Answer :
Answer:
Given data:
- Length of the solenoid, \( l = 0.711 \) m
- Radius of the solenoid, \( r = 0.0334 \) m
- Current through the solenoid, \( I = 4.06 \) A
- Stored energy in the solenoid, \( U = 4460 \) J
1. Calculate the cross-sectional area \( A \) of the solenoid:
The cross-sectional area \( A \) of a cylinder is given by \( A = \pi r^2 \).
\( A = \pi \times (0.0334)^2 \)
\( A \approx 0.00351 \) \( \text{m}^2 \)
2. Use the formula for stored energy in a solenoid:
The energy stored \( U \) in a solenoid is given by:
\[ U = \frac{1}{2} \mu_0 n^2 A l I^2 \]
Rearrange to solve for \( n \):
\[ n = \sqrt{\frac{2U}{\mu_0 A l I^2}} \]
3. Substitute the given values into the formula:
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A
- Length of the solenoid, \( l = 0.711 \) m
- Current through the solenoid, \( I = 4.06 \) A
- Cross-sectional area, \( A \approx 0.00351 \) \( \text{m}^2 \)
- Stored energy, \( U = 4460 \) J
4. Calculate \( n \):
\[ n = \sqrt{\frac{2 \times 4460}{4\pi \times 10^{-7} \times 0.00351 \times 0.711 \times (4.06)^2}} \]
5. Simplify and solve for \( n \):
Plug in the values and perform the calculation:
\[ n = \sqrt{\frac{8920}{5.585 \times 10^{-9}}} \]
\[ n = \sqrt{1.597 \times 10^{12}} \]
\[ n \approx 1.264 \times 10^6 \, \text{turns/m} \]
Therefore, the number of turns per unit length \( n \) of the solenoid is approximately \( \boxed{1.26 \times 10^6} \).
