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

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$10 \ m/s$

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(A) The magnetic force acting on a moving charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the magnetic force is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic field on the particle is zero $(W = \int \vec{F} \cdot d\vec{r} = 0)$.According to the work-energy theorem,the change in kinetic energy is equal to the work done by all forces.Since the work done is zero,the kinetic energy remains constant $(K_f = K_i)$.Therefore,the speed of the particle remains constant throughout its motion.Given the initial speed $v_0 = 10 \ m/s$,the speed at any point in time,including when it passes through the $y-$axis for the third time,will be $10 \ m/s$.

$A$ particle is projected with a velocity of $10 \ m/s$ along the $y-$axis from the point $(2, 3)$. $A$ uniform magnetic field of $(3\hat{i} + 4\hat{j}) \ T$ exists in the space. What is its speed when the particle passes through the $y-$axis for the third time? (Neglect gravity)

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