The velocity and acceleration vectors of a particle moving in a magnetic field are given by:
\( \vec{v}=3\hat{\imath}+4\hat{\jmath} \) and \( \vec{a}=2\hat{\imath}+\hat{\jmath} \) respectively, Then:
(A) the magnetic field is along z-direction
(B) v = 3
(C) the kinetic energy of the particle is constant
(D) the acceleration is in x-z plane
(E) the particle is moving in x-y plane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C), (E)
The magnetic force \( \vec{F} = q(\vec{v}\times\vec{B}) \) is always perpendicular to both \( \vec{v} \) and \( \vec{B} \). It changes the direction of the velocity but never its magnitude. Judge each statement.
(A) the magnetic field is along z-direction — Correct. The velocity and the acceleration (hence the force) both lie in the x-y plane, so the field that acts on the particle is perpendicular to that plane, that is along the z-direction.
(B) v = 3 — Incorrect. The speed is \( |\vec{v}| = \sqrt{3^{2}+4^{2}} = 5 \). The number 3 is only the x-component.
(C) the kinetic energy of the particle is constant — Correct. The magnetic force does no work because it is perpendicular to the velocity, so the speed and the kinetic energy stay constant.
(D) the acceleration is in x-z plane — Incorrect. \( \vec{a} = 2\hat{\imath}+\hat{\jmath} \) has an x- and a y-component and no z-component, so it lies in the x-y plane.
(E) the particle is moving in x-y plane — Correct. Both \( \vec{v} \) and \( \vec{a} \) have no z-component, so the particle does not leave the x-y plane.