Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

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

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
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)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon