Supporting Local Efforts: Tackling the Drugs Problem
Susan Scally, Head of Drugs Policy Unit, discusses the National Drugs Strategy at the conference. Key objectives include harm reduction, coordination, and partnerships with communities to combat substance misuse.
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PHY 712 Electrodynamics 10-10:50 AM MWF Olin 107 Plan for Lecture 18: Finish reading Chap. 7; start Chap. 8 A. Summary of results for plane waves B. Electromagnetic waves in an ideal conductor C. TEM electromagnetic modes 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 1
02/28/2014 PHY 712 Spring 2014 -- Lecture 18 2
Review: Electromagnetic plane waves in isotropic medium with real permeability and permittivity: . ( ( ) ( ) ,t c electromag plane for vector Poynting ) ( ) ( ) k r ic n ct = = 2 2 E r E , t e n c 0 n ( ) ,t r k k = = B r E r E ,t netic waves : 2 E 2 n 1 k k 2 0 c = = S E 0 2 avg electromag plane for density Energy netic waves : 1 2 = E u 0 2 avg 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 3
Reflection and refraction between two isotropic media z k x ki kR i R Reflectanc transmitt e, ance : 2 2 S z S z ' i ' ' cos E E n = = = = 0 0 R T R R S z S z ' cos E E n i 0 + 0 i i i = Note that R T 1 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 4
Reflection and refraction between two isotropic media -- continued z ( ( ) ,t r B each wave For : ( ) ( ) ) k r ic n ct = = 2 2 E r E , t e n c 0 n ( ) ,t r ( ) ,t r k k k = = E E x c ki kR i R Matching condition interface at : = 2 2 2 ' cos ' sin n n n i ' n 1 If , ' n for sin , n i i 0 n refracted longer no field propagates medium in ' ' Total internal reflection: 2 sin i = = 2 2 2 ' cos ( E sin ' 1 n i n i n i n 2 sin i 0 2 ) sin i n 1 z ( ) ( ) c 2 k sin i r i n ct = E r ' , ' t e e 0 || c 0 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 5
For s-polarization cos ' cos n i n ' 2 cos E E n i ' = = 0 0 R E E + + cos ' cos cos ' cos n i n n i n 0 0 i i ' ' = 2 2 2 Note that : ' cos ' sin n n n i For p-polarization ' cos cos n i n ' 2 cos E E n i ' = = 0 0 R E E + + ' cos cos ' cos cos n i n n i n 0 0 i i ' ' = 2 2 2 Note that : ' cos ' sin n n n i 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 6
Special case: normal incidence (i=0, =0) ' n n ' 2 E E n ' = = 0 0 R E E + ' + ' n n n n 0 0 i i ' ' Reflectanc transmitt e, ance : 2 ' n n 2 E ' = = 0 R R E + ' n n 0 i ' 2 2 ' ' 2 ' E n n n = = 0 T ' ' E n n + ' n n 0 i PHY 712 Spring 2013 -- Lecture 19 ' PHY 712 Spring 2014 -- Lecture 18 02/28/2014 7 7
Extension to complex refractive index n= nR + i nI = = = + Suppose , ' real, ' ' ' n n n in R I Reflectanc normal at e incidence : 2 ' n n ( ( ) ) ( ( ) ) 2 2 2 + ' ' E n n n ' = = = 0 R R R I 2 2 + + E ' ' n n n + ' n n 0 i R I ' Note that for ' ' : n n n I R R 1 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 8
Fields near the surface on an ideal conductor isotropic an for Suppose = E = D and E J E medium : b : H Maxwell' equations s in terms of = = E H 0 0 H E = = + E H E b t t 2 Plane : = = 2 F F E H 0 , b 2 t t E wave form for ( ) ( ) ( ) k = = + k r i i t E r E k , whe re t e n in 0 R I c ( ) ( ) ( ) k k r = r / in c i t / E r E , t e e R 0 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 9
Fields near the surface on an ideal conductor -- continued : system our For / 1 2 2 2 c = + + 1 1 b n R b / 1 2 2 2 c = + 1 1 b n I b c c e 2 ) i + k E r 1 ( For 1 n n R I ( ) k r k r i t = / / i E r E , t e 0 1 n ( ) ( ) ( ) k E r = = H r , , , t t t c 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 10
Fields near the surface on an ideal conductor -- continued 1 For 1 n n R I 2 c c 1 c ( ) i = = + = + In this limit, 1 c n in R I 0 0 ( ) ( ) k r k r i t = / / i E r E , t e e 0 i ( ) ( ) ( ) = = D r E r E r , , , t t t r|| + 1 n i ( ) ( ) ( ) k E r k E r = = H r , , , t t t c + 1 n c i ( ) ( ) ( ) ( ) k E r k E r = = = B r H r 0 z , , , , t t t t 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 11
