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(C) The net force on a charged particle moving with constant velocity must be zero. The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} 	ime

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$(A), (B)$ and $(D)$ only

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(C) The net force on a charged particle moving with constant velocity must be zero. The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$. For $\vec{F} = 0$,we have $\vec{E} + \vec{v} 	imes \vec{B} = 0$.Case $(A)$: If $\vec{E} = 0$ and $\vec{B} = 0$,then $\vec{F} = 0$. The velocity remains constant.Case $(B)$: If $\vec{E} = 0$ and $\vec{B} 
eq 0$,the particle can move with constant velocity if $\vec{v}$ is parallel or anti-parallel to $\vec{B}$ (since $\vec{v} 	imes \vec{B} = 0$).Case $(C)$: If $\vec{E} 
eq 0$ and $\vec{B} = 0$,the electric force $q\vec{E}$ would cause acceleration,so the velocity cannot remain constant. Thus,$(C)$ is not possible.Case $(D)$: If $\vec{E} 
eq 0$ and $\vec{B} 
eq 0$,the particle can move with constant velocity if $\vec{E} = -(\vec{v} 	imes \vec{B})$. This is possible (e.g.,in a velocity selector).Therefore,cases $(A), (B),$ and $(D)$ are possible.

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