(i) Induced emf, ε = - \(\frac{dφ}{dt}\) = -\(\frac{d}{dt}\) (BA cos ωt)
= BA ω sin ωt
As B, A, ω are same for both loops, so induced emf is same in both loops.
(ii) Current induced, I = \(\frac{ε}{R}\) = \(\frac{ε}{pl/A}\) = \(\frac{εA}{pl}\)
As area A, length l and emf ε are same for both loops but resistivity ρ is less for copper, therefore current I induced is larger in copper loop.