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 charged particle carrying charge 1 μC is moving with velocity (2i^+3j^+4k^) m s-1. If an external magnetic field of (5i^+3j^-6k^)×10-3T exists in the region where the particle is moving then the force on the particle is F×10-9 N . the vector F is :
PhysicsMagnetic Effects of CurrentJEE MainJEE Main 2020 (03 Sep Shift 1)
Options:
  • A -0.30i^+0.32j^-0.09k^
  • B -30i^+32j^-9k^
  • C -300i^+320j^-90k^
  • D -3.0i^+32j^-0.9k^
Solution:
2630 Upvotes Verified Answer
The correct answer is: -30i^+32j^-9k^

Using the equation of force as,

F=qV×B

F=10-6×10-3i^j^k^23453-6

=-30i^+32j^-9k^×10-9 N

 

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.