A particle of mass M and positive charge Q, moving with a constant velocity u → 1 = 4 i ^ ms - 1 ,…

A particle of mass M and positive charge Q, moving with a constant velocity u→1=4i^ms-1, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity u→2=23i^+j^ms-1. The correct statement (s) is (are)
  1. The direction of the magnetic field is -z direction.
  2. The direction of the magnetic field is +z direction.
  3. The magnitude of the magnetic field  5 0 π M 3 Q units.
  4. The magnitude of the magnetic field  10 0 π M 3 Q units.

Solution

the angular displacementθ,
θ = ω t →1

ω = angular velocity, t = time taken by the the particle emerges on the other side 

t = 10 ms = 10×10-3 S

angular velocity of particle in a magnetic field ω,ω = qBm→2

q = charge, B = magnetic field, m = mass of the particle.
substitute equation 2 in equation 1
θ = qB m t
Angle between velocity vectors is θ=π6
Solving above we get B = 5 0 π m 3 q

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Asked in: JEE Advanced 2013 (Paper 1)

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