The emf induced in the second coil is given by:
EMF=−Mdidt \text{EMF} = -M \frac{di}{dt} EMF=−Mdtdi
Substitute i=i0sinωt i = i_0 \sin \omega t i=i0sinωt:
didt=i0ωcosωt \frac{di}{dt} = i_0 \omega \cos \omega t dtdi=i0ωcosωt
Thus, the maximum emf (when cosωt=1 \cos \omega t = 1 cosωt=1) is:
EMFmax=Mi0ω=(0.002)×(5)×(50π)=π2 V \text{EMF}_{\text{max}} = Mi_0 \omega = (0.002) \times (5) \times (50\pi) = \frac{\pi}{2} \, \text{V} EMFmax=Mi0ω=(0.002)×(5)×(50π)=2πV
Therefore, α=2\alpha = 2α=2.