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(D) When a conductor moves in a magnetic field,a motional electromotive force $(EMF)$ is induced across its ends,which is given by $\varepsilon = \int

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in both $AD$ and $BC$

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(D) When a conductor moves in a magnetic field,a motional electromotive force $(EMF)$ is induced across its ends,which is given by $\varepsilon = \int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.This induced $EMF$ corresponds to an induced electric field $\vec{E} = \vec{v} 	imes \vec{B}$.In this problem,the square loop $ABCD$ moves with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the loop.For any segment of the loop,if the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,an electric field $\vec{E} = \vec{v} 	imes \vec{B}$ is induced along the conductor.Since the segments $AD$ and $BC$ are perpendicular to the velocity $\vec{v}$ and the magnetic field $\vec{B}$ is uniform and perpendicular to the plane,the cross product $\vec{v} 	imes \vec{B}$ is non-zero for both segments $AD$ and $BC$.Therefore,an electric field is induced in both $AD$ and $BC$.

$A$ metallic square loop $ABCD$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced:

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