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(B) The magnetic force acting on a moving charge is given by $\overrightarrow{F} = q_0(\overrightarrow{v} 	imes \overrightarrow{B})$.Since the force

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$v$

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(B) The magnetic force acting on a moving charge is given by $\overrightarrow{F} = q_0(\overrightarrow{v} 	imes \overrightarrow{B})$.Since the force $\overrightarrow{F}$ is always perpendicular to the velocity vector $\overrightarrow{v}$,the work done by the magnetic force on the charge is zero $(W = \overrightarrow{F} \cdot \overrightarrow{d} = 0)$.According to the work-energy theorem,the change in kinetic energy is equal to the work done,which implies that the kinetic energy remains constant.Since kinetic energy $K = \frac{1}{2}mv^2$,a constant kinetic energy implies that the speed $v$ of the charge remains constant over time.Therefore,the speed of $q_0$ after one second remains $v$.

$A$ charge $q_0$ moving with velocity $\overrightarrow{v}$ in a magnetic field of induction $\overrightarrow{B}$ experiences a force $\overrightarrow{F}$. The angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $\theta$. The speed of $q_0$ after one second will be

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