第3届泛珠三角物理竞赛试题(综合)答案

Answers Part IQ1. The plane should follow the parabola 飞机须沿抛物线运动。

2001cos , sin 2x 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()12212ml l I m ml =+= (2 points)There are two ways to find the torque.Method-1方法-1The 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) 22l lT mg mg mg θθ=-+=-. (2 points)Method-2方法-2Given 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 lU mg mg θθ=-.3 2U 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220033222222()2()2()R IR MdB 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 σσε±=. In such case, the electric displacement is D σ=. 001()DP D E D vB εεεεε-=-=-=. (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()DP D E D vB εεεεε-=-=-=. 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0001()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 Jare 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 isproportional to r 2, J must be proportional to 1/r 2. So let rrK J ˆ2= , where Kis 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rrK J ˆ2= ，K 为待定常数。

In medium-1介质-1,121111KE J r σσ==, and the voltage drop from R 1 to R 2 is 1112111()V K R R σ=- (1 point) 介质-1,121111KE J rσσ==，从R 1 到 R 2 的电压为1112111()V K R R σ=- Likewise, in medium-2,2221KE r σ=, and the voltage drop from R 2 to R 3 is 2223111()V K R R σ=-. (1 point) 同样，在介质-2, 2221KE rσ=，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 1,21,21,2D J εσ=, so the charge on the inner, outer, and boundary shells are 114K επσ,224K επσ-, and 21214K εεπσσ⎛⎫- ⎪⎝⎭, respectively.介质-1里电位移1,21,21,2D J εσ=, 因此在各面上的电荷为114K επσ,224K επσ-,21214K εεπσσ⎛⎫- ⎪⎝⎭。

(b) Due to symmetry, the electric field is of the form 2ˆK E r r=, so 1311V K R R ⎛⎫=- ⎪⎝⎭, and 3113R R K V R R -=⋅. (2 point)由对称性可知，电场为2ˆK E r r =，因此电压为1311V K R R ⎛⎫=- ⎪⎝⎭，得3113R R K V R R -=⋅。

The current densities in the two hemispheres are 11J E σ= and 22J E σ=. Thetotal 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 halfand 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2101111012()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 d h dPP P gdh g dhρρ+-=⇒=- (1 point) 775055775Constant (),()P PPV P C where C C ρρρ=⇒=⇒== (1 point)Combine these two equations, 合并两式得2700075)721()(P gh P P g C P dh dP ρ-=⇒-= (2 point)502002(1)7gh P ρρρ=-, and 0002(1)7gh T T P ρ=-. (2 point)b) 2525C o n s t a n t ,C o n s t a nt TTV ρ== (1 point) The density at 40︒C is2255521.183131.18()313293293ρρ=⇒=⨯ (1 point) 40︒C 的空气密度为2255521.183131.18()313293293ρρ=⇒=⨯ The fraction of water vapor at 40︒C at sea level 40︒C 的水蒸汽分压为,155.3590%760η=⨯ (1 point) The fraction of water at 5︒C at high altitude 5︒C 的水蒸汽分压为,272)313278(7605.6=η (1 point)Rain 下雨量= 52127255.356.5313()(90%) 1.18()0.07760293278760()313V kg ηηρ-⨯⨯=⨯-⨯⨯= (1 point)高度：52 1.189.8278313(1)35007 1.0310hh 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)]XX Y Y P D E E X E Y εεεε=-=-+- (1 point) **200111Re()Re()()cos()222X Y T P E D E E kz εεε=⨯=⨯=-∆, (1 point)Where其中k cω∆≡. (1 point)v) 220000011()cos()()sin()22d X Y X Y T E kz dz E kd k εεεεεε=-∆=-∆⎰∆. (1 point) Maximum occurs when 最大值在22kd d kππ∆=⇒=∆. (1 point)Part IIQ1. (6 points)a) ,.......3,2,1,==⇒=n a n k n a k x x ππ (1 point) b) 22,1,2,3,......y x mk b m k m bππ=⇒== (3 points) c)1210W =⇔= 1212101239123910102102m b nm b ππ--⇒<⇒>⨯≈⨯ (2 points)Q2. (22 points)i) 22123m I m l l =+(2 points) 222120()()3m I F t l m l l kl ωθθω=⋅⇔+=-⇒= (2 points)ii) 1211()c o s ()3m lm l x F t l xkl ω+=- (2 points)111212111111111c o s()()c o s ()c o s ()c o s ()3x A t m m A t Ft A t k ωφωωφωωφ=+⇒-++=-+11121210,()3F A mk m φω⇒==-+ (2 points)iii) 121122()c o s ()c o s ()3m m x F t F t xk ωω+=+- (2 points) 1122121222112122cos cos ,()()33x A t A t F F A A mmk m k m ωωωω=+⇒==-+-+ (2 points)iv) c o s nn n x A t ω=∑ 212()3nn n F A mk m ω⇒=-+ (2 points)v) t i ti ti o u t e V CiR C L C L C iR eV iL R Ci iL R eV V ωωωωωωωωωω0220011+--=+++==222220222()()||(1)()outR 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22101ωωC L LC =⇒=- (1 point) (vi) For a given L , only the signalwith 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 212()3n m k m ω=+. (1 point) So as L is varied, one finds a particular L max at which the signal peaks, and 12max ()3mk 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,212()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 ()3mk 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 33r q Ze R= that is inside the sphere r will have a non-zero netforce on the nucleus. (2 points)只有33r q Ze R=这么多的负电荷对原子核的合力不为零。