Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are v=3i^+4j^ and a=2i^+xj^. Select the correct options.
This question has multiple correct options
A
x=−1.5
B
x=3
C
Magnetic field has a component along the z-direction
D
Kinetic energy of the particle is constant
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Solution
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Correct options are A) , C) and D)
Under the influence of magnetic field only, speed and hence, kinetic energy remains constant.
As F⊥v
So F⋅v=0⇒ma⋅v=0
⇒m(2i^+xj^)⋅(3i^+4j^)=0
⇒6+4x=0⇒x=−1.5
Let the magnetic field is B=ai^+bj^+ck^, then from F=qv×B
⇒m(2i^+xj^)=q(3i^+4j^)×(ai^+bj^+ck^)
⇒2mi^−1.5mj^=q(3bk^−3cj^−4ak^+4ci^)
4c=2m⇒c=0
Hence, magnetic field has a component along z-direction.
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