TUTORIAL 9. Static magnetic field

Transcription

1 TUTOIAL 9 Static magnetic field

2 Vecto magnetic potential Null Identity % & %$ A # Fist postulation # " B such that: Vecto magnetic potential Vecto Poisson s equation The solution is: " Substitute it into B % $ A % $ B ( J, we have %$ $ A ( J % A %% & A" ) % $ % $ A * %% & ") % A ( J A A % & A % A )( J Coulomb condition Fo simplicity, we put, then we have, is a vecto Poisson s Equation. A ( J, dv+ 4- V +

3 Biot-Savat Law Fo line cuent density A thin wie with coss-sectional aea S, and the cuent flow is entiely along the wie A Biot-Savat Law ( J, dv+ 4- V + % " I dl & a B # C " 4$ J dv= JSdl= Idl % " I dl & B # C " 3 4$ Fomula fo detemining B caused by a cuent I in a closed path and is obtained by taking the cul of A o A J is the line cuent density fo 1 D condition # % I dl" 4$ C " (Wb/m)

4 To find B fom A " c Pimed souce coodinates & Unpimed field coodinates dl (x,y,z ) " # (x,y,z) a ˆ = a ˆ x (x " x)+ a ˆ y (y " y)+ a ˆ z (z " z) % I dl" A # 4$ C " 3 % I dl " -, 1 $ / f-, G 4 (-f ), G (Wb/m) % I + dl ( #. " # -, ) & C 4$ 4 C B % $ A We can do this is because the integal is ove the pimed coodinates (i.e., while the cul take the deivatives of the unpimed coodinates (i.e., ) that descibe the fields (magnetic flux density). % " I dl & a B # C " 4$ O % " I dl & B # C " 3 4$ *

5 Example 1 az dl az a dl L1 L3 dl = a ˆ z dz d = ˆ a " ˆ a z z d l " = ˆ db = µ I 4" (d B = 4 µ = a ˆ I $ 4" # " db a # dz l # ) 3 ( + z ) dz 3 l = a ˆ " bd" = a ˆ z z " a ˆ b l " = ˆ d B = µ I B = a bzd#+ a ˆ z b d# 4" (d l # ) 3 µ = a ˆ I z 4" # " db b ( b + z ) 3 d$ L=L1

6 Example (a) Detemine the magnetic flux density at a point on the axis of a solenoid with adius b and length L, and with a cuent I in its N tuns of closely wound coil. (b) Show that the esult educes to when L appoaches to infinity. B " a z ni L

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8 Execise (suggested answe) Note that N/L is added hee because we ae consideing multiple tuns, instead of one. % L x # a " a dx 3 $ x x # a L

9 Summation J (cuent density) Ampèe s Law " A ( ) = #µ J ( ) Biot-Savat Law B (Magnetic field intensity) B = " # A A Vecto magnetic potential

10 Magnetic dipole v A small cicula loop of adius b that caies cuent I is called magnetic dipole. + + I Electons and nucleus of an atom spin on thei own axes ceate cetain magnetic dipole moments. z The vecto magnetic potential due to it is : A = µ m " a 4# m is the magnetic dipole moment. x I y m = a Z I"b

11 Magnetization G.?;1,4;4,+-.4)(,: Magnetization vecto M is to detemine the quantitative change of magnetic flux density caused by the pesence of magnetic mateial. J m = " # M J ms = M " an Unde a constant B-field, the intoduction of M esults in a modification of A. % ( $" M ( A dv # 4 V & 4 & S M " a n ds The effect of the magnetization vecto is equivalent to both a volume cuent density and a suface cuent density.

12 Magnetic filed intensity The application of an extenal magnetic field causes the changing of magnetic flux density. To account fo this change, we intoduce the equivalent volume cuent density. Define the fouth new fundamental field quantity, the magnetic field intensity H, such that 1 $# B " J $ m # B " J % J " J % $ # M + + $# ) * B B H ", ( (, M & " J M (A/m) J is known as cuent density of fee cuent $# H " J (A/m ) Stokes s Theoem Ampee s law: - H. d l " C I

13 Example 3 Conside an infinite cylinde made of magnetic mateial. This cylinde is centeed along the z-axis, has a adius of m, and a pemeability of 4". Inside the cylinde thee exists a magnetic flux density: B ( ) = 8µ " ˆ a # Detemine the magnetization cuent Jsm flowing on the suface of this cylinde, as well as the magnetization cuent Jm flowing within the volume of this cylinde.

