香港物理竞赛试题
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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 ˆ2= , where K