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Question

A particle of charge of q and mass m starts moving from the origin under the action of an electric field ¯¯¯¯E=E0^i and ¯¯¯¯B=B0^i with a velocity ¯¯¯v=v0^j. The speed of the particle will become 2v0 after a time

A
t=2mv0qE
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B
t=2Bqmv0
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C
t=3mv0qE
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D
t=3Bqmv0
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Solution

The correct option is C t=3mv0qE
First consider the effect of magnetic field on the particle
Magnetic force on the particle is given by
F=q(v×B)
The force will be along -z direction and the particle will perform uniform circular motion in y-z plane.



Now, the electric field will try to accelerate the charged particle in the x-direction at each instant.
Acceleration due to electric field =qEm

So,due to the net effect of both forces, particle will be moving in a helical path with increasing pitch.



Since change in magnitude of speed is only in the x-direction (due to electric field) and initial velocity in the x-direction was zero, we can use the first equation of motion to get
At time t,
vx=ux+axt=0+qEmt=qEmt
And there will be no change in the velocity in the y-z plane i.e it will be v0

Net velocity at time t,v=v20+v2x
Putting v=2v0,
v20+v2x=4v20
​​​​​​​v20+(qEmt)2=4v20
(qEmt)2=3v20
qEmt=3v0
t=3mv0qE

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