question_answer

A metallic rod of length is tied to a suing of length \[(3\ell )\] and made to rotate with constant angular velocity \[(\omega )\] on horizontal table with one end of the string is fixed. There is a vertical magnetic field B in the region, the emf induced accross the ends of the rod is:

A)  \[B\ell V\]          

B)                     \[\frac{B{{\ell }^{2}}\omega }{2}\]

C)  \[\frac{7B{{\ell }^{2}}\omega }{2}\]

D)     \[\frac{5B{{\ell }^{2}}\omega }{2}\]

Correct Answer: C

Solution :

\[emf\,=B/{{V}_{average}}\,=Bl\left( \frac{{{V}_{1}}+{{V}_{2}}}{2} \right)\]             \[=Bl\,\frac{7\omega l}{2}=\frac{7}{2}\,B{{l}^{2}}\omega \]