Two metallic wires of identical dimensions are connected in series. If $$\sigma_1$$ and $$\sigma_2$$ are the conductivities of these wires respectively, the effective conductivity of the combination is:

We have two metallic wires of identical dimensions (same length $$l$$ and same cross-sectional area $$A$$) connected in series, with conductivities $$\sigma_1$$ and $$\sigma_2$$ respectively.

The resistance of each wire is given by $$R = \frac{l}{\sigma A}$$, since resistivity $$\rho = \frac{1}{\sigma}$$. So the resistances are $$R_1 = \frac{l}{\sigma_1 A}$$ and $$R_2 = \frac{l}{\sigma_2 A}$$.

In series, the total resistance is $$R_{total} = R_1 + R_2 = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A}\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right) = \frac{l}{A} \cdot \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$.

The combination of two wires in series can be treated as a single wire of total length $$2l$$ and the same cross-sectional area $$A$$. If the effective conductivity is $$\sigma_{eff}$$, then $$R_{total} = \frac{2l}{\sigma_{eff} A}$$.

Equating the two expressions: $$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \cdot \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$.

Simplifying: $$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$, which gives $$\sigma_{eff} = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$$.

Hence, the correct answer is Option B.