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 CurrentJEE Main
Options:
  • A 2 m s-1
  • B 20 m s-1
  • C 0.2 m s-1
  • D 100 m s-1
Solution:
1085 Upvotes Verified Answer
The correct answer is: 2 m s-1
Net force on the particle should be zero.


            


qE = 1 0 - 6 × 2 0 0 = 2 × 1 0 - 4 N

mg = 2 × 1 0 - 5 × 9.8 = 1.96 × 1 0 - 4 N

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

qE = mg + qvB 2 × 1 0 - 4 = 1.96 × 1 0 - 4 + 1 0 - 6 v × 2

⇒    v = 2 m/s

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.