香港物理竞赛试题

Answers Part I

Q1. The plane should follow the parabola 飞机须沿抛物线运动。

2001

cos , sin 2

x t y v t gt νθθ==- (3 points)

Q2 (6 points)

I T ω

= The center of the rod will not move in the horizontal direction 杆中心在水平方向不动。

22272()12212

ml l I m ml =+=

(2 points)

There are two ways to find the torque.

Method-1方法-1

The forces acting upon the rod are shown. The torque to the center of the rod is 由如图力的分析，可得

3(2) 22

l l

T mg mg mg θθ=-+=-. (2 points)

Method-2方法-2

Given a small angle deviation θ from equilibrium, the potential energy is 给定一个角度的小位移θ ，势能为

233(1cos )24l l

U mg mg θθ=- .

3 2

U l T mg θθ?=-=-? (2 points)

Finally , 最后得273122ml mgl θθω=?=

(2 points)

Q3 (6 points)

(a) The bound current density on the disk edge is 盘边的束缚电流密度为

K M n M =-?=-

, (1 point)

The bound current is 束缚电流I Jd Md ?==-, (1 point)

The B-field is 磁场为22003322222

2

()2()2()

R I

R Md

B z h R h R h μμ==

=-

++ (1 point)

(b) The bound current density is K M n M =-?=-

, which is on the side wall of the cylinder. (1 point)

柱侧面上的束缚电流密度为K M n M =-?=-

The problem is then the same as a long solenoid. Take a small Ampere loop we get 00B K M μμ== inside;

为求一长线圈的磁场，取一小闭合路径，得介质内00B K M μμ== (1 point) Outside 介质外 B = 0 (1 point)

Q4 (5 points)

Each unit charge in the slab experiences the Lorentz force 0 vB y -

. (1 point) The problem is then the same as a dielectric slab placed between two parallel conductor plates that carry surface charge density 0

, and vB σ

σε±=. In such case, the electric displacement is D σ=. 001

(

)D

P D E D vB εεεε

ε

-=-=-=. (2 point)

介质内单位电荷受力0 vB y -

。问题变成两电荷面密度为0

, and vB σσε±=的

导电板间充满介质。因此D σ=. 001

(

)D P D E D vB εεεε

ε

-=-=-=. Finally, the bound surface charge is 01

(

)b P vB εσεε

-==. The upper surface carries positive bound charge, and the lower surface carries negative charge. (1 point)

最后得束缚电荷密度01

()b P vB εσεε

-==，上表面带正电，下表面带负电。 The electric field is 0001()b E y vB y σεεε-==

, which is along the y-direction

(opposite to the Lorentz force). (1 point)

电场为000

1()b E y vB y σεεε

-==

，与Lorentz 力方向相反。

Q5. (10 points)

(a) Because of the spherical symmetry, the E-field and the current density J

are all along the radial direction. In steady condition, the electric current I through any spherical interfaces must be equal. Since the area of the sphere is

proportional to r 2

, J must be proportional to 1/r 2

. So let r r

K J ?2= , where K

is a constant to be determined, and the expression holds in both media. (1 point)

由对称性可知，电场和电流密度J

须沿半径方向。稳态时，流过每个包住

球心的球面的电流相等，因此J 与1/r 2

成正比。设在两介质里r r K J ?2= ，K 为待定常数。

In medium-1介质-1,121

11

1K

E J r

σσ=

=

, and the voltage drop from R 1 to R 2 is 11

12

1

11

(

)V K R R σ=

- (1 point)

介质-1,12

1

11

1K

E J r

σσ=

=

，从R 1 到 R 2 的电压为1112111()V K R R σ=- Likewise, in medium-2,22

21K

E r σ=

, and the voltage drop from R 2 to R 3 is 22

23

1

11(

)V K R R σ=

-. (1 point)

同样，在介质-2, 2221K

E r

σ=，从R 2 到 R 3 的电压为2223111()V K R R σ=-

The total voltage drop between R 1 and R 3 is 总电压 V = V 1 + V 2.

So 得

11222311111111()()K V R R R R σσ??=-+-????

The current is 电流为2114()4I R J R K ππ==.

