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Question: Answered & Verified by Expert
A small particle of mass m and charge Q is dropped in uniform horizontal magnetic field B. The maximum vertical displacement of particle is given by h=nm2g2Q2B2 . Find the value of n.
PhysicsMagnetic Effects of CurrentJEE Main
Solution:
1489 Upvotes Verified Answer
The correct answer is: 4
mgh=12mv2 v=2gh

At lowest position, velocity becomes horizontal.

   vx=v=2gh



Fsinθ=max

QvBsinθ=max

QBvsinθ=max

   QBVy=max

  QBdydt=mdvxdt

   QB0hdy=m0vxdvx

   QBh=mvx

   QBh=m2gh

   h=2m2gQ2B2

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