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(D) The magnetic force acting on a charged particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the force $\vec

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Momentum and energy remain unchanged

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(D) The magnetic force acting on a charged particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$,the work done by the magnetic field on the particle is $W = \int \vec{F} \cdot d\vec{s} = \int \vec{F} \cdot \vec{v} dt = 0$.According to the work-energy theorem,the change in kinetic energy is equal to the work done,so the kinetic energy remains constant.Since the speed of the particle remains constant,the magnitude of its momentum $(p = mv)$ also remains constant.Therefore,both the energy and the magnitude of the momentum remain unchanged.

When a magnetic field is applied in a direction perpendicular to the direction of cathode rays,then their

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