(1 point)

The electric displacement in media are 1,2

1,21,2

D J εσ=

, so the charge on the inner, outer, and boundary shells are 114K επσ,224K ε

πσ-, and 21214K εεπσσ??- ???

, respectively.

介质-1里电位移1,21,21,2

D J εσ=

, 因此在各面上的电荷为114K επσ,224K ε

πσ-,

21214K εεπσσ??- ???

。

(b) Due to symmetry, the electric field is of the form 2?K

E r r

= , so 1

311V K R R ??=- ???, and 31

13R R K V R R -=?. (2 point)

由对称性可知，电场为2?K E r r = ，因此电压为1311V K R R ??

=- ???

，得31

13

R R K V R R -=?

。 The current densities in the two hemispheres are 11J E σ= and 22J E σ=

. The

total current is 22111121122()2()2()I R J R R J R K πππσσ=?+?=+. (2 point)

上下半部的电流密度为11J E σ= ，22J E σ=

。 总电流为

22111121122()2()2()I R J R R J R K πππσσ=?+?=+.

The total free charge is 2101111012()2Q R E R K πεεπεε=?= on the upper half and 2022Q K πεε= on the lower half of the inner shell. On the outer shell the charges are negative of the corresponding ones of the inner shell. (1 point)

内球面上半部总自由电荷为2

1011

11012()2Q R E R K πεεπεε=?=，下半部总自由电荷为2022Q K πεε=。外球面的电呵与内球面相反。

Q6. (12 points)

a) h h dh dP

P P gdh g dh

ρρ+-=?

=- (1 point)

775

05

5

7

75

Constant (),()P P PV P C where C C ρρρ=?=?== (1 point)

Combine these two equations, 合并两式得

270

007

5

)721()(P gh P P g C P dh dP ρ-=?-= (2 point)

5

0200

2(1)7gh P ρρρ=-, and 0002(1)7gh T T P ρ=-.

(2 point)

b) 2

5

25

Constant,

Constant T

TV ρ

==

(1 point)

The density at 40?C is

2255

5

21.183131.18()313293293

ρ

ρ=?=? (1 point)

40?C 的空气密度为

2

2555

21.183131.18()313293293

ρ

ρ=?=? The fraction of water vapor at 40?C at sea level 40?C 的水蒸汽分压为,

155.35

90%760

η=

? (1 point)

The fraction of water at 5?C at high altitude 5?C 的水蒸汽分压为,

2

72)313

278

(

7605.6=

η (1 point)

Rain 下雨量

= 5

2127

2

55.35

6.5313()(90%) 1.18()0.07760

293278760()

313

V kg ηηρ-??=?-

??= (1 point)

高度：5

2 1.189.8278313(1)35007 1.0310

h

h m ???=-

?=?? (1 point)

Q7. (8 points)

i) E p

? (1 point)

ii) E P

? (1 point)

iii) 00(1) 0D E P P E P E εεε=+?=-??=

(1 point) iv) 0000[(1)(1)]X X Y Y P D E E X E Y εεεε=-=-+-

(1 point)

**200111Re()Re()()cos()222

X Y T P E D E E kz εεε=

?=?=-? , (1 point)

Where

其中k c

ω

?≡. (1 point)

v) 2

200000

11()cos()()sin()22d X Y X Y T E kz dz E kd k εεεεεε=-?=

-???. (1 point) Maximum occurs when 最大值在2

2kd d k

π

π

?=

?=

?. (1 point)

Part II

Q1. (6 points)

a) ,.......3,2,1,==

?=n a n k n a k x x π

π (1 point) b) 22,1,2,3,......y x m

k b m k m b

ππ=?==

(3 points)

c) 12

10W == 12121012391239

10102102m b nm b ππ

--?

