EAD234A_Lecture_5_revised

EAD234A_Lecture_5_revised - Amperes Law d l 2(d l 1 aR12 0...

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Ampere’s Law ˆ ( ) ∫ ∫ × × = 12 2 12 1 2 2 1 0 12 4 CC R R a l d l d I I F π μ 2 1 ˆ Magnetic Flux Density ( ) × × = 12 2 12 1 1 0 2 2 12 4 R R a l d I l d I F × = 2 1 12 2 2 B l d I 2 C I B l Id F d × =
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Biot and Savart Law ( ) FJ B d r ±± ± ± 0 Jr d r μ × ± ± 3 4 Bd r π = × R l d I 0 C R r B 3 4 ) (
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Linear Pinch: Early Magnetic Fusion Concept
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Torque Tm B = × GG G
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Ampere’s Law J B 0 μ = × encl C I l d B 0
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Vector Magnetic Potential B is solenoidal A B B × = = 0 ( ) r J μ 0 ( ) v d R r B V × = π 4 ( ) r A
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Straight Wire = L R l d I r A π μ 4 ) ( 0 Idl A 0 4 dA r = Logarithmic divergence for infinite wire Start with finite length and calculate A and B and take the limit
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Straight Wire (cont.) /2 1 22 2 00 ln L L ll II dl A ρ μμ ′′ ⎡⎤ ++ ⎣⎦ == ⎨⎬ 1 2 0 2 44 L l π πρ + ⎩⎭ ( ) ( ) 1/2 0 /2 1 1 4 / 2 LL I μ + + ⎢⎥ = 0 4 2 I L for L ±
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Two Parallel Long Wires A is zero in the vertical plane that passes through e dashed line and points the dashed line and points upward on the left and downward on the right
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Two Parallel Long Wires (cont.) Use previous results: 00 ln b II LL μ μρ ⎡⎤ = () 2 2 0 2 22 ln ab a A xD y I π ρρ ρ == ⎢⎥ ⎣⎦ +− = 4 x y + 0 2 I A Dy y B ∂− + 0 2 2 11 x ba y I A B x −∂ = =− 2 0 y z y B = Midway 0 2 ,0 , 0 xy z I BB B D = =
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Two Parallel Currents
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Two Parallel Currents
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Two Parallel Currents (cont) Force on a length L of conductor 2: 0 0 2 L II FI B d L I B L L R I μ π ′′ == = 2 7 21 0 II R =⇒ LR Take I= 1 MA and R=1 cm 2x10 7 N/m !!
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Current Sheet Use the previous result () 1 2 2 00 ln / x z z K A xx y r d x w h e r e K J μ ⎡⎤ ′′ =− + = Δ 2 2 x π ⎣⎦
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Current Sheet (cont) 1 2 0 / x K d μ + ( ) 2 1 2 00 0 2 ln 2 z x xx A y r dx Let u x x π ⎡⎤ ′′ =− ⎢⎥ ⎣⎦ () 2 22 1 1 12 11 2 2 ln 2 ln ln tan tan z K A u y du K y xx yy y y −− =+ ⎛⎞ =−− + + + + ⎜⎟ 2 yy ⎝⎠
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Current Loop
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Magnetic Flux = Ψ s d B B S S C Self Inductance 2 0 NA L μ = total d L I Ψ = l dI 22 0 2 Nr L R =
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