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A particle is released from the origin w...

A particle is released from the origin with a velocity `vhate(i)`. The electric field in the region is `Ehate(i)` and magnetic field is `Bhat(k)`. Then

A

If `v=0 and E=0` then particle will execute a circular path.

B

If `v=0, E!=0` then particle is positively charged then it executes clockwise

C

If `v=0` and `E != 0` then the particle will execute a cycloidal motion in the x-y plane.

D

If `vgt0` and `E!= 0` and the particle is positievely charged and `Bgt0` then the particle wil excute anticlockwise anticlock wise circle as seen from +z direction.

Text Solution

Verified by Experts

The correct Answer is:
C, D
if `v=0,E=0` then no force will act on particle, hence partcle remains at rest
(b) here `v=0, E!=0`, due to E the particle will gain velocity in x direction. Due to this velocity it will experience force due to magnetic field along negative y axis. As a result of it, the particle will move in curved path that will not be a circle.
(c) If `v=0,E!=0`, then particle will execute cycloidal motion.
(d) If `E=0`, then path will be circular.
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Knowledge Check

  • The electric field acts positive x -axis. A charged particle of charge q and mass m is released from origin and moves with velocity vec(v)=v_(0)hatj under the action of electric field and magnetic field, vec(B)=B_(0)hati . The velocity of particle becomes 2v_(0) after time (sqrt(3)mv_(0))/(sqrt(2)qE_(0)) . find the electric field:

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