A particle having charge q and mass m starts moving from the origin under . . .

Question : A particle having charge q and mass m starts moving from the origin under the action of an electric field E=E₀ i and B=B₀i with a velocity of v=v₀ j. The speed of the particle will become √5v₀/2 after a time 

a) mv₀/qE

b) mv₀/2qE

c) √3mv₀/2qE

d) √5mv₀/2qE

Doubt by Jayant

Ans : b) is the correct answer.

Solution : 

Here we need to understand that the particle is moving in both Electric and Magnetic Fields.

The Electric field will provide the straight line motion while the magnetic filed will provide the circular motion so the resultant path of the charge particle will be helical in nature.

E=E₀ i
B=B₀i 

v=v₀ j.

The total velocity of the charge particle at any instant t is given by 
v = √(v_x²+v_y²+v_z²)

Here 
v_y²+v_z² = v₀² — (1) [Given]

v = √(v_x²+v_y²+v_z²)

v² = v_x²+v_y²+v_z²

(√5v₀/2)² = v_x² + v₀²
5v₀2/4 = v_x² + v₀²

v_x² = 5v₀2/4 - v₀²

v_x² = v₀²/4

v_x = v₀/2

Now 

v_x = u_x+at
v_x = 0 +(qE/m)t
v₀/2 = qEt/m
mv₀/2qE = t

t = mv₀/2qE.

Hence, b)  is the correct option.