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# tsl208 - B Â cos â€ž qBt m Â-1 â€“ v y t = E B sin â€ž 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 z 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 (
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Unformatted text preview: B Â» cos â€ž qBt m Â«-1 â€“ , v y ( t ) = E B sin â€ž qBt m Â« â€¢ Solution for position of particle: x ( t ) = E B Z t Â» cos â€ž qBt m Â«-1 â€“ dt = Em qB 2 sin â€ž qBt m Â«-Et B y ( t ) = E B Z t sin â€ž qBt m Â« dt = Em qB 2 Â» 1-cos â€ž qBt m Â«â€“ â€¢ Path of particle in ( x, y )-plane: cycloid qB 2 q m E y x B 2mE mE Ï€ 2 qB 2 t = 2 Ï€ m qB tsl209 â€“ p.2/2...
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