼𝛼max − 𝛼𝛼min ) of the interval. Note that 𝛼𝛼min , 𝛼𝛼max , and, consequently, 𝜎𝜎 are dependent on 𝑝𝑝2 . Let 𝑝𝑝c be the threshold value of 𝑝𝑝2 at which 𝜎𝜎 starts to become independent of 𝑝𝑝2. In the limit of large 𝑘𝑘, give an estimate of 𝑝𝑝c . Express your answer in terms of 𝑚𝑚, 𝑀𝑀, and 𝑝𝑝1. [1.1 points] Assume 𝑀𝑀=3m and in the limit of large 𝑘𝑘, plot 𝜎𝜎 as a function of 𝑝𝑝2 for a given 𝑝𝑝1. In the plot, specify the range of σ and p2. [1.1 points]

Part B. Waves on a String Consider an elastic string stretched between two fixed ends A and B, as shown in Fig. 3.

The linear mass density of the string is 𝜇𝜇. The speed of propagation for transverse waves in the string is 𝑐𝑐. Let the length AB be 𝐿𝐿. The string is plucked sideways and held in a triangular form with a maximum height ℎ ≪ 𝐿𝐿 at its middle point. At time 𝑡𝑡 = 0, the plucked string is released from rest. All effects due to gravity may be neglected.

(c) Find the period of vibration 𝑇𝑇 for the string. [0.5 point] Plot the shape of the string at 𝑡𝑡 = 𝑇𝑇/8. In the plot, specify lengths and angles which serve to define the shape of the string. [1.7 points]

(d) Find the total mechanical energy of the vibrating string in terms of 𝜇𝜇, 𝑐𝑐, ℎ, and 𝐿𝐿. [0.8 point]

Fig. 3

A B

ℎ

𝐿𝐿


Question Number 1

Part C. An Expanding Universe Photons in the universe play an important role in delivering information across the cosmos.

However, the fact that the universe is expanding must be taken into account when one tries to extract information from these photons. To this end, we normally express length and distance using a universal scale factor 𝑎𝑎(𝑡𝑡) which depends on time 𝑡𝑡. Thus the distance 𝐿𝐿(𝑡𝑡) between two stars stationary in their respective local frames is proportional to 𝑎𝑎(𝑡𝑡):

𝐿𝐿(𝑡𝑡) = 𝑘𝑘𝑎𝑎(𝑡𝑡), (1) where 𝑘𝑘 is a constant and 𝑎𝑎(t) accounts for the expansion of the universe. We use a dot above a symbol of a variable to denote its time derivative, i.e. 𝑎(t) = 𝑑𝑑𝑎𝑎(𝑡𝑡)/𝑑𝑑𝑡𝑡, and let 𝑣𝑣(𝑡𝑡) ≡ 𝐿(𝑡𝑡). Taking time derivatives of both sides of Eq. (1), one obtains the Hubble law:

𝑣𝑣(𝑡𝑡) = 𝐻𝐻(𝑡𝑡)𝐿𝐿(𝑡𝑡), (2) where 𝐻𝐻(𝑡𝑡) = 𝑎(𝑡𝑡)/𝑎𝑎(𝑡𝑡) is the Hubble parameter at time 𝑡𝑡. At the current time 𝑡𝑡0, we have

𝐻𝐻(𝑡𝑡0) = 72 km s−1 Mpc−1, where 1 Mpc = 3.0857 × 1019 km = 3.2616 × 106 light-year.

Assume the universe to be infinitely large and expanding in such a way that 𝑎𝑎(𝑡𝑡) ∝ exp(𝑏𝑏𝑡𝑡),

where 𝑏𝑏 is a constant. In such a universe, the Hubble parameter is a constant equal to 𝐻𝐻(𝑡𝑡0). Moreover, it can be shown that the wavelength 𝜆𝜆 of photons travelling in the universe will be stretched in proportion to the expansion of the universe, i.e.

𝜆𝜆(𝑡𝑡) ∝ 𝑎𝑎(𝑡𝑡). Now suppose that photons making up a Lyman-alpha emission line were emitted at 𝑡𝑡e by

a star that was stationary in its local frame and that we as observers are stationary in our local frame. When these photons were emitted, their wavelength was 𝜆𝜆(𝑡𝑡e) = 121.5 nm. But when they reach us now at 𝑡𝑡0, their wavelength is red-shifted to 145.8 nm. (e) As these photons traveled, the universe kept expanding so that the star kept receding from

us. Given that the speed of light in vacuum 𝑐𝑐 has never changed, what was the distance 𝐿𝐿(𝑡𝑡e ) of the star from us when these photons were emitted at 𝑡𝑡e? Express the answer in units of Mpc. [2.2 points]

(f) What is the receding velocity 𝑣𝑣(𝑡𝑡0) of the star with respect to us now at 𝑡𝑡0? Express the answer in units of the speed of light in vacuum 𝑐𝑐. [0.8 point]


Question Number 1

Appendix The following formula may be used when needed:

𝑒𝑒𝛽𝛽𝑥𝑥𝑏𝑏

𝑎𝑎𝑑𝑑𝑥𝑥 =

1𝛽𝛽

(𝑒𝑒𝛽𝛽𝑏𝑏 − 𝑒𝑒𝛽𝛽𝑎𝑎 ).

Theoretical Competition 25 April 2010 Student Code Page No. Total No.

ANSWER SHEET

Question Number 1

Theoretical Question 1 Particles and Waves

Part A. Inelastic scattering and compositeness of particles

Do not write in any box marked with a solidus (oblique stroke, / ).

(a)(i) 𝑄𝑄 in terms of 𝑚𝑚, 𝑀𝑀, 𝑝𝑝1, 𝑝𝑝2𝑥𝑥 , and 𝑝𝑝2𝑦𝑦

(ii) Plot of a condition relating 𝑝𝑝1, 𝑝𝑝2𝑥𝑥 , and 𝑝𝑝2𝑦𝑦 for an elementary target as a curve with

𝑝𝑝2𝑥𝑥 -intercepts specified. Label regions with 𝑄𝑄 < 0, 𝑄𝑄 = 0, 𝑄𝑄 > 0.

Region(s) of 𝑄𝑄 allowed by a stationary composite target in its ground state before scattering.

Expression of 𝑄𝑄 =

0.2 pt / / / / / / /

0.7 pt / / / / / / /

Allowed 𝑄𝑄 region(s):

0.2 pt / / / / / / /

𝑝𝑝2𝑥𝑥

𝑝𝑝2𝑦𝑦


ANSWER SHEET

Question Number 1

(b)(i) The equation relating 𝑥𝑥 to 𝑄𝑄, 𝜃𝜃, 𝑑𝑑0, 𝑚𝑚, 𝑘𝑘, 𝑀𝑀, 𝑝𝑝1 and 𝑝𝑝2.

(ii) Threshold value 𝑝𝑝𝑐𝑐 of 𝑝𝑝2.

Plot of 𝜎𝜎 versus 𝑝𝑝2 for given 𝑝𝑝1 and 𝑀𝑀 = 3𝑚𝑚 with range of 𝜎𝜎 and 𝑝𝑝2 specified.

0.7 pt / / / / / / /

𝑝𝑝𝑐𝑐 =

1.1 pt / / / / / / /

1.1 pt / / / / / / /

𝑝𝑝2

𝜎𝜎


ANSWER SHEET

Question Number 1

Part B. Waves on a string (c) Period of vibration 𝑇𝑇 for the string.

Shape of the string at 𝑡𝑡 = 𝑇𝑇/8 (specify important lengths and angles).

(d) The total mechanical energy of the vibrating string.

𝑇𝑇 =

0.5 pt / / / / / / /

1.7 pt / / / / / / /

0.8 pt / / / / / / /


ANSWER SHEET

Question Number 1

Part C. The expanding universe (e) Distance (in units of Mpc) of the star from us.

