1 1 Displacement Current, Maxwell’s Equations, Wave Equations 2 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? 3 Maxwell’s Equations One Last Modification: Displacement Current2 4 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 = 0 5 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 6 Displacement Current: Idis=ε0ddtE ⋅ˆn daS∫∫=ε0dΦEdtIf surface S2 encloses all of the electric flux due to the charged plate then Idis = I !3 7 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” 8 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. 9 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∫∫4 10 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∫∫11 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) 12 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 charge Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time.5 13 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 P 14 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)15 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)6 16 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∫∫∫17 Maxwell’s Equations in Vacua 18 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)?7 19 How Do Maxwell’s Equations Lead to EM Waves? 20 Wave Equation Start with Ampere-Maxwell Eq and closed oriented loop B ⋅ dsC∫=µ0ε0ddtE ⋅ˆn da∫21 Wave Equation B ⋅ dsC∫= Bz(x,t)l − Bz(x + Δx,t)l −Bz(x + Δx,t) − Bz(x,t)Δx=µ0ε0∂Ey(x + Δx / 2,t)∂t −∂Bz∂x=µ0ε0∂Ey∂tSo in the limit that dx is very small: Apply it to red rectangle: B ⋅ dsC∫=µ0ε0ddtE ⋅ˆn da∫Start with Ampere-Maxwell Eq: µ0ε0ddtE ⋅ˆn da∫=µ0ε0l Δx∂Ey(x + Δx / 2,t)∂t⎛⎝⎜⎞⎠⎟8 22 Group Problem: Wave Equation Use Faraday’s Law and apply it to red rectangle to find the partial differential equation in order to find a relationship between ∂Ey/ ∂x and ∂Bz/ ∂t E ⋅ dsC∫= −ddtB ⋅ˆn da∫23 Group Problem: Wave Equation Sol. E ⋅ dsC∫= −ddtB ⋅ˆn da∫ E ⋅ dsC∫= Ey(x + Δx,t)l − Ey(x,t)l Ey(x + dx,t) − Ey(x,t)dx= −∂Bz∂t ∂Ey∂x= −∂Bz∂t −ddtB ⋅ˆn da∫= −ldx∂Bz∂tUse Faraday’s Law: So in the limit that dx is very small: and apply it to red rectangle: 24 1D Wave Equation for Electric Field ∂Ey∂x= −∂Bz∂t (1) −∂Bz∂x=µ0ε0∂Ey∂t(2)Take x-derivative of Eq.(1) and use the Eq. (2) ∂2Ey∂x2=∂∂x∂Ey∂x⎛⎝⎜⎞⎠⎟=∂∂x−∂Bz∂t⎛⎝⎜⎞⎠⎟= −∂∂t∂Bz∂x⎛⎝⎜⎞⎠⎟=µ0ε0∂2Ey∂t2 ∂2Ey∂x2=µ0ε0∂2Ey∂t29 25 1D Wave Equation for E ∂2Ey∂x2=µ0ε0∂2Ey∂t2This is an equation for a wave. Let Ey= f (x − vt) ∂2Ey∂x2= f '' x − vt( )∂2Ey∂t2= v2f '' x − vt( )⎫⎬⎪⎪⎭⎪⎪⇒ v =1µ0ε026
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