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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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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.