A particle of mass`M` and positive charge `Q`, moving with a constant velocity ` vec(u_(1)) = 4 hat(i) 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 `vec(u_(2)) = 2 (sqrt(3 hat(i) + hat(j))) ms^(-1)` . The correct statement(s) is (are)

Options

A. the direction of magnetic field is `-z` direction
B. the direction of magnetic field is `+z` direction
C. the magnitude of magnetic field is `(50piM)/(3Q)` units
D. the magnitude of magnetic field is `(100piM)/(3Q)` units

Answer

A, C

Explanation

Refer to figure, component of final velocity of particle is in positive y-direction. The centre of circular path of particle in magnetic field is present on positive y-direction. So magnetic field is present in negative z-direction.

If `theta` is the angle of deviation of the particle with x-axis while emerging from magnetic field, then
`tan theta=(v_y)/(v_x)=(2)/(2sqrt3)=1/sqrt3=tanpi/6` or `theta=pi/6`
Angular velocity of rotation of particle in magnetic field, `omega=(QB)/(M)`
Time taken by particle to cross the magnetic field is
`t=(theta)/(omega)=(pi//6)/(QB//M)=(Mpi)/(6QB)`
or `B=(Mpi)/(6Qt)=(Mpi)/(6Qxx(10xx10^-3))=(50Mpi)/(3Q)`