Question

A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed omegha on a horizontal table with one end of the string fixed .If there is a vertical magnetic field B in the region,the emf induced across the ends of rod is
(a)2B omegha l^3/2 (B)3B Omegha l^3 /2
(c)4B omegha l^2 /2 (c)5b omegha l^3 /2

Answer 1

In the options above, there seems to be  L³ , it is not correct.  It should be L².
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   The metallic rod is tied to one end of a rod and the other end is left free.  The other end of the string is fixed at the center of a horizontal table.  The rod rotates on the horizontal table about an axis perpendicular to it and passing through its center.  The angular speed = ω.    Assume that there is a metallic frame on which the ends of the rod are travelling.  The ends of the rod are always on the metallic points.  So a circular loop is established. 

   Magnetic field  B is perpendicular to the metallic rod.  Area covered by the rod and the loop:  A.   To find A, we take the radius of the revolution of the center of the rod, which is 2L + L/2 = 2.5 L.    Then the angle of revolution theta = ω * time t.    The r * theta gives the length through which the rod moves.  Its length is L. 

         A = [ 2.5 L * (ω t) ] L = 2.5 L² ω t
 EMF induced = -dΦ / dt  = - d/dt [Φ] = - d/dt [B A]
           = - d/dt [ B 2.5 L² ω t ]
           = - 2.5 B L² ω

option D

Answer 2

The correct answer is option c