The inductance \( L \) of a solenoid is given by \( L = \frac{\mu_0 N^2 A}{l} \). We are given \( L = 0.016 \, \text{H} \), \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \), and \( l = 0.81 \, \text{m} \). The cross-sectional area \( A \) is \( \pi r^2 \).
First, calculate the cross-sectional area \( A \): \[ A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4} \, \text{m}^2 \] Substitute the known values into the inductance formula: \[ 0.016 = \frac{4\pi \times 10^{-7} \times N^2 \times 4\pi \times 10^{-4}}{0.81} \] To solve for \( N^2 \), rearrange the formula: \[ N^2 = \frac{0.016 \times 0.81}{4\pi \times 10^{-7} \times 4\pi \times 10^{-4}} \] \[ N^2 = \frac{0.01296}{16\pi^2 \times 10^{-11}} \] \[ N^2 = \frac{0.01296}{1.57896 \times 10^{-9}} \] \[ N^2 \approx 8219479.5 \] Now, find \( N \) by taking the square root: \[ N \approx \sqrt{8219479.5} \] \[ N \approx 2866 \]
Therefore, the solenoid has approximately 2866 turns.