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Solve problem 15 if the length of rod is...

Solve problem `15` if the length of rod is `2L` and resistance `2r` and it is rotating about its center. Both ends of the rod now touch the conducting ring.

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`I = (epsilon)/(R + (r )/(2)) = ((1)/(2)BomegaL^(2))/((R + r )/(2))`
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CENGAGE PHYSICS-ELECTROMAGNETIC INDUCTION-Exercise 3.2
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