tsl208 - B cos „ qBt m -1 – v y t = E B sin „ qBt...

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Charged Particle in Crossed Electric and Magnetic Fields (1) Release particle from rest. Force: ~ F = q ( ~ E + ~v × ~ B ) (1) F x = m dv x dt = - qv y B dv x dt = - qB m v y (2) F y = m dv y dt = qv x B + qE dv y dt = qB m v x + qE m Ansatz: v x ( t ) = w x cos( ω 0 t ) + u x , v y ( t ) = w y sin( ω 0 t ) + u y Substitute ansatz into (1) and (2) to find w x , w y , u x , u y , ω 0 . (1) - ω 0 w x sin( ω 0 t ) = - qB m w y sin( ω 0 t ) - qB m u y (2) ω 0 w y cos( ω 0 t ) = qB m w x cos( ω 0 t ) + qB m u x + qE m u y = 0 , u x = - E B , ω 0 = qB m , w x = w y w Initial condition: v x (0) = v y (0) = 0 w = E B q m E y x B tsl208 – p.1/2
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Charged Particle in Crossed Electric and Magnetic Fields (2) Solution for velocity of particle: v x ( t ) = E
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Unformatted text preview: B » cos „ qBt m «-1 – , v y ( t ) = E B sin „ qBt m « • Solution for position of particle: x ( t ) = E B Z t » cos „ qBt m «-1 – dt = Em qB 2 sin „ qBt m «-Et B y ( t ) = E B Z t sin „ qBt m « dt = Em qB 2 » 1-cos „ qBt m «– • Path of particle in ( x, y )-plane: cycloid qB 2 q m E y x B 2mE mE π 2 qB 2 t = 2 π m qB tsl209 – p.2/2...
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This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

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tsl208 - B cos „ qBt m -1 – v y t = E B sin „ qBt...

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