第3届泛珠三角物理竞赛试题(综合)答案

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 σ

σε±=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 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