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Motion of a charged particle

Question

A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity ν . A uniform electric field E  and a uniform magnetic field B exist everywhere. The velocity ν , electric field E  and magnetic field B  are given in columns 1,2 and 3, respectively. The quantities  E0,B0  are positive in magnitude. 
Column 1Column 2Column 3
(I) Electron with ν=2E0B0x^(i) E=E0z^(P) B=B0x^
(II) Electron with ν=E0B0y^(ii) E=E0y^(Q) B=B0x^
(III) Proton with ν=0(iii) E=E0x^(R) B=B0y^
(IV) Proton with ν=2E0B0x^(iv) E=E0x^(S) B=B0z^

In which case would the particle move in a straight line along the negative direction of y-axis  (i.e., move along -y^)? 
 

Moderate
Solution

Proton released from rest at origin 

Due to electric field in negative y direction, proton accelerates and acquire a velocity along negative y direction

Felectric=qEay=qE0j^mVy=qE0tj^m

F magnetic =qVy×By 

Vy is antiparallel to ByF magnetic =0

Thus, due to of electrostatic force only, the proton moves along negative y-direction.



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