(f) The receding velocity (in units of 𝑐𝑐) of the star.

𝐿𝐿(𝑡𝑡e) =

2.2 pt / / / / / / /

𝑣𝑣(𝑡𝑡0) =

0.8 pt / / / / / / /

Theoretical Competition 25 April 2010 Page 1 of 4 __________________________________________________________________________________________

Question Number 2

Theoretical Question 2

Strong Resistive Electromagnets Resistive electromagnets are magnets with coils made of a normal metal such as copper or

aluminum. Modern strong resistive electromagnets can provide steady magnetic fields higher than 30 tesla. Their coils are typically built by stacking hundreds of thin circular plates made of copper sheet metal with lots of cooling holes stamped in them; there are also insulators with the same pattern. When voltage is applied across the coil, current flows through the plates along a helical path to generate high magnetic fields in the center of the magnet.

In this question we aim to assess if a cylindrical coil (or solenoid) of many turns can serve as a magnet for generating high magnetic fields. As shown in Fig. 1, the center of the magnet is at O. Its cylindrical coil consists of 𝑁𝑁 turns of copper wire carrying a current 𝐼𝐼 uniformly distributed over the cross section of the wire. The coil’s mean diameter is 𝐷𝐷 and its length along the axial direction 𝑥𝑥 is ℓ. The wire’s cross section is rectangular with width 𝑎𝑎 and height 𝑏𝑏. The turns of the coil are so tightly wound that the plane of each turn may be taken as perpendicular to the 𝑥𝑥 axis and ℓ = 𝑁𝑁𝑎𝑎. In Table 1, data specifying physical dimensions of the coil are listed.

In assessing if such a magnet can serve to provide high magnetic fields, two limiting

factors must not be overlooked. One is the mechanical rigidity of the coil to withstand large Lorentz force on the field-producing current. The other is that the enormous amount of Joule heat generated in the wire must not cause excessive temperature rise. We shall examine these two factors using simplified models.

The Appendix at the end of the question lists some mathematical formulae and physical data which may be used if necessary.

Part A. Magnetic Fields on the Axis of the Coil Assume 𝑏𝑏 ≪ 𝐷𝐷

so that one may regard the wire as a thin strip of width 𝑎𝑎. Let O be the

origin of 𝑥𝑥 coordinates. The direction of the current flow is as shown in Fig. 1.

ℓ = 12.0 cm 𝐷𝐷 = 6.0 cm 𝑎𝑎 = 2.0 mm 𝑏𝑏 = 5.0 mm

Table 1

Figure 1 𝑎𝑎

𝑥𝑥 𝑂𝑂

ℓ/2

I 𝐼𝐼

𝐷𝐷

𝑎𝑎

ℓ/2

𝑏𝑏

𝑏𝑏


Question Number 2

(a) Find the 𝑥𝑥-component 𝐵𝐵(𝑥𝑥) of the magnetic field on the axis of the coil as a function of 𝑥𝑥 when the steady current passing through the coil is 𝐼𝐼. [1.0 point]

(b) Find the steady current 𝐼𝐼0 passing through the coil if 𝐵𝐵(0) i s10.0 T. Use data given in Table 1 when computing numerical values. [0.4 point]

Part B. The Upper Limit of Current In Part B, we assume the length ℓ of the coil is infinite and 𝑏𝑏 ≪ 𝐷𝐷. Consider the turn of

the coil located at 𝑥𝑥 = 0. The magnetic field exerts Lorentz force on the current passing through the turn. Thus, as Fig. 2 shows, a wire segment of length Δ𝑠𝑠 is subject to a normal force Δ𝐹𝐹n which tends to make the turn expand.

(c) Suppose that, when the current is 𝐼𝐼, the mean diameter of the expanded coil remains at a

constant value 𝐷𝐷′ larger than 𝐷𝐷, as shown in Fig. 2. Find the outward normal force per unit length 𝛥𝛥𝐹𝐹n /𝛥𝛥𝑠𝑠. [1.2 point] Find the tension 𝐹𝐹t acting along the wire. [0.6 point]

(d) Neglect the coil’s acceleration during the expansion. Assume the turn will break when the wire’s unit elongation (i.e. tensile strain or fractional change of the length) is 60 % and tensile stress (i.e. tension per unit cross sectional area of the unstrained wire) is 𝜎𝜎b =455 MPa. Let 𝐼𝐼b be the current at which the turn will break and 𝐵𝐵b the corresponding magnetic field at the center O. Find an expression for 𝐼𝐼b and then calculate its value. [0.8 point] Find an expression for 𝐵𝐵b and then calculate its value. [0.4 point]

Part C. The Rate of Temperature Rise When the current 𝐼𝐼 is 10.0 kA and the temperature 𝑇𝑇 of the coil is 293 K, assume that the

resistivity, the specific heat capacity at constant pressure, and the mass density of the wire of the coil are, respectively, given by 𝜌𝜌e = 1.72 × 10−8 Ω ∙ m , cp = 3.85 × 102 J/(kg ∙ K) and 𝜌𝜌𝑚𝑚 = 8.98 × 103 kg ∙ m−3. (e) Find an expression for the power density (i.e. power per unit volume) of heat generation in

the coil and then calculate its value. Use data in Table 1. [0.5 point]

Figure 2 𝑥𝑥

Δ𝑠𝑠

𝐷𝐷′/2

𝑂𝑂

𝐼𝐼 𝑏𝑏

𝐹𝐹t 𝐹𝐹t

Δ𝐹𝐹n


Question Number 2

(f) Let 𝑇 be the time rate of change of temperature in the coil. Find an expression for 𝑇 and then calculate its value. [0.5 point]

Part D. A Pulsed-Field Magnet If the large current needed for a strong magnet lasts only for a short time, the temperature

rise caused by excessive Joule heating may be greatly reduced. This idea is employed in a pulsed-field magnet.

Thus, as shown in Fig. 3, a capacitor bank of capacitance 𝐶𝐶 charged initially to a potential 𝑉𝑉0 is used to drive the current 𝐼𝐼 through the coil. The circuit is equipped with a switch 𝐾𝐾. The inductance 𝐿𝐿

and resistance 𝑅𝑅

of the circuit are assumed to be entirely due to

the coil. The construct and dimensions of the coil are the same as given in Fig. 1 and Table 1. Assume 𝑅𝑅, 𝐿𝐿, and 𝐶𝐶

to be independent of temperature and the magnetic field is the same as

that of an infinite solenoid with ℓ → ∞.

(g) Find expressions for the inductance 𝐿𝐿 and resistance 𝑅𝑅. [0.6 point]

Calculate the values of 𝐿𝐿 and 𝑅𝑅. Use data given in Table 1. [0.4 point] (h) At time 𝑡𝑡 = 0, the switch 𝐾𝐾 is thrown to position 1 and the current starts flowing.

For 𝑡𝑡 ≥ 0, the charge 𝑄𝑄(𝑡𝑡) on the positive plate of the capacitor and the current 𝐼𝐼(𝑡𝑡) entering the positive plate are given by

𝑄𝑄(𝑡𝑡) =𝐶𝐶𝑉𝑉0

sin𝜃𝜃0𝑒𝑒−𝛼𝛼𝑡𝑡 sin(𝜔𝜔𝑡𝑡 +𝜃𝜃0), (1)

𝐼𝐼(𝑡𝑡) =𝑑𝑑𝑄𝑄𝑑𝑑𝑡𝑡

= (−𝛼𝛼

cos𝜃𝜃0)𝐶𝐶𝑉𝑉0

sin𝜃𝜃0𝑒𝑒−𝛼𝛼𝑡𝑡 sin𝜔𝜔𝑡𝑡 , (2)

in which 𝛼𝛼 and 𝜔𝜔 are positive constants and 𝜃𝜃0 is given by

tan𝜃𝜃0 =𝜔𝜔𝛼𝛼

, 0 < 𝜃𝜃0 <𝜋𝜋2

. (3)

Note that, if 𝑄𝑄(𝑡𝑡) is expressed as a function of a new variable 𝑡𝑡′ ≡ (𝑡𝑡 + 𝜃𝜃0/𝜔𝜔), then 𝑄𝑄(𝑡𝑡′) and its time derivative 𝐼𝐼(𝑡𝑡) are identical in form except for an overall constant factor. The time derivative of 𝐼𝐼(𝑡𝑡) may therefore be obtained similarly without further differentiations.

