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### lecture10

Course: ECE 3030, Fall 2007
School: Cornell
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10 Lecture Perfect Metals in Magnetism and Inductance In this lecture you will learn: Some more about the vector potential Magnetic field boundary conditions The behavior of perfect metals towards time-varying magnetic fields Image currents and magnetic diffusion Inductance ECE 303 Fall 2007 Farhan Rana Cornell University The Vector Potential - Review In electroquasistatics we had: r E = 0 Therefore...

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10 Lecture Perfect Metals in Magnetism and Inductance In this lecture you will learn: Some more about the vector potential Magnetic field boundary conditions The behavior of perfect metals towards time-varying magnetic fields Image currents and magnetic diffusion Inductance ECE 303 Fall 2007 Farhan Rana Cornell University The Vector Potential - Review In electroquasistatics we had: r E = 0 Therefore we could represent the E-field by the scalar potential: In magnetoquasistatics we have: r E = - r r . B = . o H = 0 r r r B = o H = A ( ) ( ) Therefore we can represent the B-field by the vector potential: A vector field can be specified (up to a constant) by specifying its curl and its divergence Our definition of the vector potential A is not yet unique we have only specified its curl For simplicity we fix the divergence of the vector potential A to be zero: r r r .A = 0 ECE 303 Fall 2007 Farhan Rana Cornell University 1 Magnetic Flux and Vector Potential Line Integral - Review The magnetic flux through a surface is the surface integral of the B-field through the surface r r B-field = B . da r r = o H . da Since: r r r B = o H = A We get: = B . da r r r r = A . da r r = A . ds ( ) r ds Stoke's Theorem A closed contour The magnetic flux through a surface is equal to the line-integral of the vector potential along a closed contour bounding that surface ECE 303 Fall 2007 Farhan Rana Cornell University Vector Potential of a Line-Current Consider an infinitely long line-current with current I in the +z-direction The H-field has only a -component Using Ampere's Law: y ^ Iz r r 2 A = - o J H (2 r ) = I I H = 2 r r x Work in cylindrical co-ordinates If the current has only a z-component then the vector potential also only has a z-component which, by symmetry, is only a function of the distance from the line-current But H = r A o =- 1 Az (r ) o r Az (r ) I =- o r 2 r Integrating from ro to r : Az (r ) - Az (ro ) = o I ro ln 2 r ECE 303 Fall 2007 Farhan Rana Cornell University 2 Vector Potential of a Line-Current Dipole r Consider two infinitely long equal and opposite line-currents, as shown r y r- d x ^ -Iz ^ +Iz The vector potential can be written as a sum using superposition: r+ r I r I r Az (r ) = o ln o - o ln o 2 r+ 2 r- = o I r- ln 2 r+ The final answer does not depend on the parameter ro Question: where is the zero of the vector potential? Points for which r+ equals r- have zero potential. These points constitute the entire y-z plane ECE 303 Fall 2007 Farhan Rana Cornell University H-Field of a Line-Current Dipole y H x d Something to Ponder Upon Poisson equation: Vector Poisson equation (only the zcomponent relevant for this problem) =- o 2 2 Az = - o J z Vector potential of a line-current dipole: Potential of a line-charge dipole: (r ) = r 2 o r ln - r + r I r Az (r ) = o ln - 2 r+ Notice the Similarities ECE 303 Fall 2007 Farhan Rana Cornell University 3 Magnetic Field Boundary Conditions - I The normal component of the B-field at an interface is always continuous Maxwell equation: oH1 oH2 r r . B = . o H = 0 The net magnetic flux coming into a closed surface must equal the magnetic flux coming out of that closed surface (since there are no magnetic charges to generate or terminate magnetic field lines) Therefore: oH2 = oH1 ECE 303 Fall 2007 Farhan Rana Cornell University Magnetic Field Boundary Conditions - II The discontinuity of the parallel component of the H-field at an interface is related to the surface current density (units: Amps/m) at the interface K H1 H2 - H1 = K H2 This follows from Ampere's law: r r H = J or r r r r H . ds = J . da K The line integral of the magnetic field over a closed contour must equal the total current flowing through the contour H2 L - H1 L = K L H2 - H1 = K H1 H2 L ECE 303 Fall 2007 Farhan Rana Cornell University 4 Perfect Metals and Magnetic Fields - I A perfect metal can have no time varying H-fields inside it Note: Recall that in magnetoquasistatics one can have time varying H-fields (as long as the time variation is slow enough to satisfy the quasistatic conditions) The argument goes in two steps as follows: A time varying H-field implies an E-field (from the third equation of magnetoquasistatics) r r . o H (r , t ) = 0 r r r H (r , t ) = J (r , t ) r r r r o H (r , t ) E (r , t ) = - t Since a perfect metal cannot have any E-fields inside it (time varying or otherwise), a perfect metal cannot have any time varying H-fields inside it ECE 303 Fall 2007 Farhan Rana Cornell University Perfect Metals and Magnetic Fields - II At the surface of a perfect metal there can be no component of a time varying H-field that is normal to the surface r H (t ) perfect metal The argument goes as follows: The normal component of the H-field is continuous across an interface So if there is a normal component of a time varying H-field at the surface of a perfect metal there has to be a time varying H-field inside the perfect metal Since there cannot be any time varying H-fields inside a perfect metal, there cannot be any normal component of a time varying H-field at the surface of a perfect metal ECE 303 Fall 2007 Farhan Rana Cornell University 5 Perfect Metals and Magnetic Fields - III Time varying currents can only flow at the surface of a perfect metal but not inside it r K (t ) perfect metal The argument goes as follows: Time varying currents produce time varying H-fields So if there are time varying currents inside a perfect metal, there will be time varying