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=(sqrt2mv_0)/(qE)` |
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Answer» Correct Answer - d `vecE` is parallel to `vecB` and `vecv` is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t is `v=sqrt(v_x^2+v_y^2+v_z^2)....(i)` Here, `v_y^2+v_z^2=v_0^2` and `v=2v_0 implies v_x=sqrt3v_0` `v_xa_xt implies sqrt3v_0=(qE)/mt implies t=(sqrt3mv_0)/(qE)` |
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