A straight rod of length L is rotating about an axis perpendicular to it and passing through a point O lying at a distance of L/3 from one end of the rod. A uniform magnetic field B exists parallel to the axis of rotation. Emf induced between the ends of the rod is

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$B ω L^{2} / 6$

zero

$8 B ω L^{2} / 25$

$3 B ω L^{2} / 10$

answer is C.

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Detailed Solution

For rotating rod $ε = \frac{B l^{2} ω}{2}$
$ε = B {(\frac{2 l}{3})}^{2} \frac{ω}{2} - B {(\frac{l}{3})}^{2} \frac{ω}{2} = \frac{B l^{2} ω}{6}$

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