Find 𝛼𝛼 and 𝜔𝜔 in terms of 𝑅𝑅, 𝐿𝐿, and 𝐶𝐶. [0.8 point] Calculate the values of 𝛼𝛼 and 𝜔𝜔 when 𝐶𝐶

is 10.0 mF. [0.4 point]

Figure 3 𝐶𝐶

𝐾𝐾 𝐿𝐿

1 2

𝑉𝑉0

𝑅𝑅


Question Number 2

(i) Let 𝐼𝐼m be the maximum value of |𝐼𝐼(𝑡𝑡)| for 𝑡𝑡 > 0. Find an expression for 𝐼𝐼m . [0.6 point] If 𝐶𝐶 = 10.0 mF, what is the maximum value 𝑉𝑉0b of the initial voltage 𝑉𝑉0 of the capacitor bank for which 𝐼𝐼m

will not exceed 𝐼𝐼b found in Problem (d)? [0.4 point]

(j) Suppose the switch 𝐾𝐾 is moved instantly from position 1 to 2 when the absolute value of the current | 𝐼𝐼(𝑡𝑡)| reaches 𝐼𝐼m . Let ∆𝐸𝐸 be the total amount of heat dissipated in the coil from 𝑡𝑡 = 0 to ∞ and ∆𝑇𝑇 the corresponding temperature increase of the coil. Assume the initial voltage 𝑉𝑉0 takes on the maximum value 𝑉𝑉0b obtained in Problem (i) and the electromagnetic energy loss is only in the form of heat dissipated in the coil. Find an expression for ∆𝐸𝐸 and then calculate its value. [1.0 point] Find an expression for ∆𝑇𝑇 and then calculate its value. Note that the value for ∆𝑇𝑇 must be compatible with the assumption of constant 𝑅𝑅 and 𝐿𝐿. [0.4 point]

Appendix

1. 𝑑𝑑𝑥𝑥

(𝐷𝐷2 + 𝑥𝑥2)3/2

𝐿𝐿

0=

1𝐷𝐷2

𝐿𝐿(𝐷𝐷2 + 𝐿𝐿2)1/2

2. sin(𝛼𝛼 ± 𝛽𝛽) = sin𝛼𝛼 cos𝛽𝛽 ± cos𝛼𝛼 sin𝛽𝛽

3. permeability of free space 𝜇𝜇0 = 4𝜋𝜋 × 10−7 T ∙ m/A

------------------------------------------------------ END ------------------------------------------------


ANSWER SHEET

Question Number 2

Theoretical Question 2 Strong Resistive Electromagnets

Part A. Magnetic fields on the axis of the coil


(a) 𝑥𝑥-component 𝐵𝐵(𝑥𝑥) of the magnetic field on the axis (in terms of 𝑎𝑎, 𝐷𝐷, 𝐼𝐼, ℓ, 𝜇𝜇0).

Expression of 𝐵𝐵(𝑥𝑥) =

1.0 pt / / / / / / /

(b) The current 𝐼𝐼0 when 𝐵𝐵(0) = 10.0 T (expressed in terms of 𝑎𝑎, 𝐷𝐷, 𝐵𝐵(0), ℓ, 𝜇𝜇0).

Expression of 𝐼𝐼0 =

Value of 𝐼𝐼0 =

0.4 pt / / / / / / /

Part B. The upper limit of current (c) The outward normal force per unit length 𝛥𝛥𝐹𝐹n/𝛥𝛥𝛥𝛥 (in terms of 𝑎𝑎, 𝐷𝐷′ , 𝐼𝐼, 𝜇𝜇0).

Expression of 𝛥𝛥𝐹𝐹n

𝛥𝛥𝛥𝛥=

1.2 pt / / / / / / /

The tension 𝐹𝐹t along the wire (in terms of 𝑎𝑎, 𝐷𝐷′ , 𝐼𝐼, 𝜇𝜇0).

Expression of 𝐹𝐹t =

0.6 pt / / / / / / /


ANSWER SHEET

Question Number 2

(d) The current 𝐼𝐼b at which the turn will break (expressed in terms of 𝑎𝑎, 𝑏𝑏, 𝐷𝐷, 𝜎𝜎b , 𝜇𝜇0).

Expression of 𝐼𝐼b =

Value of 𝐼𝐼b =

0.8 pt / / / / / / /

The magnetic field 𝐵𝐵b at 𝑂𝑂 when the current is 𝐼𝐼b (expressed in terms of 𝑎𝑎, 𝐼𝐼b , 𝜇𝜇0).

Expression of 𝐵𝐵b =

Value of 𝐵𝐵b =

0.4 pt / / / / / / /

Part C. Rate of temperature rise

(e) The power density of heat generation in the coil.

Expression:

Value:

0.5 pt / / / / / / /

(f) The time rate of change 𝑇 of temperature in the coil.

Expression of 𝑇 =

Value of 𝑇 =

0.5 pt / / / / / / /


ANSWER SHEET

Question Number 2

Part D. A pulsed-field magnet

(g) Expressions for the inductance 𝐿𝐿 and resistance 𝑅𝑅.

Expression of 𝐿𝐿 =

Expression of 𝑅𝑅 =

0.6 pt / / / / / / /

Values of inductance 𝐿𝐿 and resistance 𝑅𝑅.

Value of 𝐿𝐿 =

Value of 𝑅𝑅 =

0.4 pt / / / / / / /

(h) Expressions for 𝛼𝛼 and 𝜔𝜔 (in terms of 𝑅𝑅, 𝐿𝐿, and 𝐶𝐶).

Expression of 𝛼𝛼 =

Expression of 𝜔𝜔 =

0.8 pt / / / / / / /

Values of 𝛼𝛼 and 𝜔𝜔.

Value of 𝛼𝛼 =

Value of 𝜔𝜔 =

0.4 pt / / / / / / /


ANSWER SHEET

Question Number 2

(i) Expression of 𝐼𝐼m (in terms of 𝛼𝛼, 𝜔𝜔, 𝜃𝜃0, 𝑉𝑉0 and 𝐶𝐶).

Expression of 𝐼𝐼m =

0.6 pt / / / / / / /

Maximum initial voltage 𝑉𝑉0b for which 𝐼𝐼m will not exceed 𝐼𝐼b of Problem (d).

Value of 𝑉𝑉0b =

0.4 pt / / / / / / /

(j) The total amount of heat ∆𝐸𝐸 dissipated in the coil (in terms of 𝛼𝛼, 𝜔𝜔, 𝜃𝜃0, 𝑉𝑉0b and 𝐶𝐶).

Expression of ∆𝐸𝐸 =

Value of ∆𝐸𝐸 =

1.0 pt / / / / / / /

The temperature increase ∆𝑇𝑇 of the coil.

Expression of ∆𝑇𝑇 =

Value of ∆𝑇𝑇 =

0.4 pt / / / / / / /


Question Number 3

Theoretical Question 3 Electron and Gas Bubbles in Liquids

This question deals with physics of two bubble-in-liquid systems. It has two parts: Part A. An electron bubble in liquid helium Part B. Single gas bubble in liquid

Part A. An Electron Bubble in Liquid Helium When an electron is planted inside liquid helium, it can repel atoms of liquid helium and

form what is called an electron bubble. The bubble contains nothing but the electron itself. We shall be interested mainly in its size and stability.

