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 $\_\_\_\_\_ \pi \mathrm{cm}.$

We are given the following parameters for a solenoid:

Number of turns: $N = 200$

Radius: $r = 10$ cm (not needed for calculation)

Magnetic field: $B = 2.9 \times 10^{-4}$ T

Current: $I = 0.29$ A

Length of solenoid: $L$ (to be found in terms of $\pi$)

Step 1: Use the Magnetic Field Formula for a Solenoid

The magnetic field inside a solenoid is given by:

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

where:

$\mu_0 = 4\pi \times 10^{-7}$ T·m/A (permeability of free space),

$N$ is the total number of turns,

$L$ is the length of the solenoid,

$I$ is the current.

Rearranging for $L$:

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

Step 2: Substitute Given Values

$$L = \left( 4\pi \times 10^{-7} \right) \times \frac{200 \times 0.29}{2.9 \times 10^{-4}}$$

$$L = (4\pi \times 10^{-7}) \times (2 \times 10^5)$$

 $$L = 8\pi \times 10^{-2} \text{ m}$$ 

$$L = 8\pi \times 10 \text{ cm}$$ 

$$L = 8\pi \text{ cm}$$ 

Final Answer:

$\mathbf{8}$