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PHY481: Electromagnetism Constant currents and magnetic fields Lecture 24 Carl Bromberg - Prof. of Physics Current, and current density I is charge flow/unit time vdrift = drift velocity (~1mm/s) of charge carriers I = qnL vdrift (1 D) q = charge of charge carriers (+ or -) nL = no. of charge carriers per unit length Ohm’s law: I = V R nV = no. of charge carriers per unit volume vth = thermal velocity vth >> vdrift λ = mean free path Drude model: Resistance/L Conductivity σR mvth R = L q2n λ L nV q λ L σR = = RA mvth 2 I = qnL vdrift q nL λ = V mL vth Charge density ρc (x) Current densities: Current in = current out ∂ρc ∂t = 0 ρR = 1 σ R No charge buildup Linear current I = qnL v(x) (1 D) Surface current density K(x) = qnS v(x) (2 D) Current through line dI = K ⋅ ê ⊥ d (1 D) Lecture 24 2 Resistivity ρR Volume current density J(x) = qnV v(x) (3 D) Current through area dI = J(x) ⋅ dA (1 D) Carl Bromberg - Prof. of Physics J(x) dA dA 1 Conductors, “free” charge, continuity equation, Macroscopic Resistance Conductors Microscopic - Local forms of Ohm’s law Resistivity ρR Conductivity σR σ R = 1 ρR J(x) = σ R E(x) J(x) = E(x) ρ R I =V R ρ and σ are an intrinsic property of a material “Wire” with area A, and resistivity ρR L 1 R = ∫ dR = ∫ ρ R ( z ′ )dz ′ A0 “Free” charge Uniform resistivity Surface free charge resistivity d ρ R dz = 0 none in volume: resistor constant Iz Cu ++ ρ R ≈ 0 +++ –– ––– With constant ρR L R= resistivity ρR A Changing resistivity Surface & volume free charge: Cu ρR ≈ 0 ∫ ρC d x = 0 no net charge 3 --> ρ free = ε I ⎛ d ρR ⎞ ⎜ ⎟ A ⎝ dz ⎠ Continuity equation ∇⋅J = − Lecture 24 ∂ρc ∂t Integrate over volume V, and use Gauss’s theorem Rate of change of Flux of J through d 3 J ⋅ dA = − ∫ ρc d x charge in d3x ∫ 3 S surface S (of d x) dt V -> Charge conservation Carl Bromberg - Prof. of Physics 2 Resistors Each resistor has the value R. What is the total resistance, RT, of this infinite set of resistors? One resistor: RT = R Four resistors: RT = R+ (3R in parallel) N resistors: RT = not obvious The trick! Lecture 24 Carl Bromberg - Prof. of Physics 3 Magnetic force and field Magnetic force: F = qv × B Circular motion: v0 ⊥ B v0 ⊥ B Crossed E and B: E ⊥ B Perpendicular to direction of motion Magnetic forces do no work ΔKE = 0 Centripetal acceleration caused by magnetic force. ω = v R = qB m Helical motion. Helix has radius and pitch Exists an un-deflected velocity v = E × B / B Currents make magnetic fields: Biot-Savart Law µ0 Id × r̂ µ0 Id × ( x − x ′ ) dB = 4π r z Id k̂ R r Id x B(x) = 2 dB = Id φ̂ r̂ R̂ y 4π ∫wire x − x′ 3 Wire carrying current Current element Lecture 24 Carl Bromberg - Prof. of Physics 4 2 Magnetic field of long wire carrying current µ0 B(x) = 4π ∫wire Id × ( x − x ′ ) x − x′ (the hard way) x = R R̂ + z k̂ 3 Magnetic field does not depend on z. Any z will do. Pick z = 0 to simplify dz ′ R + z′ x ′ = z ′ k̂ k̂ × r̂ = k̂ × R̂ cosψ = φ̂ cosψ cosψ = R 2 d = dz ′ k̂ z Id = Idz ′ k̂ r̂ = R̂ cosψ − k̂ sin ψ µ0 I ∞ dz ′ k̂ × r̂ B(x) = ∫ ( R2 + z′2 ) 4π −∞ µ0 IR ∞ B(x) = ∫( 4π −∞ x ′ = z ′ k̂ ) 2 32 φ̂ ∞ ( R2 + z ′ 2 )1 2 dz ′ x r = ( R + z′ ψ y x=R r̂ 2 2 ∞ z′ 2 ⎡ ⎤ = = ∫ ( 2 2 )3 2 ⎢ 2 2 2 ⎥ R2 −∞ R + z ′ ⎣ R R + z ′ ⎦ −∞ µ0 I B(R) = φ̂ 2π R Lecture 24 Carl Bromberg - Prof. of Physics 5 2 ) Ampere’s Law ∇⋅B = 0 Magnetic fields are “solenoidal” (always true) ∇ × B = µ0 J Ampere’s Law (constant currents only. Maxwell’s equations have another term) Integrate over a surface element: ∫ (∇ × B ) ⋅ n dA = µ0 ∫ J ⋅ dA S Apply Stokes’s theorem: S ∫ B ⋅d = µ0 ∫ J ⋅ dA C S dI = J ⋅ dA ∫ B ⋅d = µ0 Iencl C Ampere’s Law and symmetries Straight wire (the easy way) ∫ B ⋅d = µ0 Iencl C Right hand rule #2 Lecture 24 Bφ 2π R = µ0 I B(R) = µ0 I φ̂ 2π R Carl Bromberg - Prof. of Physics 6 More Ampere’s law applications Infinite current sheet Infinite current slab Lecture 24 Carl Bromberg - Prof. of Physics 7 General field equations for B(x) Constant vs. changing currents Always true ∇⋅B = 0 ∇ ⋅ ( ∇ × B ) = 0 = µ0∇ ⋅ J Constant currents only. ∇⋅J = 0 Maxwell’s equations have another term: proportional to Rate of change of electric field ∇ × B = µ0 ( J + ε 0 dE dt ) ∇ ⋅ ( ∇ × B ) = µ0 ( ∇ ⋅ J + ε 0 d ( ∇ ⋅ E ) dt ) = 0 ∇ ⋅ J = − d ρc dt ∇ ⋅ E = ρc ε 0 Continuity equation General B field equation µ0 B(x) = 4π ∫wire ⎡µ B(x) = ∇ × ⎢ 0 ⎣ 4π Lecture 24 Id × ( x − x ′ ) x − x′ 3 3 J( x ′ )d x ′ ⎤ ∫ x − x′ ⎥⎦ ( x − x′ ) x − x′ 3 = −∇ 1 x − x′ B(x) = ∇ × A(x) 1 ⎞ ⎛ ⎛ J( x ′ ) ⎞ ∇ × J( x ) = ∇ × ′ ⎜⎝ ⎟ ⎜⎝ ⎟ x − x′ ⎠ x − x′ ⎠ 3 µ J( x ′ )d x ′ A(x) = 0 ∫ 4π x − x′ Carl Bromberg - Prof. of Physics Vector potential 8