電磁場與電磁波第15講邊界條件電感磁能

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1、Field and Wave Electromagnetic 電磁場與電磁波 2015. 11.10 2 1. The Magnetic Dipole 2. Magnetization and Equivalent Current Densities 2 ( A m )m I S 0 3 ( 2 c os si n )4 R mB a a R 2 00 22 s in 44 RI b m aAa RR 1 0 l im ( A /m ) nv k k v m M v A / mm s nJ M a2 ( A /m )mJM Review 3 3. Magnetic Field Intensit

2、y and Relative Permeability 0 2 0 0 0 ( ) ( 1 ) ( W b /m )mr B HM B H M H H H HJ 0B Differential form 0S B d s C H d l I Integral form Postulates of Magnetostatics in magnetic material 4 Main topic 靜磁場 1. 靜磁場的邊界條件 2. 電感和電感器 3. 靜磁場能量 5 1. 靜磁場的邊界條件 HJ 0B 0S B d s C H d l I 1 2 B2 H1 B1 H2 an2 Js 6 E2

3、E1 2 1 at w h a c d b an2 h S 2 1 an2 D 1 D2 s D 0E 0C E d l S D d s Q 2 1 2 1 2 2 1 2 1 2 ( ) 0 o r ) o r n t t n s n n s a E E E E a D D D D （ 1 2 1 t 2 t 1 t 2 t d d d d d 0 d d ( ) 0 b c d a l a b c d bd ac E w E w EE 12 12 E l E l E l E l E l E l E l E w E w 21 1 2 1 222 2 1 2 1 n 2 n d ( ) ( s

4、S b o tto m s id e to p nnto p b o tto m ns D d S D d S D d S S D d S D d S D a S D a S a D D S D D S S DS 7 1 2 2 1 2 0n n nB B a B B 1 2 2 1 2 snt t n sH H a H HJJ (a) 在分界面處 B的法向分量是連續(xù)的 . For linear isotropic media, we have n22 n11 HH (b) 磁場 H 的切向分量在跨越存在面電流的分界面時 ,是不連續(xù)的 . 在跨越幾乎所有物理媒質(zhì)的邊界時 , H 的切向分量都是

5、連續(xù)的 ; 只有 當(dāng)分界面為理想導(dǎo)體或超導(dǎo)體時 ,它才不連續(xù) . 8 Example 1. A loop magnetic core with a gap is closely wound by a coil with N turns, as shown in the figure. When the coil carries a current I, and the leakage magnetic flux outside the coil is neglected, find the magnetic flux density and the magnetic field intensi

6、ty in the core and the gap. Solution: Since the leakage magnetic flux is neglected, the direction of the magnetic flux density is around the circle, and it is perpendicular to the end faces of the gap. From the boundary condition, we know that the magnetic flux density Bg in the gap is equal to Bf i

7、n the core, i.e. fg0fg HHBB 9 Since r0 a , the magnetic field in the core can be considered to be uniform. Using Amperes circuital law in media, and taking the circle of radius r0 as the integral path, then we have d NI Hl NIdrBdB ) 2( 0f 0 g Considering , we have fg BB 0gf 0 0 ( 2 ) NI d r d eBB Th

8、en g g 0 0 0 ( 2 ) NI d r d e BH In the gap 0ff 0 0 ( 2 ) NI d r d e BH In the core 10 2. 電感和電感器 互磁通 12 (Mutual flux) 2 12 1 2 ( W b)S B ds Faradays law of electromagnetic induction Biot-Savart law 2 0 3 0 ( T ) 44 R CC II a RB d l d l RR 1 2 1 2 1LI 其中 ,比例系數(shù) L12 稱為回路 C1 和 C2之間的互感 , with the unit he

9、nry (H). In case C2 has N2 turns, the flux linkage 12 (磁鏈數(shù) )due to 12 1 2 2 1 2 ( W b )N 11 12 12 12 1 12 1 ( W b) or ( H )L I L I 由上式 ,兩個電路之間的互感是一電路通以單位電流時 ,另一電路 所交鏈的磁通鏈 . 這個關(guān)系只適用于 線性媒質(zhì) . 12 12 1 (H )dL dI L12 更一般的定義為 : 1 1 1 1 1 1 1 2NN I1 所產(chǎn)生的某些磁通量只與回路 C1 自身交鏈 ,而不與 C2交鏈 .the total flux linkage (磁