We use ∆𝑓𝑓 to denote the uncertainty of a quantity 𝑓𝑓. The components of an electron’s position vector 𝑞 = (𝑥𝑥,𝑦𝑦, 𝑧𝑧) and momentum vector 𝑝 = (𝑝𝑝𝑥𝑥 ,𝑝𝑝𝑦𝑦 ,𝑝𝑝𝑧𝑧) must obey Heisenberg’s uncertainty relations ∆𝑞𝑞𝛼𝛼∆𝑝𝑝𝛼𝛼 ≥ ℏ/2 , where ℏ is the Planck constant divided by 2 𝜋𝜋 and 𝛼𝛼 = 𝑥𝑥,𝑦𝑦, 𝑧𝑧.

We shall assume the electron bubble to be isotropic and its interface with liquid helium is a sharp spherical surface. The liquid is kept at a constant temperature very close to 0 K with its surface tension 𝜎𝜎 given by 3.75 × 10−4 N ∙ m−1 and its electrostatic responses to the electron bubble may be neglected.

Consider an electron bubble in liquid helium with an equilibrium radius 𝑅𝑅. The electron, of mass 𝑚𝑚, moves freely inside the bubble with kinetic energy 𝐸𝐸k and exerts pressure 𝑃𝑃e on the inner side of the bubble-liquid interface. The pressure exerted by liquid helium on the outer side of the interface is 𝑃𝑃He .

(a) Find a relation between 𝑃𝑃He , 𝑃𝑃e , and 𝜎𝜎. [0.4 point] Find a relation between 𝐸𝐸k and 𝑃𝑃e . [1.0 point]

(b) Denote by 𝐸𝐸0 the smallest possible value of 𝐸𝐸k consistent with Heisenberg’s uncertainty relations when the electron is inside the bubble of radius 𝑅𝑅. Estimate 𝐸𝐸0 as a function of 𝑅𝑅. [0.8 point]

(c) Let 𝑅𝑅e be the equilibrium radius of the bubble when 𝐸𝐸k = 𝐸𝐸0 and 𝑃𝑃He = 0. Obtain an expression for 𝑅𝑅e and calculate its value. [0.6 point]

(d) Find a condition that 𝑅𝑅 and 𝑃𝑃He must satisfy if the equilibrium at radius 𝑅𝑅 is to be locally stable under constant 𝑃𝑃He . Note that 𝑃𝑃He can be negative. [0.6 point]

(e) There exists a threshold pressure 𝑃𝑃th such that equilibrium is not possible for the electron bubble when 𝑃𝑃He is less than 𝑃𝑃th . Find an expression for 𝑃𝑃th . [0.6 point]


Question Number 3

Part B. Single Gas Bubble in Liquid — Collapsing and Radiation In this part of the problem, we consider a normal liquid, such as water. When a gas bubble in a liquid is driven by an oscillating pressure, it can show dramatic

responses. For example, following a large expansion, it can collapse rapidly to a small radius and, near the end of the collapse, emit light almost instantly. In this phenomenon, called single-bubble sonoluminescence, the gas bubble undergoes cyclic motions which typically consist of three stages: expansion, collapse, and multiple after-bounces. In the following we shall focus mainly on the collapsing stage.

We assume that, at all times, the bubble considered is spherical and its center remains stationary in the liquid. See Fig 1. The pressure, temperature, and density are always uniform inside the bubble as its size diminishes. The liquid containing the bubble is assumed to be isotropic, nonviscous, incompressible, and very much larger in extent than the bubble. All effects due to gravity and surface tension are neglected so that pressures on both sides of the bubble-liquid interface are always equal.

Radial motion of the bubble-liquid interface As the bubble’s radius 𝑅𝑅 = 𝑅𝑅(𝑡𝑡) changes with time 𝑡𝑡, the bubble-liquid interface will move

with radial velocity 𝑅 ≡ 𝑑𝑑𝑅𝑅/𝑑𝑑𝑡𝑡. It follows from the equation of continuity of incompressible fluids that the liquid’s radial velocity 𝑟 ≡ 𝑑𝑑𝑟𝑟/𝑑𝑑𝑡𝑡 at distance 𝑟𝑟 from the center of the bubble is related to the rate of change of the bubble’s volume 𝑉𝑉 by

𝑑𝑑𝑉𝑉𝑑𝑑𝑡𝑡

= 4𝜋𝜋𝑅𝑅 2𝑅 = 4𝜋𝜋𝑟𝑟2𝑟. (1)

This implies that the total kinetic energy 𝐸𝐸k of the liquid with mass density 𝜌𝜌0 is

𝐸𝐸k =12 𝜌𝜌0

𝑟𝑟0

𝑅𝑅(4𝜋𝜋𝑟𝑟2𝑑𝑑𝑟𝑟)𝑟2 = 2𝜋𝜋𝜌𝜌0𝑅𝑅4𝑅2

1𝑟𝑟2

𝑟𝑟0

𝑅𝑅𝑑𝑑𝑟𝑟 = 2𝜋𝜋𝜌𝜌0𝑅𝑅4𝑅2

1𝑅𝑅−

1𝑟𝑟0 (2)

where 𝑟𝑟0 is the radius of the outer surface of the liquid.

𝑃𝑃0 0T

𝑟𝑟0

𝑅𝑅

𝑟𝑟

𝜌𝜌0

Fig. 1 𝑃𝑃

𝑇𝑇


Question Number 3

(f) Assume the ambient pressure 𝑃𝑃0 acting on the outer surface 𝑟𝑟 = 𝑟𝑟0 of the liquid is constant. Let 𝑃𝑃 = 𝑃𝑃(𝑅𝑅) be the gas pressure when the radius of the bubble is 𝑅𝑅. Find the amount of work 𝑑𝑑𝑑𝑑 done on the liquid when the radius of the bubble changes from 𝑅𝑅 to 𝑅𝑅 + 𝑑𝑑𝑅𝑅. Use 𝑃𝑃0 and 𝑃𝑃 to express 𝑑𝑑𝑑𝑑. [0.4 point] The work 𝑑𝑑𝑑𝑑 must be equal to the corresponding change in the total kinetic energy of the liquid. In the limit 𝑟𝑟0 → ∞, it follows that we have Bernoulli’s equation in the form

12 𝜌𝜌0 𝑑𝑑𝑅𝑅m 𝑅2 = (𝑃𝑃 − 𝑃𝑃0)𝑅𝑅n𝑑𝑑𝑅𝑅. (3)

Find the exponents m and n in Eq. (3). Use dimensional arguments if necessary. [0.4 point]

Collapsing of the gas bubble From here on, we consider only the collapsing stage of the bubble. The mass density of the

liquid is 𝜌𝜌0 = 1.0 × 103 kg ∙ m−3, the temperature 𝑇𝑇0 of the liquid is 300 K and the ambient pressure 𝑃𝑃0 is 1.01 × 105 Pa. We assume that 𝜌𝜌0, 𝑇𝑇0, and 𝑃𝑃0 remain constant at all times and the bubble collapses adiabatically without any exchange of mass across the bubble-liquid interface.

The bubble considered is filled with an ideal gas. The ratio of specific heat at constant pressure to that at constant volume for the gas is 𝛾𝛾 = 5/3. When under temperature 𝑇𝑇0 and pressure 𝑃𝑃0, the equilibrium radius of the bubble is 𝑅𝑅0 = 5.00 μm.

Now, this bubble begins its collapsing stage at time 𝑡𝑡 = 0 with 𝑅𝑅(0) = 𝑅𝑅i = 7𝑅𝑅0, 𝑅(0) = 0, and the gas temperature 𝑇𝑇i = 𝑇𝑇0. Note that, because of the bubble’s expansion in the preceding stage, 𝑅𝑅i is considerably larger than 𝑅𝑅0 and this is necessary if sonoluminescence is to occur.

