A rod of length 60 cm rotates with a uniform | vedclass

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(C) The motional electromotive force $(EMF)$ induced in a rod rotating in a magnetic field is given by $\varepsilon = \frac{1}{2} B \omega r^2$,where

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(C) The motional electromotive force $(EMF)$ induced in a rod rotating in a magnetic field is given by $\varepsilon = \frac{1}{2} B \omega r^2$,where $r$ is the length of the rotating segment.In this case,the rod rotates about its center $O$. The two halves of the rod,$OA$ and $OB$,each of length $L = 30 \ cm = 0.3 \ m$,act as two separate rods rotating about one end.The induced $EMF$ in segment $OA$ is $\varepsilon_{OA} = V_O - V_A = \frac{1}{2} B \omega L^2$.The induced $EMF$ in segment $OB$ is $\varepsilon_{OB} = V_O - V_B = \frac{1}{2} B \omega L^2$.Since the magnetic field is parallel to the axis of rotation,the motional $EMF$ is actually zero because the velocity vector $\vec{v}$ is always perpendicular to the magnetic field $\vec{B}$ only if the field is perpendicular to the plane of rotation. However,if the magnetic field is parallel to the axis of rotation,the force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is directed along the rod. But for a rod rotating about its center in a field parallel to the axis,the flux through the area swept by the rod is zero. Alternatively,since the potential at both ends $A$ and $B$ relative to the center $O$ is the same,the potential difference between the ends $A$ and $B$ is $V_A - V_B = (V_O - V_B) - (V_O - V_A) = 0 - 0 = 0 \ V$.

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