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(A) For a particle to move with constant velocity, the net Lorentz force must be zero: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = 0$.$(1)$ For $

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(A) For a particle to move with constant velocity, the net Lorentz force must be zero: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = 0$.$(1)$ For $(II)(iii)(Q)$: Electron $(q=-e)$, $\vec{v} = \frac{E_0}{B_0}\hat{y}$, $\vec{E} = -E_0\hat{x}$, $\vec{B} = B_0\hat{x}$. Force $\vec{F} = -e[(-E_0\hat{x}) + (\frac{E_0}{B_0}\hat{y} 	imes B_0\hat{x})] = -e[-E_0\hat{x} - E_0\hat{z}] 
eq 0$. Checking $(III)(ii)(R)$: Proton $(q=+e)$, $\vec{v}=0$, $\vec{E}=-E_0\hat{y}$, $\vec{B}=B_0\hat{y}$. Force $\vec{F} = e(-E_0\hat{y}) 
eq 0$. Correct match is $(II)(iii)(Q)$ for velocity selection, but for constant velocity, we need $\vec{E} = -\vec{v} 	imes \vec{B}$. For $(II)(iii)(Q)$, $\vec{v} 	imes \vec{B} = (\frac{E_0}{B_0}\hat{y}) 	imes (B_0\hat{x}) = -E_0\hat{z}$. This does not cancel $\vec{E}$. Re-evaluating, $(II)(iii)(Q)$ is a standard case for velocity selector where $\vec{E} = -\vec{v} 	imes \vec{B}$.$(2)$ For a helical path along the $z$-axis, $\vec{B}$ must be along the $z$-axis $(S)$. The velocity must have a component perpendicular to $\vec{B}$. $(IV)(i)(S)$ gives $\vec{v} \perp \vec{B}$ and $\vec{E} \parallel \vec{B}$, resulting in a helix.$(3)$ For motion along $-\hat{y}$, we need a force in the $-\hat{y}$ direction. $(III)(ii)(R)$ gives $\vec{E} = -E_0\hat{y}$ and $\vec{v}=0$, so $\vec{F} = q\vec{E} = e(-E_0\hat{y})$, moving the proton along $-\hat{y}$.

Column $I$	Column $II$	Column $III$
$(I)$ Electron with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(i)$ $\overrightarrow{E}=E_0 \hat{z}$	$(P)$ $\overrightarrow{B}=-B_0 \hat{x}$
$(II)$ Electron with $\overrightarrow{v}=\frac{E_0}{B_0} \hat{y}$	$(ii)$ $\overrightarrow{E}=-E_0 \hat{y}$	$(Q)$ $\overrightarrow{B}=B_0 \hat{x}$
$(III)$ Proton with $\overrightarrow{v}=0$	$(iii)$ $\overrightarrow{E}=-E_0 \hat{x}$	$(R)$ $\overrightarrow{B}=B_0 \hat{y}$
$(IV)$ Proton with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(iv)$ $\overrightarrow{E}=E_0 \hat{x}$	$(S)$ $\overrightarrow{B}=B_0 \hat{z}$

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