H-fields inside a perfect metal Since there cannot be any time varying H-fields inside a perfect metal, there cannot be any time varying currents inside a perfect metal ECE 303 Fall 2007 Farhan Rana Cornell University Current Flow and Surface Current Density So how does a (time varying) current flow in perfect metal wires? there Remember cannot be any time varying currents inside a perfect metal...... y Consider an infinitely long metal wire of radius a carrying a (time varying) current I in the +z-direction The current flows entirely on the surface of the perfect metal wire in the form of a uniform surface current density K( t ) K a x K (t ) = I (t ) 2 a Can use Ampere's law to calculate the H-field outside a perfect metal wire carrying a (time varying) current I y r H K (2 r ) H (t ) = (2 a ) K (t ) a I (t ) H (t ) = K (t ) = r a x 2 r ECE 303 Fall 2007 Farhan Rana Cornell University 6 Parallel Plate Conductors Consider two infinitely long (in the z-direction) metal plates, of width W and separated by a distance d, as shown The top plate carries a (time varying) current I in the +z-direction and the bottom plate carries a (time varying) current I in the zdirection y W ^ Kz d Hx x ^ -Kz Surface current density on the top plate = K = I W If W >> d, then the H-field inside the plates in very uniform and can be calculated by using the boundary condition: H x outside metal - H x 0 inside metal =K H x outside metal = K = I W One can also use Amperes law directly - see if you can identify an appropriate contour for using Ampere's law to get the same answer ECE 303 Fall 2007 Farhan Rana Cornell University Image Currents - I Consider a perfect metal with a wire carrying a time varying current I(t ) in the +zdirection at a distance d above the perfect metal, as shown below H( t ) d perfect metal Surely this picture cannot be right............there is time varying H-field inside the perfect metal ECE 303 Fall 2007 Farhan Rana Cornell University 7 Image Currents - II So what does really happen ...... H d perfect metal Currents flow on the surface of the perfect metal that completely cancel the time varying H-field inside the perfect metal In other words, surface currents screen out the time varying H-field from the perfect metal Question: Is there a better way to understand what the resulting H-field looks like outside the perfect metal? ECE 303 Fall 2007 Farhan Rana Cornell University Image Currents - III The magnetic field outside the perfect metal can be obtained by imagining a fictitious current element that is a mirror image of the actual current element but carrying a current in the opposite direction H d perfect metal d image current ECE 303 Fall 2007 Farhan Rana Cornell University 8 Image Currents - IV Example: A current loop carrying a time varying current over a perfect metal I(t ) perfect metal d d image current ECE 303 Fall 2007 Farhan Rana Cornell University Not So Perfect Metals - I So what does really happen when a current is suddenly switched on at time t=0 Time = t = 0 d H H Time = t = If you wait "long enough" magnetic field will penetrate real metals! d ECE 303 Fall 2007 Farhan Rana Cornell University 9 Not So Perfect Metals Magnetic Diffusion Question: How long does it take for the magnetic field to penetrate into real metals? Answer: Start from these magnetoquasistatic equations: r r H = J r r oH E = - t r r J =E r r r r H H = J = E = - o t r r r H 2 . H - H = - o t r r H 2 H = o Magnetic diffusion equation t ( ) In time "t " the magnetic field will diffuse a distance "d " into the metal, where: t o d 2 2 ECE 303 Fall 2007 Farhan Rana Cornell University Inductance The magnetic flux enclosed by current carrying conductors is directly proportional to the current carried by the conductors H r r r r Magnetic Flux = = B . da = o H . da I The constant of proportionality is called the Inductance (units: Henry) I =LI or L= I L= d dI ECE 303 Fall 2007 Farhan Rana Cornell University More accurately: 10 Inductance of the Parallel Plate Conductors y From previous slide: W ^ Kz Hx = K = I W d Hx x K= I W ^ -Kz Since the structure is infinite in the z-direction, one can only talk about the inductance per unit length (units: Henry/m) (unit length means unit length in the z-direction) L = Inductance per unit length = L= Flux per unit length Total current o H x d I = o d W ECE 303 Fall 2007 Farhan Rana Cornell University Inductance of the Concentric Cylinders Consider two perfect metal concentric cylinders infinitely long in the z-direction. The inner one carries a total (time varying) current I in the +z-direction. The outer shell carries a total (time varying) current I in the -z-direction The magnetic field, by symmetry, has only a -component Using Ampere's law: (2 r ) H (r ) = I b a r H (r ) = I 2 r Since the structure is infinite in the z-direction, one can only talk about the inductance per unit length (units: Henry/m) L= I = o H (r ) dr a b I b = o ln 2 a ECE 303 Fall 2007 Farhan Rana Cornell University 11 Inductance of Two Metal Wires - I Consider two infinitely long metal wires carrying equal (time varying) currents but in opposite directions Need to solve: y d >> a d x ^ +Iz metal wires of radius a ^ -Iz As long as d >> a, the surface current density on each metal wire is circularly symmetric and uniform, and so the outside field appears as if one had line-currents at the center of each metal wire So the vector potential can be written as: Az (r ) = r o I r- ln 2 r+ Now remember the golden formula from a previous lecture for the magnetic flux: = B . da = A . ds ECE 303 Fall 2007 Farhan Rana Cornell University r r r r Inductance of Two Metal Wires - II y The line integral of the A-field along the indicated contour gives the net flux passing through the contour d >> a d = B . da = A . ds = r r r ^ +Iz r x H metal wires of radius a ^ -Iz o I r- I r ln - o ln - 2 r+ r- = d 2 r+ r- = a r+ = a r+ = d I d = o ln a L= ^ -Iz d Looking from the top........ x contour I = o d ln a z unit length ECE 303 Fall 2007 Farhan Rana Cornell University ^ +Iz 12
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