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(A) The magnetic force on a moving charged particle is given by $\overrightarrow{F} = q(\overrightarrow{v} 	imes \overrightarrow{B})$.Since the force

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

Vedclass · Answer

(A) The magnetic force on a moving charged particle is given by $\overrightarrow{F} = q(\overrightarrow{v} 	imes \overrightarrow{B})$.Since the force $\overrightarrow{F}$ is the cross product of velocity $\overrightarrow{v}$ and magnetic field $\overrightarrow{B}$,the force is always perpendicular to the velocity $(\overrightarrow{F} \perp \overrightarrow{v})$.Work done by the magnetic force is $W = \int \overrightarrow{F} \cdot d\overrightarrow{s} = \int \overrightarrow{F} \cdot \overrightarrow{v} dt$.Since $\overrightarrow{F} \perp \overrightarrow{v}$,the dot product $\overrightarrow{F} \cdot \overrightarrow{v} = 0$,so the work done is $0$.According to the work-energy theorem,the change in kinetic energy is equal to the work done. Since work done is $0$,the kinetic energy remains constant.Since kinetic energy $K = \frac{1}{2}mv^2$ is constant,the speed $v$ of the particle also remains constant.Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

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