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(A) When a charged particle enters a uniform magnetic field perpendicularly,the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ acts on it.Since

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Energy remains constant but momentum changes

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(A) When a charged particle enters a uniform magnetic field perpendicularly,the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ acts on it.Since the magnetic force is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic force is $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt = 0$.According to the work-energy theorem,the change in kinetic energy $\Delta K = W = 0$,which implies that the kinetic energy (and thus the speed) remains constant.However,because the magnetic force acts as a centripetal force,it changes the direction of the velocity vector $\vec{v}$. Since momentum $\vec{p} = m\vec{v}$ is a vector quantity,a change in the direction of velocity results in a change in momentum.Therefore,energy remains constant,but momentum changes.

If a charged particle enters perpendicularly in a uniform magnetic field,then:

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