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(C) The particle moves in a circular path in a uniform magnetic field because the magnetic force acts perpendicular to the velocity.Given charge per u

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$\frac{2 v_0}{B_0 \alpha}$

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(C) The particle moves in a circular path in a uniform magnetic field because the magnetic force acts perpendicular to the velocity.Given charge per unit mass $\frac{q}{m} = \alpha$.The radius $R$ of the circular path is given by $R = \frac{mv}{qB} = \frac{v_0}{\alpha B_0}$.Since the particle is released at the origin $(0,0,0)$ with velocity along the $x$-axis and the magnetic field is along the $-z$-axis,the Lorentz force $\vec{F} = q(\vec{v} 	imes \vec{B}) = q(v_0 \hat{i} 	imes -B_0 \hat{k}) = q v_0 B_0 \hat{j}$ acts in the $+y$ direction.The particle completes a semi-circle to reach the point $(0, y, 0)$.Therefore,the distance $y$ is equal to the diameter of the circular path.$y = 2R = 2 \left( \frac{v_0}{\alpha B_0} ight) = \frac{2 v_0}{\alpha B_0}$.

$A$ particle of charge per unit mass $\alpha$ is released from the origin with a velocity $\vec{v} = v_0 \hat{i}$ in a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. If the particle passes through $(0, y, 0)$,then $y$ is equal to

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