(g) Express the pressure 𝑃𝑃 ≡ 𝑃𝑃(𝑅𝑅) and temperature 𝑇𝑇 ≡ 𝑇𝑇(𝑅𝑅) of the ideal gas in the bubble as a function of 𝑅𝑅 during the collapsing stage, assuming quasi-equilibrium conditions hold. [0.6 point]

(h) Let 𝛽𝛽 ≡ 𝑅𝑅/𝑅𝑅i and 𝛽 = 𝑑𝑑𝛽𝛽/𝑑𝑑𝑡𝑡 . Eq. (3) implies a conservation law which takes the following form

12 𝜌𝜌0 𝛽2 + 𝑈𝑈(𝛽𝛽) = 0. (4)

Let 𝑃𝑃i ≡ 𝑃𝑃(𝑅𝑅i) be the gas pressure of the bubble when 𝑅𝑅 = 𝑅𝑅i . If we introduce the ratio 𝑄𝑄 ≡ 𝑃𝑃i/[(𝛾𝛾 − 1)𝑃𝑃0] , the function 𝑈𝑈(𝛽𝛽) may be expressed as

𝑈𝑈(𝛽𝛽) = 𝜇𝜇𝛽𝛽−5[𝑄𝑄(1 − 𝛽𝛽2) − 𝛽𝛽2(1− 𝛽𝛽3)]. (5) Find the coefficient 𝜇𝜇 in terms of 𝑅𝑅i and 𝑃𝑃0. [0.6 point]

(i) Let 𝑅𝑅m be the minimum radius of the bubble during the collapsing stage and define 𝛽𝛽m ≡ 𝑅𝑅m /𝑅𝑅i. For 𝑄𝑄 ≪ 1, we have 𝛽𝛽m ≈ 𝐶𝐶m𝑄𝑄 . Find the constant 𝐶𝐶m . [0.4 point]


Question Number 3

Evaluate 𝑅𝑅m for 𝑅𝑅i = 7𝑅𝑅0. [0.3 point] Evaluate the temperature 𝑇𝑇m of the gas at 𝛽𝛽 = 𝛽𝛽m . [0.3 point]

(j) Assume 𝑅𝑅i = 7𝑅𝑅0. Let 𝛽𝛽𝑢𝑢 be the value of 𝛽𝛽 at which the dimensionless radial speed 𝑢𝑢 ≡|𝛽| reaches its maximum value. The gas temperature rises rapidly for values of 𝛽𝛽 near 𝛽𝛽𝑢𝑢 . Give an expression and then estimate the value of 𝛽𝛽𝑢𝑢 . [0.6 point] Let 𝑢𝑢 be the value of 𝑢𝑢 at 𝛽𝛽 = 𝛽 ≡ (𝛽𝛽m + 𝛽𝛽𝑢𝑢)/2. Evaluate 𝑢𝑢. [0.4 point] Give an expression and then estimate the duration ∆𝑡𝑡m of time needed for 𝛽𝛽 to diminish from 𝛽𝛽𝑢𝑢 to the minimum value 𝛽𝛽m . [0.6 point]

Sonoluminescence of the collapsing bubble Consider the bubble to be a surface black-body radiator of constant emissivity 𝑎𝑎 so that

the effective Stefan-Boltzmann’s constant 𝜎𝜎eff = 𝑎𝑎𝜎𝜎SB . If the collapsing stage is to be approximated as adiabatic, the emissivity must be small enough so that the power radiated by the bubble at 𝛽𝛽 = 𝛽 is no more than a fraction, say 20 %, of the power 𝐸 supplied to it by the driving liquid pressure.

(k) Find the power 𝐸 supplied to the bubble as a function of 𝛽𝛽. [0.6 point] Give an expression and then estimate the value for an upper bound of 𝑎𝑎. [0.8 point]

Appendix

1. 𝑑𝑑𝑑𝑑𝑥𝑥

𝑥𝑥𝑛𝑛 = 𝑛𝑛𝑥𝑥𝑛𝑛−1

2. Electron mass 𝑚𝑚 = 9.11 × 10−31 kg

3. Planck constant ℎ = 2𝜋𝜋 ℏ = 2𝜋𝜋 × 1.055 × 10−34 J ∙ s

4. Stefan-Boltzmann’s constant 𝜎𝜎SB = 5.67 × 10−8 W ∙ m−2 ∙ K−4

END ----------------------------------------------------------------------------------------------------------------


ANSWER SHEET

Question Number 3

Theoretical Question 3 Electron and Gas Bubbles in Liquids

Part A. An electron bubble in liquid helium


(a) Relation between 𝑃𝑃He , 𝑃𝑃e , and 𝜎𝜎.

Expression of 𝑃𝑃e =

0.4 pt / / / / / / /

Relation between 𝐸𝐸k and 𝑃𝑃e .

Expression:

1.0 pt / / / / / / /

(b) The smallest possible kinetic energy 𝐸𝐸0 as a function of 𝑅𝑅.

Expression of 𝐸𝐸0 =

0.8 pt / / / / / / /

(c) The bubble’s equilibrium radius 𝑅𝑅e when 𝐸𝐸k = 𝐸𝐸0 and 𝑃𝑃He = 0.

Expression of 𝑅𝑅e =

Value of 𝑅𝑅e =

0.6 pt / / / / / / /


ANSWER SHEET

Question Number 3

(d) Condition satisfied by 𝑅𝑅 and 𝑃𝑃He for locally stable equilibrium at radius 𝑅𝑅.

Expression:

0.6 pt / / / / / / /

(e) The threshold pressure 𝑃𝑃th below which no equilibrium is possible for the bubble.

Expression of 𝑃𝑃th =

0.6 pt / / / / / / /

Part B. Single gas bubble in liquid — collapsing and radiation

(f) Work 𝑑𝑑𝑑𝑑 done on the liquid when the bubble’s radius changes from 𝑅𝑅 to 𝑅𝑅 + 𝑑𝑑𝑅𝑅.

Expression of 𝑑𝑑𝑑𝑑 =

0.4 pt / / / / / / /

Values of the exponents m and n.

m =

n =

0.4 pt / / / / / / /


ANSWER SHEET

Question Number 3

(g) Pressure 𝑃𝑃 ≡ 𝑃𝑃(𝑅𝑅) and temperature 𝑇𝑇 ≡ 𝑇𝑇(𝑅𝑅) as a function of 𝑅𝑅.

Expression of 𝑃𝑃 ≡ 𝑃𝑃(𝑅𝑅) =

Expression of 𝑇𝑇 ≡ 𝑇𝑇(𝑅𝑅) =

0.6 pt / / / / / / /

(h) The coefficient 𝜇𝜇 in terms of 𝑅𝑅i and 𝑃𝑃0.

Expression of 𝜇𝜇 =

0.6 pt / / / / / / /

(i) Values of the constant 𝐶𝐶m .

Value of 𝐶𝐶m =

0.4 pt / / / / / / /

The minimum radius 𝑅𝑅m for 𝑅𝑅i = 7𝑅𝑅0.

Value of Rm =

0.3 pt / / / / / / /

The temperature 𝑇𝑇m of the gas at 𝛽𝛽m .

Value of 𝑇𝑇m =

0.3 pt / / / / / / /


ANSWER SHEET

Question Number 3

(j) The radius 𝛽𝛽𝑢𝑢 at which the radial speed 𝑢𝑢 ≡ |𝛽| reaches its maximum value.

Expression of 𝛽𝛽𝑢𝑢 =

Value of 𝛽𝛽𝑢𝑢 =

0.6 pt / / / / / / /

The value of 𝑢𝑢 of the dimensionless radial speed 𝑢𝑢 at 𝛽𝛽 = 𝛽 ≡ 12

(𝛽𝛽m + 𝛽𝛽𝑢𝑢).

Value of 𝑢𝑢 =

0.4 pt / / / / / / /

The time duration ∆𝑡𝑡m for 𝛽𝛽 to diminish from 𝛽𝛽𝑢𝑢 to the minimum value 𝛽𝛽m .

