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Question-53686




Question Number 53686 by ajfour last updated on 25/Jan/19
Commented by ajfour last updated on 25/Jan/19
Charged particle (q,m) released  at origin. Find velocity v^�  as a  function of the z coordinate.  Electric field, E_0 k^�  ,   Magnetic field, −B_0  j^�  are present.
Chargedparticle(q,m)releasedatorigin.Findvelocityv¯asafunctionofthezcoordinate.Electricfield,E0k^,Magneticfield,B0j^arepresent.
Commented by Tinkutara last updated on 25/Jan/19
v_z =(√(2Aωz−ω^2 z^2 ));  A=(E_0 /B_0 )  ω=((qB_0 )/m)
vz=2Aωzω2z2;A=E0B0ω=qB0m
Commented by ajfour last updated on 25/Jan/19
seems correct,v_x =?
seemscorrect,vx=?
Commented by Tinkutara last updated on 25/Jan/19
v_x (t)=(E_0 /B_0 )(1−cos ωt)
vx(t)=E0B0(1cosωt)
Commented by ajfour last updated on 25/Jan/19
not right process, sir; soon v_x   develops , so you must take this  into consideration, moreover,  v^�   is asked for as a function of z,  and not time t[ v^� (t) might be still  more difficult].
notrightprocess,sir;soonvxdevelops,soyoumusttakethisintoconsideration,moreover,v¯isaskedforasafunctionofz,andnottimet[v¯(t)mightbestillmoredifficult].

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