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 charge per unit mass α  is released from the origin with velocity v = v 0 i ^  in the magnetic field
         B = - B 0 k ^   for   x 3 2 v 0 B 0 α
and    B = 0   for   x > 3 2 v 0 B 0 α
The x-coordinate of the particle at time t > π 3 B 0 α  would be
PhysicsMagnetic Effects of CurrentJEE Main
Options:
  • A 3 2 v 0 B 0 α + 3 2 v 0 t - π B 0 α
  • B 3 2 v 0 B 0 α + v 0 t - π 3 B 0 α
  • C 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α
  • D 3 2 v 0 B 0 α + v 0 t 2
Solution:
1069 Upvotes Verified Answer
The correct answer is: 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α
r = mv 0 B 0 q = v 0 B 0 α ,    x r = 3 2 = sin θ   ⇒   θ = 6 0

t OA = T 6 = π 3 B 0 α


               

Therefore x-coordinate of particle at any time t > π 3 B 0 α  will be

         x = 3 2 v 0 B 0 α + v 0 t - π 3 B 0 α cos 6 0

           = 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α

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.