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 hatj `. The speed of the particle will becomes `(sqrt(5))/(2) v_0` after a time

Options

A. `(mv_(0))/(qE_(0))`
B. `(mv_(0))/(2qE_(0))`
C. `(sqrt3mv_(0))/(2qE_(0))`
D. `(sqrt5mv_(0))/(2qE_(0))`

Answer

B

Explanation

Here `vec E and vec B` are acting along x-axis and `vec v` and is along y-axis i.e, perpendicular to both `vec E and vec B.` therefore , the path of charged particles is a with increasing speed .Speed of particles
at time is `v= sqrt(v_(x)^(2)+v_(y)^(2))`...(i)
Here `v_(y)=v_(0),v_(x)=(qE_(0))/(m)t andv=(sqrt5)/(2)v_(0)`
Putting values in (i), We get `t=(mv_(0))/(2qE_(0))`