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
B
t=2Bqmv0
C
t=√3Bqmv0
D
t=√3mv0qE
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Solution
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→E=E0^i∥→B=B0^i
→v=v0^j⊥→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=√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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