A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is 5) its radius? (The value of o is 4 × 10-7 N/A2.)

 To calculate the radius of the solenoid, we can use the formula for the self-inductance (L) of a solenoid which is given by:

\[ L = \mu_0 (N^2 A) / l \]

Where: - \( L \) is the self-inductance, - \( \mu_0 \) is the permeability of free space (\( 4 \pi \times 10^{-7} \, \text{N/A}^2 \)), - \( N \) is the number of turns in the solenoid, - \( A \) is the cross-sectional area of the solenoid, and - \( l \) is the length of the solenoid.

First, we'll rearrange the formula to solve for the area \( A \):

\[ A = \frac{L \cdot l}{\mu_0 \cdot N^2} \]

Now we can plug in the values given in the problem:

\[ A = \frac{25.0 \times 10^{-3} \, \text{H} \cdot 70.0 \times 10^{-2} \, \text{m}}{4 \pi \times 10^{-7} \, \text{N/A}^2 \cdot (3000.0)^2} \]

\[ A = \frac{1.75}{4 \pi \times 9 \times 10^{6}} \, \text{m}^2 \]

\[ A \approx \frac{1.75}{113097} \, \text{m}^2 \]

\[ A \approx 0.00001547 \, \text{m}^2 \]

Now, since the area \( A \) of a circle is given by \( A = \pi r^2 \), where \( r \) is the radius, we can solve for \( r \):

\[ r = \sqrt{A / \pi} \]

\[ r = \sqrt{0.00001547 \, \text{m}^2 / \pi} \]

\[ r \approx \sqrt{0.00000492} \, \text{m} \]

\[ r \approx 0.00222 \, \text{m} \]

\[ r \approx 2.22 \, \text{mm} \]

Therefore, the radius of the solenoid is approximately 2.22 mm.