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(D) The motional $e.m.f.$ induced in a conductor moving in a magnetic field is given by $\int (\vec v 	imes \vec B) \cdot d\vec l$.In this case,the m

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(D) The motional $e.m.f.$ induced in a conductor moving in a magnetic field is given by $\int (\vec v 	imes \vec B) \cdot d\vec l$.In this case,the magnetic field $\vec B$ is parallel to the axis of rotation.For any element $d\vec l$ along the rod,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,$\vec B$ is perpendicular to the velocity vector $\vec v$ of any element on the rod.However,the motional $e.m.f.$ is generated only if there is a component of the magnetic field perpendicular to the velocity of the conductor.In this specific configuration,the magnetic field $\vec B$ is parallel to the axis of rotation,which means it is perpendicular to the rod itself at all points.Wait,let's re-evaluate: The velocity $\vec v$ of an element at distance $x$ from the center is $\vec v = \omega x \hat{	heta}$. The magnetic field is $\vec B = B \hat{k}$ (along the axis).The force on a charge $q$ in the rod is $\vec F = q(\vec v 	imes \vec B) = q(\omega x \hat{	heta} 	imes B \hat{k}) = q \omega x B \hat{r}$.This force is directed radially outward from the axis of rotation.Since the force is directed radially,it acts to push charges toward both ends of the rod simultaneously.Because the rod is symmetric about the axis of rotation,the potential at both ends of the rod will be the same relative to the center.Therefore,the potential difference (induced $e.m.f.$) between the two ends of the rod is $V_{end1} - V_{end2} = 0 - 0 = 0$.

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

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