?≈? (2 points)

Q2. (22 points)

i) 22

123

m I m l l =+

(2 points)

222120()()3m I F t l m l l kl ω

θθω=??+=-?= (2 points)

ii) 1211()cos()3

m l

m l x F t l xkl ω+

=- (2 points)

1112

12111111111cos()

()cos()cos()cos()3

x A t m m A t F t A t k

ωφωωφωωφ=+?-++=-+ 1

112

1210,()3

F A m

k m φω?==

-+ (2 points)

iii) 1

21122()cos()cos()3

m m x F t F t xk ωω+

=+- (2 points)

11221

2

122

2

112122cos cos ,()()3

3

x A t A t

F F A A k m k m ωωωω=+?=

=

-+-+ (2 points)

iv) cos n

n n x A t ω=∑ 2

12()3

n

n n F A m

k m ω?=

-+ (2 points)

v) t i t

i t

i out e V C

iR C L C L C iR e

V iL R C

i iL R e

V V ωωωωωωωωωω02

20011

+--=+++=

=

222

2

20

222

()()||(1)()out R C L C V V

L C R C ωωωω+=-+ (1 point)

To make the denominator minimum, we should have 使分母最小

2

21

01ωωC L LC =?=- (1 point) (vi) For a given L , only the signal

with n ω=

in the answer of (iv) can pass through the filter, (1 point) and the output is proportional to A n . (1 point) By varying L one selects different n ω, and the output is proportional to the selected A n . (1 point)

From (iv), the maximum A n is the one when 21

2(

)3

n m k m ω=+. (1 point) So as L is varied, one finds a particular L max at which the signal peaks, and 12max ()3

m

k m CL =+.

(1 point)

给定 L ,

则只有频率为n ω=

, (1 point) 其大小正比于 A n . (1 point) L 的变化等于选择不同的n ω的信号. (1 point) 由(iv)知,

21

2(

)3n m k m ω=+时A n 最大. (1 point) 因此可得使输出信号最大的电感L max ，并有12max ()3m

k m CL =+.

Correct sketch is a flat line with a peak at L max 正确的简图是一平线，在L max 有一尖峰 (1 point)

Q3. (22 points) i) Only the charge 3

3r q Ze R

= that is inside the sphere r will have a non-zero net

force on the nucleus. (2 points)

只有3

3r q Ze R

=这么多的负电荷对原子核的合力不为零。

30000230041144E R Zeq rZe

ZeE E r r R Ze

πεπεπε=?=?=

(2 points)

ii) 3

0()4e Ze rZe

Zm r ZeE t R πε=-?

(2 points)

()cos(),cos()E t A t r B t ωω?==+

(1 point)

222

3

0cos()

cos()cos()4e Z e B t Zm B t ZeA t R

ω?ωω?ωπε+?-+=-? 2

2

300,1

4e eA

B Ze m R

?ωπε?==

?- (2 points)

222

30cos()1

4e Ze A

p Zer t Ze m R

ωωπε?==?-

(1 point)

iii) 2

2

2

30()()1

4e NZe P t Np E t Ze m R

ωπε==?- (2 points)

iv) 00(1)(1)()ext total ext self P CE E E E εεεε==-=-+ (2 points)

The electric field E in a uniform polarized ball with polarization vector P can be calculated by considering the surface bound charge cos P θ distributed on the ball. 均匀极化球内的电场E 可由球面的束缚电荷cos P θ求得。 220

1cos (cos )sin 43self P P

E R d d R θθθθ?πεε-=

=-

??

(2 points)

01

33

self ext P E CE ε=-

= (1 point) 00

2

022

3022

222

2

30(1)()31312

4()(), 1()4ext

ext ext e e e CE CE E NZe C Ze m R

NZe N Ze K C Ze Zm m R

εεεεεεωπεωωωωπε?=--

-?==

+?-?===-?-

3

2

00

2

4where is the resonant frequenc of an atom ()

and 2

e R Zm Ze n πεω≡= (2 points)

3

200

2

4 ()

e R Zm Ze πεω≡是原子的共振频率。 v) For air K is very small and ()εωis close to 1. So the L-L relation becomes

1

()1()K εωωε-=

(1 point)

Air density decreases with height. So refractive index decreases with height. (1 point)

Light rays are bent in such condition. (1 point)

空气的()εω接近1，K 很小，上述结果简化为0

1

()1()K εωωε-=

。空气的密

度随高度减小，因此折射率也减小，形成如图所示的光线弯曲。