Q.
A particle of charge q and mass m starts moving from the origin under the action of an electric field E→=E0i^ and B→=B0i^ with a velocity v→=v0j^. The speed of the particle will become 2v0 after a time
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
t=2mv0qE
b
t=2Bqmv0
c
t=3Bqmv0
d
t=3mv0qE
answer is D.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
E→ is parallel to B→ and v→ is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t isv=vx2+vy2+vz2 ……..(i)Here, vy2+vz2=v02and v=2v0⇒vx=3v0 vx=axt⇒3v0=qEmt⇒t=3mv0qE
Watch 3-min video & get full concept clarity