Expression of ∆𝑡𝑡m =

Value of ∆𝑡𝑡m =

0.6 pt / / / / / / /


ANSWER SHEET

Question Number 3

(k) The power 𝐸 supplied to the bubble at 𝛽𝛽.

Expression of 𝐸 =

0.6 pt / / / / / / /

The upper bound of the emissivity 𝑎𝑎.

Expression of 𝑎𝑎 =

Value of 𝑎𝑎 =

0.8 pt / / / / / / /

Experimental Competition 27 April 2010 Page 1 of 12 __________________________________________________________________________________________

Question Number 1

Experimental components Set-C for common components：

Label Items Quantity C-A Optical track (60 cm) 1 C-B Optical clamps 4 C-C-# Collimated laser diode (CLD) 1 C-D-# Sine wave generator (sine wave, DC 5V output) 1 C-E-# Variable resistor (5 kΩ) 1 C-F-# Connecting wires 4 C-F-6 Component stand 1 C-G Ruler 1 Note: “#” is the serial number for the component. This number is for examiner’s use.


Question Number 1

Set-I for Experiment-I Label Items Quantity I-H-# Black box on a 1-D translational stage 1 I-I-# Brass reed attached to a driving box* 1 I-J Screen for amplitude measurement 1 I-K-# Vertical slider (with a ruler and a magnet) 1 *The brass reed with a fixed end inside a box is attached to a piezo driven by an AC voltage.


Question Number 1

Instructions for the Sine Wave Generator:

‧ The power button, not shown in the picture, is on the right-hand side of the instrument.

‧ The “Display Panel” shows the frequency of the output sine wave. ‧ Use the “Sine Wave” BNC connector for supplying a sine wave voltage. ‧ Use the “5V DC” banana connectors for supplying a constant voltage of 5V. ‧ The frequency of the sine wave can be changed by turning the “FREQUENCY” knob,

faster for coarse adjustment and slower for fine adjustment. Ignore the fine and coarse labels next to the “FREQUENCY” knob.

‧ The amplitude of the sine wave voltage can be adjusted by turning the “AMPL ADJ” knob.

‧ The “RESET” bottom may be pushed to reset frequency to 0.00 Hz.


Question Number 1

Set-II for Experiment -II Label Items Quantity II-L-# Uncollimated laser diode (ULD) 1 II-M Holders for ULD 1 II-P-# Polarizer with indicator (PR2) 1 II-Q-# Polarizer (PR1) 1 II-R Holder for PR1 and PR2 1 II-T Holder for light filter 1 II-U Light filters 4 II-V Beam viewing box (screen) 1 II-W-# Photoconductor (PC) 1 II-X Digital multimeter 1 II-Y Digital multimeter 1 Items II-L-#, II-M, and II-V are not used.


Question Number 1

Instructions for the digital multimeter: ‧ You can turn the digital multimeter on or off by pressing the power button. ‧ Use the “VΩ” and the “COM” inlets for voltage and resistance measurements. ‧ Use the “mA” and the “COM” inlets for small current measurements. ‧ Use the function dial to select the proper function and measuring range. “V” is for

voltage measurement, “A” is for current measurement and “Ω” is for resistance measurement.

‧ Do not press the “HOLD” button, which will hold the display reading and stop the measurement function. You can release it by pressing the button again.


Question Number 1

Experiment I. Magnetic force probe Introduction

As shown in Fig. I-1, the free end of a reed can oscillate in the vertical direction when it is driven by an external oscillating force, and its frequency is determined by the external driver. If we plot the average power dissipated in the vibrating reed, which has certain damping mechanisms, as a function of frequency, we can find a maximum dissipated power at a certain frequency called the resonance frequency fR, as illustrated in Fig. I-2. The sharpness of the resonance is described by the quality factor Q as:

Q = f

fR

∆

where f∆ is the full width at half maximum of the Pav-f curve, as shown in Fig. I-2, i.e. f∆ = f2 - f1 with f1 and f2 corresponding to 2maxP on the lower side and the higher side of the resonance frequency respectively.

Fig. I-1. A vibrating reed.

Fig. I-2. Plot of the average dissipated power versus the driving frequency.

Besides the oscillating driving force, if the free end of the reed is subjected to a uniform force, its resonance frequency, amplitude and quality factor remain the same. On the other hand, under a non-uniform force, many properties of the vibrating reed, such as the resonance frequency fR, the maximum amplitude A, and the quality factor Q, may vary with the position of its free end.


Question Number 1

In this experiment, a small magnet adhered to the free end of the reed serves as a probe tip as shown in Fig. I-3, while a target magnet underneath the tip magnet produces a non-uniform magnetic field and exerts a non-uniform force on the tip magnet. When the tip magnet approaches the target magnet underneath with same pole opposing each other, then the repulsive force becomes stronger. Thus the resonance frequency fR of the reed varies with the distance between the tip magnet and the target magnet. The resonance frequency increases with decreasing separation distance between the two repulsing magnets. However, when moving the tip magnet horizontally away from the target magnet as shown in Fig. I-4, it may sense a weak attractive force at a certain distance. The resonance frequency shifts to lower values when the non-uniform force is attractive. We shall use this property, that the resonance frequency of the reed sensitively depends on the separation between the tip magnet and target magnet, to locate the hidden magnets inside a black box.

Fig. I-3. Near a pole of a target magnet, the magnetic field is non-uniform. Fig. I-4. Moving a tip magnet horizontally may cause it to sense an attractive or repulsive force.


Question Number 1

Experimental procedures

Error analysis is not required in any parts of Experiment I. Exp. I-A、Measuring the resonance frequency

Carefully take out the experimental components from Set-C and Set-I, and set up the experimental apparatus as shown in Fig. I-A-1. The schematic plot is shown in Fig. I-A-2. Connect the 5V-DC voltage source to the laser box (C-C-#). Connect the oscillating output of the sine wave generator to the driving box of the brass reed (I-I-#). Turn on the power and fix the output voltage of the sine wave generator. Direct the laser beam into the mirror at the free end of the brass reed so that the reflected beam spot on the screen (I-J) can be used to determine the vibrating amplitude of the reed.

Caution: 1) Carefully remove the paper protected cover before using the brass reed (I-I-#) for

experimental measurements. The resonance frequency of the brass reed is very sensitive to its shape, and any deformation of reed during the experiment may give an inaccurate result. 2) Do not directly look into the laser beam, which can damage your eyes.

Fig. I-A-1. Experimental setup for finding the resonance frequency.


Question Number 1

Fig. I-A-2. The schematic plot of Fig. I-A-1.

(1) Measure the amplitude A of the oscillating laser beam spot by changing the

frequency of the sine wave generator. Record the measured amplitude as a function of frequency in the data table on the answer sheet. (0.8 points)

(2) Make a proper plot on one of the supplied graph papers to determine the resonance frequency fRO and quality factor Q. Also record the obtained fRO and Q in the proper blank spaces on the answer sheet. (1.2 points)

Exp. I-B、Resonance frequency versus the external force.

In this part of experiment, the resonance frequency under the influence of a non-uniform force is investigated. The non-uniform force is provided by a small 3-mm cylindrical metallic calibration magnet MC fixed on a vertical slider (I-K-#) with its N pole pointing upward. The tip magnet MT, adhered at the free end of the oscillating reed, has its N pole pointing downward. The pole axes of both magnets should be aligned along the same vertical line.

Set up the experiment as shown in Fig. I-B-1. The schematic plot is shown in Fig. I-B-2.


Question Number 1

Fig. I-B-1. Experimental setup of finding the relation of resonance frequency with the nominal distance between two magnets, MC and MT.

Fig. I-B-2. The schematic plot of Fig. I-B-1.

