A conducting rod of length 2l is rotating wit | vedclass

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(D) The motional e.m.f. induced in a conductor moving in a magnetic field is given by $\varepsilon = \int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.In

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$Zero$

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(D) The motional e.m.f. induced in a conductor moving in a magnetic field is given by $\varepsilon = \int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.In this case, the magnetic field $\vec{B}$ is parallel to the axis of rotation.For any element $dr$ of the rod at a distance $r$ from the center $O$, the velocity vector $\vec{v}$ is perpendicular to the rod and lies in the plane of rotation.Since $\vec{B}$ is parallel to the axis of rotation, the vector $\vec{v} 	imes \vec{B}$ is directed along the rod.However, the motional e.m.f. is induced only if there is a component of the magnetic field perpendicular to the velocity and the length of the rod.Since $\vec{B}$ is parallel to the axis of rotation, the angle between $\vec{B}$ and the velocity vector $\vec{v}$ is $90^\circ$, but the magnetic field is perpendicular to the plane of rotation, meaning it is parallel to the axis of rotation. The force on the charge carriers $q(\vec{v} 	imes \vec{B})$ is directed perpendicular to both $\vec{v}$ and $\vec{B}$. Since $\vec{v}$ is in the horizontal plane and $\vec{B}$ is vertical, the force is radial. There is no force component along the length of the rod.Alternatively, the magnetic flux through the area swept by the rod is zero because the magnetic field lines are parallel to the axis of rotation and do not cut the area swept by the rod.Therefore, the induced e.m.f. between the two ends $A$ and $B$ is $Zero$.

$A$ conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. $A$ uniform magnetic field $\overrightarrow{B}$ exists parallel to the axis of rotation. The e.m.f. induced between the two ends of the rod is

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