Two coils have mutual inductance $$0.002$$ H. The current changes in the first coil according to the relation $$i = i_0 \sin \omega t$$, where $$i_0 = 5$$ A and $$\omega = 50\pi \text{ rad s}^{-1}$$. The maximum value of emf in the second coil is $$\frac{\pi}{\alpha}$$ V. The value of $$\alpha$$ is

The mutual inductance is $$M = 0.002$$ H. The current in the first coil is $$i = i_0 \sin\omega t$$ with $$i_0 = 5$$ A and $$\omega = 50\pi$$ rad/s.

The induced EMF in the second coil:

$$\varepsilon = -M\frac{di}{dt} = -M i_0 \omega \cos\omega t$$

Maximum EMF:

$$\varepsilon_{max} = M i_0 \omega = 0.002 \times 5 \times 50\pi = 0.5\pi$$ V $$= \frac{\pi}{2}$$ V

So $$\frac{\pi}{\alpha} = \frac{\pi}{2}$$, giving $$\alpha = 2$$.

The answer is $$\boxed{2}$$.