(1) On the scale of the vertical slider, read out the position 0z of the bottom plane for the

tip magnet MT without the interaction of MC by properly moving MC away from MT. Record the measured 0z in the data table. (0.2 points)

(2) Adjust the position of the magnet MC to be right underneath MT. The pole axes of both magnets should be aligned along the same vertical line. Determine the position z of the top plane of the N-pole of MC. Calculate the nominal distance d by defining zzd −= 0 . Record z and d in the data table. (Note: The equilibrium separation between the two magnets is not the same as d because the two magnets repel each other.)

(3) Determine the resonance frequency fR for the distance d by tuning the frequency of


Question Number 1

the sine wave generator until the maximum amplitude is reached, plotting amplitude versus frequency is not necessary for determining fR of each distance d. Record the determined resonance frequency fR in the data table.

(4) Change the vertical position of the magnet MC and repeat the steps (2) and (3) for a number of measurements of different distance d and the corresponding resonance frequency fR. (1.2 points)

(5) Plot a graph of fR as a function of distance d using a graph paper. Guiding by eyes, draw the best line through the data points. (1.2 points)

(6) Define ΔfR = fR - fRO, and plot ln(ΔfR) as a function of d using another graph paper. Guiding by eyes, draw the best line through the data points. (1.0 points)

Exp. I-C、Finding the positions and depths of the magnets inside a black box.

There are two magnets MA and MB buried in the black box (I-H-#) which is fixed on a 1D translational stage. The N poles of both magnets are pointed upward. Magnets MA and MB, and MC used in Exp. I-B are very close in size, shape, and magnetic properties. The depths of the magnets MA and MB may be different. Magnet MA is located at the intersection of the two lines marked on the top surface of the black box. Magnet MB is located somewhere along the longer line as shown in Fig. I-C-1. The horizontal distance between magnets MA and MB is denoted by AB .

Fig. I-C-1. Magnet MA is located beneath the intersection of the two lines marked on the top surface while the magnet MB is located somewhere along the longer line.

(1) On the scale of the vertical slider, read out the position z0 (in this part, z0 may be different from the z0 in Exp. I-B) of the bottom plane for the tip magnet MT without the interaction of the magnets inside the black box. On the scale of the vertical slider, read out the position zbox of the top plane of black box. Record z0 and zbox on the answer sheet. (0.2 points)


Question Number 1

(2) Move the black box along the longer line and observe the variation in resonance frequency fR of the reed to find the position of MB. Record the measured distances y and their corresponding resonance frequencies fR in the data table. (1.4 points)

(3) Plot fR as a function of y on a graph paper to determine the position of magnet MB. Mark the positions of magnets MA and MB on the y-axis of your graph, and write down the value of AB on the answer sheet. (1.2 points)

(4) Determine the depths dA and dB of the magnets MA and MB from the top surface of the black box using the results in Exp. I-B. Write down the values of dA and dB on the answer sheet. (1.6 points)

Experimental Competition 27 April 2010 Student Code Page No. Total No.

ANSWER SHEET

Question Number 1

Exp. I-A、Measuring the resonance frequency

(1) Measure the amplitude A of the oscillating laser beam by tuning the frequency f of the sine wave generator. Record the measured data in the data table.

f A f A


ANSWER SHEET

Question Number 1

(2) Plot a proper data in the graph paper to determine the resonance frequency fRO

and the quality factor Q. Record fRO and Q in the following blank.

fRO = ； Q=


ANSWER SHEET

Question Number 1

Exp. I-B、Resonance frequency versus the external force.

(1) Measure and record the measured data 0z in the data table.

0z =

(2) Determine the position z of the top plane of the N-pole of MC. Calculate the

nominal distance d by defining zzd −= 0 . Record z and d in the data table. (3) Determine the resonance frequency fR for the distance d by tuning the frequency of

the sine wave generator until the maximum amplitude is reached. Record the determined resonance frequency fR in the data table.

(4) Change the vertical position of the magnet MC and repeat the steps (2) and (3) for a number of measurements of different distance d and the corresponding resonance frequency fR.

z d fR ΔfR ln(ΔfR)

(5) Plot a graph of fR as a function of distance d using a graph paper. (6) Define ΔfR = fR - fRO, and plot ln(ΔfR) as a function of d using another graph paper.


ANSWER SHEET

Question Number 1

Exp. I-C、Find the positions and depths of magnets in a black box. (1) Record z0 and zbox on the answer sheet.

z0 = ; zbox =

(2) Move the black box along the longer line and observe the variation in resonance

frequency fR of the reed to find the position of MB. Record the measured distances y and their corresponding resonance frequencies fR in the data table.

y fR y fR


ANSWER SHEET

Question Number 1

(3) Plot fR as a function of y on a graph paper to determine the position of magnet MB. Mark the positions of magnets MA and MB on the y-axis of your graph, and write down the value of AB on the answer sheet.

AB =

(4) Determine the depths dA and dB of the magnets MA and MB from the top surface of

the black box using the results in Exp. I-B. Write down the values of dA and dB on the answer sheet.

dA = ; dB =


Question Number 2

Experiment II Understanding Semiconductor Lasers The purpose of this experiment is to explore the basic characteristics of semiconductor

lasers. We will measure and calculate the fraction of the linear polarization of the collimated laser beam by using a pair of polarizers and a photoconductor. Finally, we will determine the maximum value of the power increase per current increment of the collimated laser.

Safety Caution: Do not look directly into the laser beam, which can damage your eyes!!

Background Description

The photoconductor, the light-sensing device in this experiment, is made of semiconductor, which has a band gap of EG = (EC – EV) (see Fig. II-1). When the energy of the incident photons is larger than that of the band gap, the photons can be absorbed by the semiconductor to create free electrons and holes. The density of charge carriers, including electrons and holes, is then increased, and so is the conductivity of the material. In this experiment, the resistance (the inverse of conductance) is measured by using a multimeter.

EV (Valence Band Edge)

Hole

EG (Energy Gap)

EC (Conduction Band Edge)

Electron

Photon

Fig. II-1 Schematic diagram of an electron-hole pair generated by a single photon in a semiconductor.


Question Number 2

In the semiconductor laser, the light emitting device in this experiment, as the external source injects electrons and holes into the device, they can combine to emit photons as shown schematically in the Fig. II-2. Ideally, the combination of one pair of electron and hole can generate one photon. Realistically, there are also nonradiative processes through which an electron-hole pair recombines without generating a photon. Thus the number of photons generated is not equal to the number of electron-hole pairs recombined. The average fractional number of photons generated by an electron-hole pair is called the quantum efficiency.

The semiconductor laser can emit a monochromatic, partially polarized and coherent

light beam. The partially polarized light is composed of two parts – linearly polarized and

unpolarized. The light intensity due to the former is denoted by Jp and the other by Ju.

When the partially polarized light is incident upon a polarizer, the transmittance of the

linearly polarized part depends on the angle between its polarized direction and the

direction of the polarizer. But for the unpolarized part, a constant portion is allowed to pass

through the polarizer and is independent of the angle.

Hole

Fig. II-2 Schematic diagram of a single photon generated by an electron-hole pair combined in a semiconductor.

EC (Conduction Band Edge)

EV (Valence Band Edge)

Photon

Electron

EG (Energy Gap)


Question Number 2

Experiments and Procedures

Exp. II-A：Light Response of the Photoconductor

The light source used in this part should be the collimated laser diode (CLD). Use the

circuit diagram shown in Fig. IIA-1 to provide current for the CLD. The symbols used in Fig. IIA-1 are listed in Table IIA-1.

Fig. IIA-1 Circuit diagram used for the CLD.

Table IIA-1 Symbols used in Fig. IIA-1

Devices Collimated Laser Diode

5V dc-power

Variable Resistors

Ammeter (Multimeter)

Symbols

Label C-C-# C-D-# C-E-# II-X, II-Y

Operate the CLD with the maximum current. The laser intensity is detected by a

photoconductor (PC). When you shine light on a PC, the conductance increases with the light intensity. You should minimize the ambient light effect. In this experiment, we actually measure the resistance, which is the inverse of conductance. The intensity of the laser light reaching the PC may be varied by using the supplied polarizers or filters. The symbols of other optical components are given in the Table IIA-2. The partial polarization of the laser light may be observed by using the experimental setup in the Fig. IIA-2.

