大学物理课堂PPT复习资料

(电流I 均匀分布在圆柱的横截面内)内外的磁场.
Example:
∫ B ⋅ dl
L
r
r
= B ∫ dl = 2πrB = µ0i
L
B out =
µ 0i 2 πr
i πr 2 πR 2
r L2 r
r B
Choose a circular loop L2 with r < R,
Solution:
Cylindrical symmetry— —P609 Choose a circular loop L1 with r > R,
B A q
(b) Potential of the shell B
VB = Q + q 4πε 0 R2
B A q
R2
R1
Potential of the sphere A
VA = q q q + Q − + 4πε 0 R 4 πε 0 R1 4πε 0 R2
oR
Q
V A = V AB + V B = q 1 1 Q + q ( − )+ 4πε 0 R R1 4πε 0R2
?
v dE v dB
v v E = ∫ dE v v B = ∫ dB
z
ˆ µ I dl × r B = ∫ dB = ∫ 0 4π r 2
x
D
I
θ2
2)外力做功 A外 = F外 d = F电d = Q ( E电 )d Q 2d = >0 2ε 0 S
σ Q = 2ε 0 2ε 0 S
B=
µ0 I （cosθ1 − cosθ 2 ) 4πr
(Right-hand rule direction)
Example: v Find B inside a “infinitely long” solenoid (载流长直螺 线管内部的磁场)(The number of turns per unit length of the solenoid n and total number of coils N). Based r on the feature of the distribution Solution: of B, we choose a rectangular loop abcd,
v v µ m µ 0 IR 2 B= 0 3 3 2πx 2x v µ0m v B= en 2 π x3 v 说明：m 的方向与圆电流 v 的单位正法矢 e n 的方向相同.
v v m = IS e n
I
S
B=
vm en
I
v
v m
v en
S
10
2 I 1 5 4 R1 * o
R2
3
B1 = 0 µ I B2 = 0 4 R2 方向垂直于纸面向里
B= µ 0 Ni µ 0 Ni = = µ 0 ni 2 πr L
R2
R1
∫LB ⋅ dl = 2πrB = µ0 Ni
B= µ 0 Ni 2 πr
v
r
It is as same as the expression of the one of long solenoid. (2) Q Bout = 0, so the magnetic field of toroid is limited in its interior.
σ 3 S + σ 4 S = QB Considering the point PB in plate B,
σ σ σ σ EB = 1 + 2 + 3 − 4 = 0 2 ε 0 2ε 0 2ε 0 2ε 0
S
σ σ σ σ EB = 1 + 2 + 3 − 4 = 0 2ε 0 2ε 0 2ε 0 2ε 0
r R
o
2
3
3
(2) letting sphere is
U ∞ = 0 , the electrical potential outside the
U =∫
r
∞
Qdr Q = 2 4 πε 4πε 0 r 0r
Example:
A conducting sphere, of radius R and charge +q, is concentric with a spherical conducting shell of inner radius R1 and outer radius R2. This shell has a net charge of Q. (a) What are the charges on the inner and outer surfaces of the shell? (b) The potential of the sphere and the shell? Solution:
2π d1
15
2. The application of Ampere’s law It plays very important role in the theory of 电磁 r 场。One can also use Ampere’s law to calculate B with symmetrical distribution of current.
o
θ1
C I B
X
P
y
E电 =
B=
µ0I 2πr
I
B
（1）若线圈有 N 匝 B =
N µ 0 IR 2
磁偶极矩
3
（ 2 x + R）2
2 2
x = 0 B= （ 2）
R
µ0 I 2R
r
x
o
*p
v B
I
9
x >> R （ 3） µ IR 2 B= 0 3 ， x 2x µ 0 IS B= 2 π x3
1
2014/6/13
Example
平行板 电容器带电Q，间距d ，缓慢拉动至2d
Last Time…… 静电场: 取 dq
。求：1) 电容器能量变化；2) 外力做功. 解：（1）电容器能量的变化：
Q W=W ∆ 初 = 2C 末−W
2 末
磁 场： 取 Idl Biot-Savart Law
v
−
Q2 Q 2d = 2 C 初 2ε 0 S
圆电流的磁场 σdldr dI = = σvdr = σω r dr dt v×1 dB = µ0dI µ0σω = dr 2r 2
dI dr
σ R
o
σ R o rຫໍສະໝຸດ ωωdrv σ > 0, B 向外
14
µ σω R µ σω R dr = 0 B= 0 ∫ 0 2 2 v σ < 0, B 向内
13
方向垂直于纸面向外 方向垂直于纸面向外
B3 =
µ0 I 4πR
方向垂直于纸面向外
B = B1 + B2 + B3 =
3µ 0 I µ0 I + 8R 4πR
2
2014/6/13
解法 例 半径为 R 的带电薄圆盘的电荷 面密度为 σ ，并以角 速度 ω 绕通过盘心垂 直于盘面的轴转动， 求圆盘中心的磁感强 度.
σ 3 S + σ 4 S = QB
σ1 σ 2 σ 3 σ 4
PB
σ 2 = −σ 3
σ1 = σ4 =
d
QA − QB 2S
d
Think that if the plate B is connected to the ground, then σ1= σ4 = 0
QA + QB 2S
σ 2 = −σ 3 =
Look at the description given in textbook of P608. 例 如图载流长直导线的电流为 I ，试求 通过矩形面积的磁通量. µI 解 B= 0 v B 2 πx dx µ0I dΦ = BdS = ld x 2π x l I v v µ Il d2 dx Φ = ∫S B ⋅ dS = 0 ∫d d1 d2 2π 1 x o x x Φ = µ 0 Il ln d 2
conservation of charges
Example: Two parallel-conducting-plates with same area (S>>d2). They charged QA and QB respectively. Find the surface density at each face of the plates under electrostatic equilibrium.
Main steps: (1) Symmetry analyses (由I的分布，分析分布的对称性). (2) Choose a loop (使回路上各处B相等，方向特殊，从而 可从回路积分中提出B). (3) Put into the law and calculate B.
r Find the B inside and outside a long straight wire of radius R and uniformly distributed current I over a cross section of the wire 长直圆柱形载流圆柱体
Solution: Assume surface density is σ
QA
Plot a Gauss’s surface S', we have r r σ 2 = −σ 3 ∫∫ E ⋅ dS = 0
σ 1 S + σ 2 S = QA
QA
QB
S
σ1 σ 2 σ 3 σ 4