Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are v=3i^+4j^ and a=2i^+xj^. Select the correct alternative(s) :
This question has multiple correct options
A
x=1.5
B
x=3
C
magnetic field is along z-direction
D
kinetic energy of the particle is constant
Hard
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Updated on : 2022-09-05
Solution
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Correct options are A) , C) and D)
Force & velocity are perpendicular to each other in magnetic field So V⋅a=0 (3i^+4j^)⋅(2i^+xj^)=0 6+4x=0 x=−1.5 No work is done by magnetic field. So, KE of particle is constant
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