A

A

Ω


Question Number 2

Table IIA-2 Symbols of optical components. Devices Collimated Laser

Diode Polarizers Photoconductor Light filter

Symbols

Optional Label C-C-# II-P-#

II-Q-# II-W-# II-U

Rotating P1, one should observe that the PC resistance varies. Adjust P1 so that the PC

resistance reaches a minimum. If the observed minimum resistance happens in a range, say °10 or larger of the rotation angle of P1, then the PC is saturated. In this case light filter(s)

should be used to avoid PC saturation near the maximum light intensity. Fix the P1 position according to the description in previous paragraph. Characterize the

conductance of the PC versus the relative light intensity following the experimental setup shown in the Fig. IIA-3.

Fig.IIA-2 Experimental setup for the preparation

PC P1 F L

P1 P2 PC D F L


Question Number 2

(1) Define Pθ to be the relative angle between polarization axes of P1 and P2. By varying the angle Pθ from °0 to °180 in step of °5 . Record the measured PC resistance and

Pθ in the data table. Transform the measured PC resistance values into conductance values and record them in the data table. No error analysis is required. (1.2 points)

(2) Plot the PC conductance values as a function of Pθ on a graph paper. No error analysis is

required. (1.2 points)

CLD P1

P2

PC

PC P1 P2 F L

Fig.IIA-3 Experimental setup for the PC characterization


Question Number 2

Exp. II-B：The Fraction of Linearly Polarized Laser Light

The light source used in this part should be the collimated laser diode (CLD) with a 15

mA current from the dc power supply. The task in this part is to determine the fraction β of the laser light that is linearly polarized by using the setup in Fig. IIA-2, which is the same as the previous section. No error analysis is required in this part.

minmax

minmax

)()()(

JJJJ

PolarizedLinearlyJdUnpolarizeJPolarizedLinearlyJ

+−

=+

=β

Jmax and Jmin are the maximum and minimum light intensity detected by PC while rotating P1.

(1) Find the maximum and minimum values of PC resistance (Rmax and Rmin) by rotating P1

360 °. Transform Rmax and Rmin into the minimum and maximum values of PC conductance Cmin and Cmax. Record the data in the data table. (0.8 points)

(2) Utilizing the conductance versus Pθ graph in Exp. II-A-(2) to determine the relative

intensities Jma x and Jm in corresponding to Cmax and Cmin. Write down the result on the

answer sheet. (1.6 points)

(3) Calculate β and write down the result on the answer sheet. (0.2 points)

Fig.IIA-2 Experimental setup for the preparation

PC P1 F L


Question Number 2

Exp. II-C： The Differential Quantum Efficiency of the

Collimated Laser Diode The task of this part is to characterize the relative light intensity versus the current through

the collimated laser diode (CLD) and determine the Differential Quantum Efficiency η, which will be defined below. Control the current of CLD in the range between 5 mA and 20 mA. Make sure that the PC is not saturated when the current is close to 20 mA. Filters or polarizers can be used to avoid saturation.

(1) Control the CLD current and measure the corresponding PC resistance values. Record the

data in the data table. Transform your data and plot the PC conductance versus CLD current on a graph paper. No error analysis is required. (1.3 points)

(2) Based on the graph of step (1), choose a region (∆I ~ 3 mA) centered around the

maximum slope. By using the conductance versus Pθ graph in Part II-A-(2), transform and record the data of this region in the table of step (1) into the relative light intensity (J). Plot the relative light intensity (J) versus CLD current (I) on a graph paper. No error analysis is required. (0.8 points)

(3) The maximum radiating power of the CLD is assumed to be exactly 3max =P .0 mW.

Extract the maximum slope from the graph in step (2) and transfer it to the value of

max

PGI

∆≡∆

, which is the maximum ratio of the increased amount of radiating power and

the increase amount of input current. Write down your analysis and the calculated value G on the answer sheet. Estimate the error of G. Do not include the error of the Pmax. Write down your analysis and the calculated value ∆G on the answer sheet. (2.0 points)

(4) The Quantum Efficiency equals the probability of one photon being generated per

electron injected. From a particular bias current of the laser, a small increment of electrons injected would cause a corresponding increment of photons. The Differential Quantum Efficiency η is defined as the ratio of the increased number of photons and the increased number of injected electrons. Determine η of your CLD by using the value of G obtained in step (3). Write down your analysis and the calculated value η on the answer sheet.


Question Number 2

Estimate the error of η. Write down your analysis and the calculated value ∆η on the answer sheet. (Laser wavelength = 650 nm. Planck’s constant = 346.63 10 J s−× ⋅ . Light

speed = 83.0 10 m s× ) (0.9 points)


ANSWER SHEET

Question Number 2

Exp. II-A：Light response of the photoconductor

(1) Define Pθ to be the relative angle between polarization axes of P1 and P2. By varying the angle Pθ from °0 to °180 in step of °5 . Record the measured PC resistance (R) and Pθ in the data table. Transform the measured R values into conductance (C) values and record them in the data table.

Pθ R C


ANSWER SHEET

Question Number 2

(2) Plot the PC conductance values as a function of Pθ on a graph paper.


ANSWER SHEET

Question Number 2

Exp. II-B：The fraction of the linearly polarized laser light

(1) Find the maximum and minimum values of PC resistance (Rmax and Rmin) by rotating P1

360 °. Transform Rmax and Rmin into the minimum and maximum values of PC

conductance Cmin and Cmax. Record the data in the data table.

Rmin

Rmax

Cmax

Cmin

(2) Utilizing the conductance versus Pθ graph in Exp. II-A-(2) to determine the relative

intensities Jma x and Jmin corresponding to Cmax and Cmin. Write down the result.

Jmin= , Jma x=


ANSWER SHEET

Question Number 2

(3) Calculate β and write down the result on the answer sheet.

β =


ANSWER SHEET

Question Number 2

Exp. II-C： The differential quantum efficiency of the collimated laser

diode (1) Control the CLD current (I) and measure the corresponding PC resistance (R) values.

Record the data in the data table. Transform your data and plot the PC conductance (C) versus CLD current on a graph paper.

I R C


ANSWER SHEET

Question Number 2


ANSWER SHEET

Question Number 2

(2)Based on the graph of step (1), choose a region (∆I ~ 3 mA) centered around the maximum slope. By using the conductance versus Pθ graph in Part II-A-(2), transform and record the data of this region in the table of step (1) into the relative light intensity (J). Plot the relative light intensity (J) versus CLD current (I) in a graph paper.

(3) The maximum radiating power of the CLD is assumed to be exactly 3max =P mW.

Extract the maximum slope from the graph in step (3) and transfer it to the value of

MaxIPG∆∆

≡ , which is the maximum ratio of the increased amount of radiating power

and the increase amount of input ampere. Write down your analysis and the calculated value G on the answer sheet. Estimate the error of G. Do not include the error of the Pmax. Write down your analysis and the calculated value ∆G on the answer sheet.

M a xIPG∆∆

= = , =∆G


ANSWER SHEET

Question Number 2

(4) The Quantum Efficiency equals the probability of one photon being generated per

electron injected. From a particular bias current of the laser, a small increment of

electrons injected would cause a corresponding increment of photons. The

Differential Quantum Efficiency η is defined as the ratio of the increased number of

photons and the increased number of injected electrons. Determine η of your CLD by

using the value of G obtained in step (3). Write down your analysis and the calculated

value η on the answer sheet. Estimate the error of η. Write down your analysis and the

calculated value ∆η on the answer sheet. (Laser wavelength = 650 nm. Planck’s

constant = sJ ⋅× −341063.6 . Light speed = sm8100.3 × )

η = , η∆ =

· theoretical competition 25 april 2010 page 1 of 5...

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