Fields near the surface on an ideal conductor -- continued ( i t t + = = H r k E r ) ( ) k r k r i t = / / i E r E , t e e 0 ( ) ( ) ( ) = = D r E r E r , , , t r|| 1 n i ( ) ( ) ( ) k E r , , , t t t c + 1 E n c i 0 z ( ) ( ) ( ) ( ) k E r k E r = = = B r H r , , , , t t t t Note that the field is larger than field so we can write: H ( i ) ( ) k r k r i t = / / i H r H , t e e 0 1 ( ) ( ) k H r = E r , , t t 2 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 12
Boundary values for ideal conductor ( ) , t e = E r ( ) k r k r i t / / i E e 0 k + 1 n i E0 ( ) ( ) ( ) k E r k E r = = H r , , , t t t c At the boundary of an ideal conductor, the E and H fields decay in the direction normal to the interface, the field directions are in the plane of the interface. Waveguide terminology TEM: transverse electric and magnetic (both E and H fields are perpendicular to wave propagation direction) TM: transverse magnetic (H field is perpendicular to wave propagation direction) TE: transverse electric (E field is perpendicular to wave propagation direction) 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 13
TEM waves Transverse electric and magnetic (both E and H fields are perpendicular to wave propagation direction) In the free space or within non a - conducting medium; the " normal" electromag netic modes areTEM : ( ) ( ) ( ( ) r ,t ) k r ic n ct = n = 2 2 E r E , t e n c 0 ( ) r ,t ( ) r ,t k k = = B E E c k k = = E B 0 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 14
Wave guides Coaxial cable TEM modes Simple optical pipe TE or TM modes 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 15
Comment on HW #11 1. Consider an infinitely long wire with radius a, oriented along the z axis. There is a steady uniform current inside the wire. Specifically the current is along the z-axis with the magnitude of J0for a and zero for > a, where denotes the radial parameter of the natural cylindrical coordinates of the system. a. Find the vector potential (A) for all . b. Find the magnetic flux field (B) for all . Solution to problem using PHY 114 ideas In this case, it is convenient to solve part b first. Top view for < a Top view for > a J0 B B 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 16
Comment on HW #11 -- continued Top view for < a Top view for > a B B J dA = B d J dA = B d 0 0 = 2 2 B J = 2 2 B J a 0 0 0 0 J 2 J a = 0 0 B = 0 0 B 2 J 2 J a = = B A 0 0 2 = = B A 0 0 2 2 ( ) 2 2 J a ( ) 0 0 = 2 A z ln 2 / J a a = 0 0 A z 4 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 17
Comment on HW #11 -- continued Alternative treatment using differential equ J = ations: z for for a a 0 0 2 A 0 ( ) for for J a a A 1 0 0 0 = z 2 J + 0 0 for C a ( ) = 1 A 4 C z + ( ) ln for continuity C a 2 3 Choosing constants from = = req uirements: 2 2 J J a + 0 0 0 0 for a J0 4 J a 4 ( ) A z 2 ( / ) ln 0 0 for a a 2 ( ) A B z 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 18
Comment on HW #12 A sphere of radius a carries a uniform surface charge distribution The sphere is rotated about a diameter with constant angular velocity . Find the vector potential A and magnetic field B both inside and outside the sphere. r = J r ( ) r r = 3 A r 0 ( ) . d r 4 | | ( ) for otherwise + r a r a r ( ) J r 0 l 1 4 l r = * r ( ) r Note that: ( ) Y Y lm lm + 1 l r r | | 2 1 r lm r r = * r ( ) r r r and: ( ) . d Y Y 1 lm lm l m 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 19
Comment on HW #12 -- continued a J r ( ) r r 4 r r r = = ' ( 3 3 A r 0 0 ( ) ' ) d r r dr r a 2 4 | | 4 3 r 0 for a r a = r A r 0 3 ( ) 4 a r for r a 3 2 for a r a = ) ( B r 0 3 ( ) ( ) 4 a r r r 3 for r a 3 02/28/2014 PHY 712 Spring 2014 -- Lecture 18 20
