InterviewSolution
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InterviewSolution
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A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec E = E_0 hat i and vec B = B_0 hat i` with a velocity `vec v = v_0 hat j`. The speed of the particle will become `2v_0` after a time.A. `t=(2mv_(0))/(qE)`B. `t=(2Bq)/(mv_(0))`C. `t=(sqrt3Bq)/(mv_(0))`D. `t=(sqrt3mv_(0))/(qE)` |
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Answer» Correct Answer - D `vecE` is parallel to `vecB` and `vecv` is perpendicular both. Therefore, path of the particle is a helix w3ith increasing pitch. Speed of particle at any time `t` is `v=sqrt(v_(x)^(2)+v_(y)^(2)+v_(z)^(2))….(1)` Here ` v_(y)^(2)+v_(z)^(2)=v_(0)^(2)` and `v=2v_(0)` Substituting the va,ues in equation (1) we get ` t-(sqrt3mv_(0))/(qE)` |
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