The magnetic intensity at the centre of a long current carrying solenoid is found to be $$1.6 \times 10^3$$ A m$$^{-1}$$. If the number of turns is 8 per cm, then the current flowing through the solenoid is ______ A.

We are given that the magnetic intensity at the centre of a long current-carrying solenoid is $$H = 1.6 \times 10^3$$ A/m, and the number of turns is $$n = 8$$ turns/cm.

Convert turns per cm to turns per metre: $$ n = 8 \text{ turns/cm} = 8 \times 100 = 800 \text{ turns/m} $$

Use the formula for magnetic intensity in a solenoid. For a long solenoid, the magnetic intensity inside is:

$$ H = nI $$

where $$n$$ is turns per unit length and $$I$$ is the current.

Solve for current: $$ I = \frac{H}{n} = \frac{1.6 \times 10^3}{800} = \frac{1600}{800} = 2 \text{ A} $$

The current flowing through the solenoid is 2 A.