wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eo^iandB=Bo^i with the velocity v=vo^j. The speed of the particle will become 2vo after a time

A
t=2mvoqE
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
t=2Bqmvo
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
t=3Bqmvo
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
t=3mv0qE
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D t=3mv0qE
Given :A particle of charge q and mass m starts moving from the origin under the action of an electric field .
E=E0^i and B=B0^i with velocity v=v0^j. Final speed of particle becomes 2v0

Solution :
Here, E=E0^i and B=B0^i are acting along x-axis and v=v0^j is along y-axis i.e. perpendicular to both E=E0^i and B=B0^i . Therefore, the path of charged particle is a helix with increasing speed. Speed of particle at time t is

v0=v2x+v2y eq-1
Here vy=v0 , vx=qE0tm v=2v0
Putting values in (i),(i), we get
t=3vomqEo
The Correct Opt: D


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Charge Motion in a Magnetic Field
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon