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- t=2mv0qE
- t=√3Bqmv0
- t=√3mv0qE
- t=2Bqmv0

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Solution

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→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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