Questions
A particle of mass 2 x 105 kg moves horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of 200 NC-1 acting upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both and . If g is 9.8 m s-2 and the charge on the particle is 10-6 C, then find the velocity of charge particle so that it continues to move horizontally
detailed solution
Correct option is A
Net force on the particle should be zero.qE=10-6×200=2×10-4 Nmg=2×10-5×9.8=1.96×10-4 NSince qE > mg, so magnetic forc e qvB should act downward to balance the forcesqE=mg+qvB⇒2×10-4=1.96×10-4+10-6v×2 ⇒ v=2 m/sTalk to our academic expert!
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A charged particle having charge 10-6 C and mass of 10-10 kg is fired at the middle of the plate making an angle 30° with plane of the plate. Length of the plate is 0. 17 m and it is separated by 0.1 m. Electric field E= is present between the plates. Just outside the plates magnetic field is present. Find the magnitude of the magnetic field (in mT)perpendicular to the plane of the figure, if it has to graze the plate at C and A (in mT) parallel to the surface of the plate.(Neglect gravity)
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