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(D) The motional $e.m.f.$ induced in a small element $dx$ of a rod moving with velocity $v$ in a magnetic field $B$ is given by $de = B v dx$.Since th

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Zero

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(D) The motional $e.m.f.$ induced in a small element $dx$ of a rod moving with velocity $v$ in a magnetic field $B$ is given by $de = B v dx$.Since the rod is rotating with angular speed $\omega$ about its perpendicular bisector,the velocity of an element at distance $x$ from the center is $v = \omega x$.Thus,$de = B (\omega x) dx$.The magnetic field $\vec B$ is parallel to the axis of rotation. The velocity vector $\vec v$ of any point on the rod is perpendicular to the radius vector (in the plane of rotation),and the magnetic field $\vec B$ is perpendicular to this plane of rotation.However,the motional $e.m.f.$ is induced only when the velocity vector $\vec v$ has a component perpendicular to the magnetic field $\vec B$. Here,$\vec v$ is always in the plane of rotation,and $\vec B$ is perpendicular to this plane. Thus,$\vec v \perp \vec B$ is satisfied for all points.But for a rod rotating about its center,the induced $e.m.f.$ between the center and one end is $\int_0^l B \omega x dx = \frac{1}{2} B \omega l^2$.Since the rod is symmetric about the center,the potential at both ends is the same relative to the center. Therefore,the potential difference between the two ends is $V_{end1} - V_{end2} = (V_{center} + \frac{1}{2} B \omega l^2) - (V_{center} + \frac{1}{2} B \omega l^2) = 0$.Hence,the induced $e.m.f.$ between the two ends is zero.

$A$ 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 the two ends of the rod is

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