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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 timet=2mv0qEt=√3Bqmv0t=√3mv0qEt=2Bqmv0

A
t=2mv0qE
B
t=2Bqmv0
C
t=3Bqmv0
D
t=3mv0qE
Solution
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#### →E=E0^i∥→B=B0^i→v=v0^j⊥→E →v⊥→BHence, 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=v20Given v=2v0⇒(2v0)2=v2x+v20⇒(vx)2=3(v0)2⇒vx=√3v0⇒axt=√3v0⇒(qEm)t=√3v0⇒t=√3v0mqE

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