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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.For $y < 0$,the magnetic field $\vec{B}$ is out of the pla

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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.For $y When the proton crosses $y = 0$ and enters the region $y > 0$,the magnetic field $\vec{B}$ is into the plane (negative $z$-direction). The velocity $\vec{v}$ is now in the $+y$-direction. Using the right-hand rule for $\vec{v} 	imes \vec{B}$,the force $\vec{F}$ points in the $-x$-direction. Thus,the proton continues to curve in the same direction,completing a full circular path in the $xy$-plane.Comparing this with the given options,the trajectory is a full circle,which matches the visual representation in the provided solution image.

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