A solenoid of radius 2.50cm has 400turns and a length of 20.0cm. The inductance (L) of this solenoid is ________.

Options

A. 4mH
B. 5mH
C. 3mH
D. 2mH

Answer

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

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

where:
- \(L\) = inductance in henries (H)
- \(\mu_0\) = permeability of free space = \(4\pi \times 10^{-7} \, \text{H/m}\)
- \(N\) = number of turns in the solenoid
- \(A\) = cross-sectional area of the solenoid
- \(l\) = length of the solenoid in meters

**Step 1: Convert the given dimensions to SI units.**
- Radius \(r = 2.50 \, \text{cm} = 2.50 \times 10^{-2} \, \text{m}\)
- Length \(l = 20.0 \, \text{cm} = 20.0 \times 10^{-2} \, \text{m} = 0.20 \, \text{m}\)
- Number of turns \(N = 400\)

**Step 2: Calculate the cross-sectional area \(A\) of the solenoid.**
\[ A = \pi r^2 = \pi (2.50 \times 10^{-2})^2 \]
\[ A = \pi (6.25 \times 10^{-4}) \approx 1.9635 \times 10^{-3} \, \text{m}^2 \]

**Step 3: Substitute the values into the inductance formula.**
\[ L = \frac{(4\pi \times 10^{-7}) (400)^2 (1.9635 \times 10^{-3})}{0.20} \]

**Step 4: Calculate \(N^2\).**
\[ N^2 = 400^2 = 160000 \]

**Step 5: Substitute \(N^2\) back into the formula.**
\[ L = \frac{(4\pi \times 10^{-7}) (160000) (1.9635 \times 10^{-3})}{0.20} \]

**Step 6: Calculate the numerator.**
\[ 4\pi \times 10^{-7} \times 160000 \approx 2.0106 \times 10^{-1} \]
\[ 2.0106 \times 10^{-1} \times 1.9635 \times 10^{-3} \approx 3.951 \times 10^{-4} \]

**Step 7: Divide by the length.**
\[ L = \frac{3.951 \times 10^{-4}}{0.20} \approx 1.9755 \times 10^{-3} \, \text{H} \]

**Step 8: Convert to millihenries.**
\[ L \approx 1.98 \times 10^{-3} \, \text{H} = 1.98 \, \text{mH} \approx 2 \, \text{mH} \]

Thus, the inductance \(L\) of the solenoid is approximately \(2 \, \text{mH}\).