10、鏈數(shù) ) with C1 caused by I1 is 21 0 1221 dd 4 llL R ll 12 0 2112 dd 4 llL R ll Neumann formula 12 11 11 1 (H ),L I 回路 C1 的自感定義為在回路本身通以單位電流時所產(chǎn)生的磁鏈 ,即 : 11 11 1 (H ) .dL dI 對線性媒質(zhì)而言 ,更一般的表達(dá)式 L11 為 : 一個回路或電路的自感 ,取決于構(gòu)成這個回路或電路的導(dǎo)體的 幾何形狀 和 物理排列 以及 媒質(zhì)的磁導(dǎo)率 .在線性媒質(zhì)中 ,自感與回路 或電路的電流無關(guān) .事實(shí)上 ,無論回路或電路是開路還是閉合的 ,也無 論它是否靠

11、近另一回路或電路 ,其自感總是存在的 . 13 排列成適當(dāng)形狀 (例如由導(dǎo)線纏繞而成的線圈 )以提供一定數(shù)量自 感的導(dǎo)體稱為電感器 . 就象電容器可以儲存電能一樣 ,電感器能夠儲存 磁能 . The procedure for determining the self-inductance of an inductor is as follows: 1. 對于給定的幾何形狀選擇適當(dāng)?shù)淖鴺?biāo)系 . 2. 假設(shè)導(dǎo)線中的電流為 I. 3. 如果存在對稱性 ,可以根據(jù)安培環(huán)路定理由 I 求 B; 如果不存在對稱性 , 則可以用 Biot-Savart law. 4. 用積分的方法由 B求出每一圈所交鏈的

12、磁通 5. Find the flux linkage by multiplying by the number of turns. 6. Find L by taking the ratio L=/I. ( W b )S B d s 14 EXAMPLE 6-14 P181 15 Example. Calculate the mutual inductance (互感 ) between an infinitely long straight line and a rectangular coil. The line and the coil are at the same plane, a

13、nd in vacuum. a b dr r D 0 I1 I2 z S2 Solution: Select cylindrical coordinate system, and let the infinitely long straight line to be at the z-axis. The magnetic flux density produced by current I1 is then 01 1 2 IBa r The magnetic flux linkage 12 with current I2 by current I1 is 21 2 1S B d S 16 We

14、 have 012 12 1 l n 02 a DbL ID If the flowing direction of the current I2 is counter clockwise, then the B1 and dS are opposite, and L12 0. a b dr r D 0 I1 I2 z S2 If the flowing direction of the current I2 is clockwise, dS and B1 have the same direction. Then 0 1 0 1 12 1 d l n 2 2 Db D I a I a Dbr

15、 rD 0 1 0 1 0 1 12 0 11 ( ) l n 2 2 2 D b a SD I I I a Dbdrdz dz dr r r D 17 02 2 2 IBa r 1 02 21 2 1 2S IB ds a a dzdr r 02 21 3 02 0 02 2 1 2 3 l n 1 2 d b r d b d I dzdr r I dr dz r I b d b b d 例題 :求無限長導(dǎo)線與三角形回路之間的互感 002 21 2 1 0 02 02 ( ) ta n 6 0 2 ta n 6 0 () 2 3 l n 1 2 bd d I B d s a a b d r

16、d r r I b d r dr r I b d b b d 21 12 2 L I d db I1 2I 60 r dr z x 0 18 3. 磁能 (Magnetic Energy) If an impressed source (外源 )is applied to a circuit(回路 ), a current will be generated in the circuit. In the process of establishing the current, the reaction magnetic flux (感應(yīng)磁通 )in the circuit will resis

17、t (阻止 )the increment of the current. Assume the current is increased very slowly(非常慢 ) so that radiation loss can be neglected, all energy provided by the impressed source will be stored(儲存 ) in the magnetic field around the circuit. Based on the work done by the impressed source, the energy stored

18、in the magnetic field can be calculated. In order to overcome the back electromotive force due to the reaction magnetic flux and to maintain the current, the impressed source has to do work(做功 ). 19 考慮一個自感為 L1的單獨(dú)閉合回路 ,其初始電流為 0,將電流源與回 路連接 ,電流從 0增加到 I1. 從物理學(xué)中知道回路中將產(chǎn)生感應(yīng)電動勢并 阻礙電流的變化 . 要克服這個感應(yīng)電動勢 ,必須有外源作

