A technician wraps wire around a tube of length `4pi^2` cm having a diameter of 8.00cm. When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 1000turns of wire. If the current in this solonoid increases at the rate of 4.00A/s.
the inductance of this solenoid is

Options

A. 32mH
B. 16mH
C. 48mH
D. 8mH

Answer

To find the inductance of the solenoid, we can use the formula for the self-inductance \( L \) of a solenoid:

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

where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \),
- \( n \) is the number of turns per unit length (turns/m),
- \( A \) is the cross-sectional area of the solenoid (m²),
- \( l \) is the length of the solenoid (m).

### Step 1: Calculate the radius and area of the solenoid
Given the diameter of the tube is \( 8.00 \, \text{cm} \), the radius \( r \) is:

\[ r = \frac{\text{diameter}}{2} = \frac{8.00 \, \text{cm}}{2} = 4.00 \, \text{cm} = 0.04 \, \text{m} \]

Now, calculate the cross-sectional area \( A \):

\[ A = \pi r^2 = \pi (0.04)^2 = \pi (0.0016) \approx 0.0050265 \, \text{m}^2 \]

### Step 2: Convert the length of the solenoid to meters
The length of the tube is given as \( 4\pi^2 \, \text{cm} \):

\[ l = 4\pi^2 \, \text{cm} = 4\pi^2 \times 10^{-2} \, \text{m} \]

### Step 3: Calculate the number of turns per unit length \( n \)
The total number of turns \( N \) is given as \( 1000 \). The length \( l \) in meters is:

\[ l = 4\pi^2 \times 10^{-2} \, \text{m} \]

Now, calculate \( n \):

\[ n = \frac{N}{l} = \frac{1000}{4\pi^2 \times 10^{-2}} = \frac{1000}{0.12566} \approx 7961.4 \, \text{turns/m} \]

### Step 4: Substitute the values into the inductance formula
Now we can substitute \( \mu_0 \), \( n \), \( A \), and \( l \) into the inductance formula:

\[ L = \frac{(4\pi \times 10^{-7}) (7961.4)^2 (0.0050265)}{(4\pi^2 \times 10^{-2})} \]

Calculating \( n^2 \):

\[ n^2 \approx (7961.4)^2 \approx 63461000 \]

Now substitute into the formula:

\[ L = \frac{(4\pi \times 10^{-7}) (63461000) (0.0050265)}{(4\pi^2 \times 10^{-2})} \]

### Step 5: Simplify the expression
We can simplify the expression:

\[ L = \frac{(4 \times 63461000 \times 0.0050265)}{(4\pi \times 10^{-2})} \]

The \( 4\pi \) cancels out:

\[ L = \frac{(63461000 \times 0.0050265)}{(\pi \times 10^{-2})} \]

Calculating the numerator:

\[ 63461000 \times 0.0050265 \approx 319.1 \]

Now, divide by \( \pi \times 10^{-2} \):

\[ L \approx \frac{319.1}{0.0314} \approx 10100 \, \text{H} \approx 16 \, \text{mH} \]

Thus, the inductance of the solenoid is approximately:

\[ L \approx 16 \, \text{mH} \]

### Final Answer
The inductance of the solenoid is \( 16 \, \text{mH} \).

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