First slide
Motion of a charged particle
Question

A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field  E=2E0i^ and B=B0i^ with a velocity v=v0j^ . The speed of the particle will becomes 2v0 after a time.

Moderate
Solution

E is parallel to B and V is perpendicular to both,  therefore path of the particle is a helix with increasing pitch .

Speed of particle at any time ‘t’ is given by
 V=Vx2+Vy2+Vz2
here,  Vy2+V22=V02   at all times because the yz components will remain same.

Only the x component will increase due to electric field.
After some time, it is given that   V=2Vo

Using,  V=Vx2+Vy2+Vz2 2Vo=Vx2+Vo2 Vx = 3Vo
 Equation of kinematics,  Vx=Ux+axt 3V0=q2EoMt
t=3MV02qEo .

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