19、功 . 假設(shè)電感兩 端的的電壓為 v1=L1di1/dt ,則所需做的功為 : 1 2 1 1 1 1 1 1 1 10 1 . 2 IW v i d t L i d i L I 對于線性媒質(zhì)有 L1=1/I1 ,則上式變?yōu)?: 2 1 1 1 1 1 11 , 22W L I I 這個功看作是以磁能形式儲存起來 . Now consider two closed loops C1 and C2 carrying currents i1 and i2, respectively. The total amount of work done in rasing the currents in l

20、oops C1 and C2 from zero to I1 and I2 , respcetively, is 20 22 2 1 1 2 2 21 1 2 11 22W L I L I L I I which is stored as magnetic energy. Where W21 is 22 21 21 1 21 1 21 1 2 21 1 200 tIdiW v I dt L I dt L I di L I I dt Generalizing this result to a system of N loops (N個回路 ) carrying currents I1 , I2

21、, In , we obtain 11 1 ( J ) 2 NN m jk j k jk W L I I Which is the energy stored in the magnetic field. For a current I flowing in a single inductor(單個電感器 )with inductance L, the stored magnetic energy is 21 (J) 2mW L I 21 It is often desirable to express the magnetic energy in terms of field quantit

22、ies B and H. We have 1 ( J) 2m VW H B dv Noting that H=B/ , we can rewrite 2 1 ( J ) 2m V BW d v and 2 1 ( J) 2m VW H d v Where wm is defined as magnetic energy density(磁能密度 ) 2 2311 ( J / m ) 2 2 2m Bw H B H 22 靜 電 場 恒定磁場 物理量 媒質(zhì)特性 場方程式 邊界條件 能量密度 力 0 E 無旋 0 B 無散 D 有散 HJ 有旋 電場強(qiáng)度 E 磁通密度 B 電通密度 D 磁場強(qiáng)度

23、H 介電常數(shù) 磁導(dǎo)率 DE BH t2t1 EE n2n1 BB 2n 1nDD 2t 1tHH m 1 2w HBe 1 2w DE 常數(shù) l WF m 常數(shù) l WF e 常數(shù) Il WF m 常數(shù) ql WF e 23 1. Introduction 2. Fundamental Postulates of Magnetostatics in free space F q E q u B Lorentzs force equation 0BJ 0B Differential form 0S B d s 0C B d l I Integral form Postulates of Magn

24、etostatics in Free Space Review 24 3. Vector Magnetic Potential 4. The Biot-Savart Law and Applications 0B B A 2 0AJ ; 020202 zzyyxx JAJAJA 0 ( W b /m ) 4 V JA d v R 00 23 ( T ) 44 R C C C II a RB d l d l d B RR 25 5. The Magnetic Dipole 6. Magnetization and Equivalent Current Densities 2 ( A m )m I

25、 S 0 3 ( 2 c os si n )4 R mB a a R 2 00 22 s in 44 RI b m aAa RR 1 0 l im ( A /m ) nv k k v m M v A / mm s nJ M a2 ( A /m )mJM 26 7. Magnetic Field Intensity and Relative Permeability 0 2 0 0 0 ( ) ( 1 ) ( W b /m )mr B HM B H M H H H HJ 0B Differential form 0S B d s C H d l I Integral form Postulate

26、s of Magnetostatics in magnetic material 27 8. Boundary Conditions for Magnetostaic Fields 9. Inductances and Inductors 1 2 2 1 2 0n n nB B a B B 1 2 2 1 2 snt t n sH H a H HJJ HJ 0B 0S B d s C H d l I 12 12 12 1 12 1 ( W b) or ( H )L I L I 1111 1 (H ),L I 磁通量；磁鏈數(shù)；自感；互感 28 10. Magnetic Energy 11 1 ( J ) 2 NN m jk j k jk W L I I 21 (J)2mW L I 1 ( J) 2m VW H B dv 2 2311 ( J / m ) 2 2 2m Bw H B H 29 homework Thank you! Bye-bye! 答疑安排 時間： 地點(diǎn)： 1401， 1403 P. 6-38