A square loop of side $$10$$ cm and resistance $$0.7 \Omega$$ is placed vertically in the east-west plane. A uniform magnetic field of $$0.20$$ T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $$1$$ s at a steady rate. Then, the magnitude of induced emf is $$\sqrt{x} \times 10^{-3}$$ V. The value of $$x$$ is _______.

Square loop in east-west plane, B in NE direction. B makes 45° with the plane of the loop.

$$\Phi = BA\cos 45° = 0.20 \times 0.01 \times \frac{1}{\sqrt{2}} = \frac{0.002}{\sqrt{2}}$$

$$\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.002/\sqrt{2}}{1} = \frac{0.002}{\sqrt{2}} = \sqrt{2} \times 10^{-3}$$ V.

$$\sqrt{x} \times 10^{-3} = \sqrt{2} \times 10^{-3}$$. $$x = 2$$.

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