1 1 W13D2: Displacement Current, Maxwell’s Equations, Wave Equations Today’s Reading Course Notes: Sections 13.1-13.4 Announcements Math Review Tuesday May 6 from 9 pm-11 pm in 26-152 Pset 10 due May 6 at 9 pm 2 3 Outline Displacement Current Poynting Vector and Energy Flow Maxwell’s Equations2 4 Maxwell’s Equations E ⋅ˆn daS∫∫=1ε0ρdVV∫∫∫(Gauss's Law)B ⋅ˆn daS∫∫= 0 (Magnetic Gauss's Law)E ⋅ dsC∫= −ddtB ⋅ˆn daS∫∫(Faraday's Law)B ⋅ dsC∫=µ0J ⋅ˆn daS∫∫(Ampere's Law quasi - static)Is there something missing? 5 Maxwell’s Equations One Last Modification: Displacement Current 6 Ampere’s Law: Capacitor Consider a charging capacitor: Use Ampere’s Law to calculate the magnetic field just above the top plate What’s Going On? B ⋅ ds∫=µ0Ienc Ienc=J ⋅ˆn daS∫∫1) Surface S1: Ienc= I 2) Surface S2: Ienc = 03 7 Displacement Current dQdt=ε0dΦEdt≡ IdisWe don’t have current between the capacitor plates but we do have a changing E field. Can we “make” a current out of that? E =Qε0A⇒ Q =ε0EA =ε0ΦEThis is called the “displacement current”. It is not a flow of charge but proportional to changing electric flux 8 Displacement Current: Idis=ε0ddtE ⋅ˆn daS∫∫=ε0dΦEdtIf surface S2 encloses all of the electric flux due to the charged plate then Idis = I !9 Maxwell-Ampere’s Law B ⋅ dsC∫=µ0J ⋅ˆn daS∫∫+µ0ε0ddtE ⋅ˆn daS∫∫=µ0(Ienc+ Idis) Ienc=J ⋅ˆn daS∫∫ Idis=ε0ddtE ⋅ˆn daS∫∫“flow of electric charge” “changing electric flux”4 10 Concept Question: Capacitor If instead of integrating the magnetic field around the pictured Amperian circular loop of radius r we were to integrate around an Amperian loop of the same radius R as the plates (b) then the integral of the magnetic field around the closed path would be 1. the same. 2. larger. 3. smaller. 11 Sign Conventions: Right Hand Rule Integration direction clockwise for line integral requires that unit normal points into page for surface integral. Current positive into the page. Negative out of page. Electric flux positive into page, negative out of page. B ⋅ dsC∫=µ0J ⋅ˆn daS∫∫+µ0ε0ddtE ⋅ˆn daS∫∫12 Sign Conventions: Right Hand Rule Integration direction counter clockwise for line integral requires that unit normal points out page for surface integral. Current positive out of page. Negative into page. Electric flux positive out of page, negative into page. B ⋅ dsC∫=µ0J ⋅ˆn daS∫∫+µ0ε0ddtE ⋅ˆn daS∫∫5 13 Concept Question: Capacitor Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is 1. Zero (No current through loop) 2. Positive 3. Negative 4. Can’t tell (need to know direction of E) 14 Concept Question: Capacitor The figures above shows a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 15 Group Problem: Capacitor A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. At time t = 0 , the plates are uncharged 1. Find the electric field E(t) at P vs. time t (mag. & dir.) 2. Find the magnetic field B(t) at P6 16 Energy Flow 17 Poynting Vector S =E ×Bµ0 Power per unit area: Poynting vector P =S ⋅ˆn daopensurface∫∫Power through a surface 18 Energy Flow: Capacitor S =E ×Bµ0What is the magnetic field?7 19 Concept Question: Capacitor The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the energy stored in the electric field is: 1. Increasing in time. 2. Constant in time. 3. Decreasing in time. 20 Energy Flow: Capacitor S = (E ×B) /µ0= (Eˆk × Bˆθ) /µ0= −(EB /µ0)ˆr! P =S ⋅ˆnoutdacylindricalshell∫∫= −( EB /µ0)ˆr ⋅ˆr dacylindricalbody∫∫= −(EB /µ0)2πRh B2πR =µ0ε0dEdtπR2⇒ B =µ0ε02dEdtR P = −EBµ0⎛⎝⎜⎞⎠⎟2πRh= −Eµ0µ0ε02dEdtR⎛⎝⎜⎞⎠⎟2πRh= −ε0EdEdtπR2h= −ddt12ε0E2⎛⎝⎜⎞⎠⎟πR2h= −duEdt⎛⎝⎜⎞⎠⎟(Volume)21 Maxwell’s Equations E ⋅⋅ˆn daS∫∫=1ε0ρdVV∫∫∫(Gauss's Law)B ⋅⋅ˆn daS∫∫= 0 (Magnetic Gauss's Law)E ⋅ dsC∫= −ddtB ⋅ˆn daS∫∫(Faraday's Law)B ⋅ dsC∫=µ0J ⋅ˆn daS∫∫+µ0ε0ddtE ⋅ˆn daS∫∫(Maxwell - Ampere's Law)8 22 Electromagnetism Review E fields are associated with: (1) electric charges (Gauss’s Law ) (2) time changing B fields (Faraday’s Law) B fields are associated with (3a) moving electric charges (Ampere-Maxwell Law) (3b) time changing E fields (Maxwell’s Addition (Ampere-Maxwell Law) Conservation of magnetic flux (4) No magnetic charge (Gauss’s Law for Magnetism) 23 Electromagnetism Review Conservation of charge: E and B fields exert forces on (moving) electric charges: Energy stored in electric and magnetic fields J ⋅ dAclosedsurface∫∫= −ddtρdVvolumeenclosed∫∫∫ Fq= q(E +v ×B) UE= uEdVall space∫∫∫=ε02E2dVall space∫∫∫ UB= uBdVall space∫∫∫=12µ0B2dVall space∫∫∫24 Maxwell’s Equations in Vacua9 25 Maxwell’s Equations 1.E ⋅ dAS∫∫=Qinε0(Gauss's Law)2.B ⋅ dAS∫∫= 0 (Magnetic Gauss's Law)3.E ⋅ dsC∫= −dΦBdt(Faraday's Law)4.B ⋅ dsC∫=µ0Ienc+µ0ε0dΦEdt(Ampere - Maxwell Law)0 0 What about free space (no charge or current)? 26 Maxwell: First Colour Photograph 27 James Clerk Maxwell is sometimes credited as being the father of additive color. He had the photographer Thomas Sutton photograph a tartan ribbon on black-and-white film three times, first with a red, then green, then blue color filter over the lens. The three black-and-white images were developed and then projected onto a screen with three different projectors, each equipped with the corresponding red, green, or blue color filter used to take its image. When brought into
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