The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 $$\times$$ $$10^{-4}$$ Tesla . If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ $$\pi$$ cm.

The magnetic field inside a solenoid is given by:

$$ B = \mu_0 n I = \mu_0 \frac{N}{L} I $$

We are given that $$B = 2.9 \times 10^{-4}$$ T, $$N = 200$$, $$I = 0.29$$ A, and $$\mu_0 = 4\pi \times 10^{-7}$$ T·m/A.

We rearrange this formula to solve for $$L$$:

$$ L = \frac{\mu_0 N I}{B} = \frac{4\pi \times 10^{-7} \times 200 \times 0.29}{2.9 \times 10^{-4}} $$

It follows that

$$ L = \frac{4\pi \times 10^{-7} \times 58}{2.9 \times 10^{-4}} = \frac{4\pi \times 58 \times 10^{-7}}{2.9 \times 10^{-4}} $$

Simplifying further gives

$$ = \frac{4\pi \times 20 \times 10^{-7}}{10^{-4}} = 4\pi \times 20 \times 10^{-3} = 80\pi \times 10^{-3} \text{ m} $$

Finally,

$$ = 8\pi \text{ cm} $$

Therefore, the answer is 8.