Two persons pull a wire towards themselves. Each person exerts a force of $$200$$ N on the wire. Young's modulus of the material of wire is $$1 \times 10^{11}$$ N m$$^{-2}$$. Original length of the wire is $$2$$ m and the area of cross section is $$2$$ cm$$^2$$. The wire will extend in length by ______ $$\mu$$m.

Given: $$F = 200$$ N, $$Y = 1 \times 10^{11}$$ N/m², $$L = 2$$ m, $$A = 2$$ cm² = $$2 \times 10^{-4}$$ m².

When two persons pull a wire from both ends, each exerting 200 N, the tension in the wire is 200 N (not 400 N, since the forces are in opposite directions maintaining equilibrium).

Using Young's modulus formula: $$Y = \frac{FL}{A\Delta L}$$

$$\Delta L = \frac{FL}{AY} = \frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}} = \frac{400}{2 \times 10^{7}} = 2 \times 10^{-5} \text{ m} = 20 \text{ }\mu\text{m}$$

The answer is $$\boxed{20}$$ μm.