Lecture 10 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 r In electroquasistatics we had: ∇ × E = 0 r Therefore we could represent the E-field by the scalar potential: E = −∇φ r r In magnetoquasistatics we have: ∇ .(B) = ∇ .(µo H) = 0 Therefore we can represent the B-field by the vector potential: r r r B = µo H = ∇ × A A vector field can be specified (up to a constant) by specifying its curl and its divergence r Our definition of the vector potential A is not yet unique – we have only specified its curl r For simplicity we fix the divergence of the vector potential A to be zero: r ∇ . A = 0 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 1 Magnetic Flux and Vector Potential Line Integral - Review The magnetic flux λ through a surface is the surface integral of the B-field through the surface r r B-field λ = ∫∫ B.da r r = µo ∫∫ H .da Since: r r r B = µo H = ∇ × A We get: r r λ = ∫∫ B.da dsr r r = ∫∫ ()∇ × A .da r r Stoke’s Theorem A closed contour = ∫ A . ds 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 y current I in the +z-direction I zˆ The H-field has only a φ-component x Using Ampere’s Law: Hφ (2π r ) = I r I ⇒ H = φ 2π r Work in cylindrical 2 r r co-ordinates ∇ A = −µo J 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 r ∇ × A 1 ∂Az (r ) ∂Az (r ) µo I But Hφ = = − ⇒ = − µo µo ∂r ∂r 2π r Integrating from ro to r : µo I ⎛ ro ⎞ Az ()r − Az ()ro = ln⎜ ⎟ 2π ⎝ r ⎠ ECE 303 – Fall 2007 – Farhan Rana – Cornell University 2 Vector Potential of a Line-Current Dipole rr Consider two infinitely long equal and y opposite line-currents, as shown r+ I zˆ + r− − I zˆ The vector potential can be written as a x sum using superposition: d r µo I ⎛ ro ⎞ µo I ⎛ ro ⎞ Az ()r = ln⎜ ⎟ − ln⎜ ⎟ The final answer does not 2π ⎝ r+ ⎠ 2π ⎝ r− ⎠ depend on the parameter ro µo I ⎛ r− ⎞ = ln⎜ ⎟ 2π ⎝ r+ ⎠ 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 z- ρ component relevant for this problem) ∇2 φ = − ∇2A = −µ J εo z o z Potential of a line-charge dipole: Vector potential of a line-current dipole: r λ ⎛ r− ⎞ r µo I ⎛ r− ⎞ φ()r = ln⎜ ⎟ Az ()r = ln⎜ ⎟ 2π εo ⎝ r+ ⎠ 2π ⎝ 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 µoH1 µoH2 r r Maxwell equation: ∇ .B = ∇ . µo 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: µoH2 = µoH1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Magnetic Field Boundary Conditions - II K 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 H1 H2 H2 − H1 = K This follows from Ampere’s law: r r r r r r ∇ × H = J or ∫ 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 H H L H2 L − H1 L = K L 1 2 ⇒ H2 − H1 = K 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 r r r ∂ µ H()rr,t ∇. µ H()rr,t = 0 ∇ × H(r ,t ) = J(rr,t ) ∇ × E()rr,t = − o o ∂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? Remember there cannot be any time varying currents inside a perfect metal…… y Consider an infinitely long metal wire of radius a K carrying a (time varying) current I in the +z-direction The current flows entirely on the surface of the perfect a x metal wire in the form of a uniform surface current density K( t ) I(t ) K()t = 2π a Can use Ampere’s law to calculate the H-field y outside a perfect metal wire carrying a (time K varying) current I r Hφ ()()()2π r Hφ t = 2π a K(t ) a a I()t x ⇒ H ()t = K ()t = φ r 2π r ECE 303 – Fall 2007 – Farhan Rana – Cornell University 6 Parallel Plate Conductors y K zˆ Consider two infinitely long (in the W z-direction) metal plates, of width W and separated by a distance d, as shown d Hx The top plate carries a (time varying) current I in the +z-direction x and the bottom plate carries a (time varying) current I in the –z- direction − K zˆ I Surface current density on the top plate = K = W If W >> d, then the H-field inside the plates in very uniform and can be calculated by using the boundary condition: 0 One can also use Amperes law Hx outside metal − Hx inside metal = K directly - see if you can identify an appropriate contour for using I Ampere’s law to get the same ⇒ Hx outside metal = K = W 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 +z- direction 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 H at time t=0 Time = t = 0 d σ ≠ ∞ H Time = t = ∞ If you wait “long enough” magnetic field will penetrate real metals! d σ ≠ ∞ 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 r r ∇ × H = J J = σ E r r ∂ µ H ∇ × E = − o ∂t r r r r ∂H ∇ × ∇ × H = ∇ × J = σ ∇ × E = −σ µ o ∂ t r r r ∂H ⇒ ∇()∇ .H − ∇2H = −σ µ o ∂ t r 2 r ∂H ⇒ ∇ H = σ µ Magnetic diffusion equation o ∂ t In time “t ” the magnetic field will diffuse a distance “d ” into the metal, where: σ µ d 2 t ≈ o 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