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Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eo^i and B=B0^i with a velocity v=v0^j. The speed of the particle will become 2v0 after a time

A
t=2mv0qE
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B
t=2Bqmv0
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C
t=3Bqmv0
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D
t=3mv0qE
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Solution

The correct option is C t=3mv0qE
E=E0^iB=B0^i

v=v0^jE

vB

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=v2x+v2y+v2z=(axt)2+v20 where v2y+v2z=v20

Given v=2v0

(2v0)2=v2x+v20

(vx)2=3(v0)2

vx=3v0

axt=3v0

(qEm)t=3v0

t=3v0mqE

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