Q.

The electric field acts along positive x-axis. A charged particle of charge q and mass m is released from origin and moves with velocity v→=v0j^ under the action of electric field and magnetic field, B→=B0i^ The velocity of particle becomes 2v0 after time 3 mv02 qE0.  Find the electric field.

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a

2 3 E0i^

b

3 2 E0i^

c

3 E0i^

d

2 E0i^

answer is D.

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Detailed Solution

v→ is perpendicular to both E→ and B→.Path of particle is helix with increasing pitchv=vx2  +  vy2  +  vz21/2Here,   vx2=qEmt2  and  vy2  +  vz2  =  v02Also,  v=2v02v02  =  q2E2t2m2 +  v02t= 3  mv0qE  = 3  mv03qE0       givenE→=2 E0i^
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