A particle of charge q and mass m starts moving from the origin under the action of an electric field E→=E0i^ and B→=B0i^ with a velocity v→=v0j^. The speed of the particle will become 2v0 after a time
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a
t=2mv0qE
b
t=2Bqmv0
c
t=3Bqmv0
d
t=3mv0qE
answer is D.
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Detailed Solution
E→ is parallel to B→ and v→ is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t isv=vx2+vy2+vz2 ……..(i)Here, vy2+vz2=v02and v=2v0⇒vx=3v0 vx=axt⇒3v0=qEmt⇒t=3mv0qE