14 M # " m H B # ) ( 1 " m H # H Volume magnetization cuent density is zeo. No magnetization cuent within the cylinde. Magnetization cuent flows on the suface.

15 "1 B1n B1 #1 E1n E1 B1t E1t " Bn Bt B # En E Et B 1n = B n µ 1 H 1n = µ H n "1 " H1n Hn a ˆ n " ( H 1 # H ) = H 1t = H 1t B 1t B = t µ 1 µ H H1 H1t Ht J s D 1n " D n = # s D 1n = D n $ 1 E 1n = $ E n #1 # D1n Dn E 1t = E t D 1t D = t " 1 " Dt D D1 D1t

16 Example 4 H 1 = 3ˆ a x + 5 ˆ a z Z "1=" J s = ˆ a y "=3" H =?

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18 eview of Electostatics and Magnetostatics Analogous elations between the quantities between electostatics and magnetostatics Electostatics Magnetostatics Electic field intensity (V/m) E H Magnetic field intensity (A/m) Electic flux density (C/m) D B Magnetic flux density (T) Pemittivity (F/m) 1 & & pemeability (H/m) Polaization vecto P -M Magnetization vecto (A/m) Chage density " J Cuent density Electic potential (V) V A Magnetic vecto potential Divegence opeato $ # $% Cul opeato Cul opeato $% $ # Divegence opeato

19 eview of Electostatics and Magnetostatics Analogous elations between the quantities between electostatics and magnetostatics Electostatics Fundamental Postulates of Magnetostatics " Q % $ E # E$ ds %( # & J B $ dl I # S B % $ D # " D$ ds # Q %( H # J S Gauss s law C C # & H $ dl # I Ampee s cicuital law %( E # E$ dl # % $ B # C B $ ds # S %( D # D $ dl # % $ H # H $ ds # C S

20 eview of Electostatics and Magnetostatics Analogous elations between the quantities between electostatics and magnetostatics Electic Potential Magnetic Vecto Potential E # "V B # " $ A V 1 # & dv% 4() v % * J A # & dv% 4( V % Discete chage V # 1 4() n q k k # 1 % k + Cuent loop * I A # 4( dl% & C %

21 eview of Electostatics and Magnetostatics Analogous elations between the quantities between electostatics and magnetostatics Columb s Law Biot-Savat Law E 1 $ dv" V " 3 4%( B $ & " I dl # C " 3 4% E-field due to discete chage E $ 1 4%( + ), " * n q k k $ 1, " k k 3

22 eview of Electostatics and Magnetostatics Analogous elations between the quantities between electostatics and magnetostatics Bounday conditions fo Electostatic field Magnetostatic field E $ " a n % 1 E # ) D # ( 1t D t Tangential components continuous 1 ( H1 $ H " J S a # n % Tangential components discontinuous when suface cuent exists a D1 $ " # & S n D Nomal components discontinuous when suface chage exists B $ B " a n 1 # ) * # 1H 1 * H Nomal components continuous & # # ( # 1E 1 ( E When and, and. S J S B # * 1t B t 1 *

23 eview of Electostatics and Magnetostatics Capacitance, esistance and inductance Capacitance esistance Inductance C Q V 1 C = capacitance between conducto 1 and Steps 1. Assume +Q and -Q on conductos.. Find E fom Q by Gauss s law o othe elations. 3. Find V 1 by V I = esistance of conducting mateial between specified equipotential sufaces Steps 1. Assume a p.d. V between conducto teminals.. Find E within conducto teminals. 1 May ty solving V fo V, then obtain E by 1 %# E$ dl & V E %&V fo homogeneous fom conducto caying mateial. -Q to that with +Q. 3. Find total cuent by 4. Obtain C fom the fomula. # J $ ds # E$ I ds S S 4. Obtain by the fomula. L = self-inductance a coil with N tuns Steps L N" I 1. Assume a cuent I in the conducting wie. Find B fom Ampee s law o Biot-Savat law 3. Find " fom B by " # B $ ds S whee S is aea ove which B exists and link with the assumed cuent. 4. Obtain L by the fomula.