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(B) Consider a rod of length $l$ rotating about an axis passing through its center,perpendicular to its length. The magnetic field $B$ is parallel to

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(B) Consider a rod of length $l$ rotating about an axis passing through its center,perpendicular to its length. The magnetic field $B$ is parallel to the axis of rotation.For any small element $dr$ at a distance $r$ from the center,the motional $emf$ induced is $d\varepsilon = Bv dr = B(\omega r) dr$.The $emf$ induced between the center and one end is $\varepsilon = \int_{0}^{l/2} B\omega r dr = B\omega [\frac{r^2}{2}]_{0}^{l/2} = \frac{Bl^2\omega}{8}$.Since the rod rotates about its center,both halves of the rod act as two batteries connected in opposition. The $emf$ induced in the first half (from center to one end) is $\frac{Bl^2\omega}{8}$,and the $emf$ induced in the second half (from center to the other end) is also $\frac{Bl^2\omega}{8}$.Because these two induced $emf$s are in opposite directions relative to the center,the net potential difference between the two ends of the rod is $\frac{Bl^2\omega}{8} - \frac{Bl^2\omega}{8} = 0$.

$A$ rod of length $l$ rotates with a uniform angular velocity $\omega$ about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $B$ with its direction parallel to the axis of rotation. The induced $emf$ between the two ends of the rod is

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$A$ metal rod of length $1\,m$ is rotated about one of its ends in a plane at right angles to a magnetic field of induction $2.5 \times 10^{-3}\,Wb/m^2$. If it makes $1800\,rpm$,calculate the induced e.m.f. between its ends in $V$.

What is motional $emf$?

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