A charged particle enters a uniform magnetic | vedclass

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Question

(C) The correct answer is $(c)$.The magnetic force $\vec{F}$ acting on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$

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Changes the direction of motion of the particle

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(C) The correct answer is $(c)$.The magnetic force $\vec{F}$ acting on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the magnetic force is always perpendicular to the velocity vector $(\vec{F} \perp \vec{v})$,the work done by the magnetic field on the particle is zero $(W = \vec{F} \cdot \vec{d} = 0)$.According to the work-energy theorem,since the work done is zero,the kinetic energy and the speed of the particle remain constant.However,because the force acts perpendicular to the velocity,it provides the necessary centripetal force to change the direction of the particle's motion,causing it to move in a circular path.

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

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