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(B) To determine the potential difference across the moving conducting rod,we use the formula for motional electromotive force $(EMF)$,which is given

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(B) To determine the potential difference across the moving conducting rod,we use the formula for motional electromotive force $(EMF)$,which is given by $\vec{\varepsilon} = \int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.According to the right-hand rule for the cross product $\vec{v} 	imes \vec{B}$:$1$. The velocity vector $\vec{v}$ is directed towards the right.$2$. The magnetic field $\vec{B}$ is directed into the page (represented by $\otimes$).$3$. The cross product $\vec{v} 	imes \vec{B}$ points in the direction perpendicular to both $\vec{v}$ and $\vec{B}$,which is upwards along the rod towards point $P$.Since the force on the free electrons (which are negatively charged) is $\vec{F} = q(\vec{v} 	imes \vec{B})$,and $q$ is negative,the electrons will experience a force in the direction opposite to $\vec{v} 	imes \vec{B}$.Therefore,the electrons will accumulate at end $Q$,making it negatively charged (lower potential),while end $P$ will become positively charged (higher potential).Thus,end $Q$ is at a lower potential.

$A$ conducting rod is moving in a uniform magnetic field as shown in the diagram. Which end is at a lower potential?

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