A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eoi^ and B=B0i^ with a velocity v=v0j^. The speed of the particle will become 2v0 after a time
A
t=qE2mv0
B
t=mv02Bq
C
t=mv03Bq
D
t=qE3mv0
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Updated on : 2022-09-05
Solution
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Correct option is D)
E=E0i^∥B=B0i^
v=v0j^⊥E
v⊥B
Hence, the path of the particle is a helix with speed remaining constant for the circular path in the yz plane.
Magnitude of velocity of particle at any time t:v=vx2+vy2+vz2=(axt)2+v02 where vy2+vz2=v02
Given v=2v0
⇒(2v0)2=vx2+v02
⇒(vx)2=3(v0)2
⇒vx=3v0
⇒axt=3v0
⇒(mqE)t=3v0
⇒t=qE3v0m
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