Answer:
The induced emf in both the coils will be same as \[e=-(d\phi /dt)=BA\omega \,\,\sin \,\omega t.\] Since, \[B,\,\,A\] and \[\omega \] are same for both cases, induced emf will be same. \[\Rightarrow \] \[\frac{{{e}_{c}}}{{{e}_{A}}}=\frac{1}{1}=1:1\]
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