Lecture_20_Part_2

Lecture_20_Part_2 - Reference Frame 0 qv r ^ B= =0 2 4 r...

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charged tape Any magnetic field? 0 ˆ 4 2 0 = × = r r v q B π μ 0 ˆ 4 2 0 × = r r v q B Reference Frame
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Two protons +e +e r 1 2 v v Electric force: E 1 r r e E q F e ˆ 4 1 2 2 0 1 2 , 21 πε = = F 21,e Magnetic field: 2 1 1 0 1 ˆ 4 r r v q B × = π μ B 1 Magnetic force: 1 2 2 , 21 B v q F m × = F 21,m 2 2 2 0 1 2 , 21 4 r v e vB q F m = = Magnetic Forces in Moving Reference Frames
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Electric force: 2 2 0 , 21 4 1 r e F e πε = Magnetic force: +e +e r 1 2 v v E 1 F 21,e B 1 F 21,m 2 2 2 0 , 21 4 r v e F m π μ = Ratio: = 2 2 0 2 2 2 0 , 21 , 21 4 1 / 4 r e r v e F F e m F 21, m F 21, e = μ 0 ε 0 ( ) v 2 1 0 0 = 3 × 10 8 ( ) 2 (m/s) 2 =c 2 it is not accidental! 2 2 , 21 , 21 c v F F e m = Magnetic Forces in Moving Reference Frames
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+e +e r 1 2 v v E 1 F 21,e B 1 F 21,m 2 2 , 21 , 21 c v F F e m = For v<<c the magnetic force is much smaller than electric force How can we detect the magnetic force on
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Lecture_20_Part_2 - Reference Frame 0 qv r ^ B= =0 2 4 r...

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