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.

Options

A. `t=(2mv_(0))/(qE)`
B. `t=(2Bq)/(mv_(0))`
C. `t=(sqrt(3)Bq)/(mv_(0))`
D. `t=(sqrt(3)mv_(0))/(qE)`

Answer

D

Explanation

`vecE=vecE_(0)I, vecB=B_(0)hai,v=v_(0)hatj`
Speed of charge particle will double due to work done by electric field.
`W_("electric field")=(qE)ximplies(qE_(0))x=1/2m((2v_(0))^(2)-v_(0)^(2)),x-(3mv_(0)^(2))/(2qE_(0))`
Acceleration of particle is `a=((qE_(0))/m),x=ut+1/2 at^(2)`
`3/2 (mv_(0)^(2))/(qE_(0))=1/2((qE_(0))/m)t^(2)impliest^(2)=(3m^(2)v_(0)^(2))/((q^(2)E_(0)^(2))),t=sqrt(3)(mv_(0))/(qE_(0))`