Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A particle of mass 2×10-5 kg moves horizontally between the plates of a parallel plate capacitor which produce an electric field of 200 N C-1 in the vertically upward direction. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both E and v. If g is 9.8 m s-2 and the charge on the particle is 10-6 C, then the velocity of the charged particle so that it continues to move horizontally is
PhysicsMagnetic Effects of CurrentNEET
Options:
  • A 2 m s-1
  • B 20 m s-1
  • C 0.2 m s-1
  • D 100 m s-1
Solution:
2919 Upvotes Verified Answer
The correct answer is: 2 m s-1
Net force on the particle should be zero.


            


qE=10-6×200=2×10-4 N

mg=2×10-5×9.8=1.96×10-4 N

Since qE>mg, so magnetic force qvB should act downwards to balance the forces.

qE=mg+qvB2×10-4=1.96×10-4+10-6v×2

⇒   v=2 m s-1

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.