The horizontal component of earth's magnetic field at a place is $$3.5 \times 10^{-5} \text{ T}$$. A very long straight conductor carrying current of $$\sqrt{2} \text{ A}$$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is ________ $$\times 10^{-6} \text{ N m}^{-1}$$.

The force per unit length on a current-carrying conductor in a magnetic field is given by $$\frac{F}{L} = I \times B \times \sin\theta$$, where $$\theta$$ is the angle between the direction of current and the magnetic field. The horizontal component of Earth’s magnetic field is $$B_H = 3.5 \times 10^{-5} \text{ T}$$ directed towards geographic North, and the current is $$I = \sqrt{2} \text{ A}$$ flowing from South-East to North-West. The angle between the North-West direction and the North direction is $$45°$$, so the angle between the current direction (SE to NW) and $$B_H$$ (pointing North) is $$\theta = 45°$$.

Substituting into the formula gives

$$\frac{F}{L} = I \times B_H \times \sin 45°$$

$$\frac{F}{L} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}}$$

$$\frac{F}{L} = 3.5 \times 10^{-5} = 35 \times 10^{-6} \text{ N/m}$$

Therefore, the correct answer is $$35$$.