InterviewSolution
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InterviewSolution
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Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are `vecv=3hati+4hatj and veca=2hati+xhatj`. Select the correct options.A. `x=-1*5`B. `x=3`C. magnetic field is along z-directionD. K.E. of particle is constant |
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Answer» Correct Answer - A::C::D (a) Since magnetic force `(F_m)` is always perpendicular to the velocity vector `vecv`, i.e., `vecF_m_|_vecv` or `veca_|_vecv` or `veca.vecv=0` or `(2hati+xhatj).(3hati+4hatj)=0`, `x=-1*5`. (c) Since `vecB` is `_|_` to the plane containing the velocity vector, in this case (XY-plane, `vecB` is along Z-direction) (d) Since work done by magnetic force is zero, So K.E. of particle remains constant. |
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