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 self-induced emf in the solenoid is

Options

A. 32mV
B. 16mV
C. 64mV
D. 24mV

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the area of the cross-section of the solenoid.
The diameter of the tube is given as 8.00 cm. To find the radius, we divide the diameter by 2:
\[ \text{Radius} (r) = \frac{\text{Diameter}}{2} = \frac{8.00 \text{ cm}}{2} = 4.00 \text{ cm} = 0.04 \text{ m} \]
Now, we can calculate the area \( A \) of the cross-section using the formula:
\[ A = \pi r^2 = \pi (0.04)^2 = \pi (0.0016) \approx 0.0050265 \text{ m}^2 \]

### Step 2: Calculate the inductance \( L \) of the solenoid.
The formula for the inductance of a solenoid is given by:
\[ L = \mu_0 \frac{n^2 A}{l} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (permeability of free space),
- \( n = 1000 \) (number of turns),
- \( A \) is the area calculated in Step 1,
- \( l = 4\pi^2 \, \text{cm} = 0.04\pi^2 \, \text{m} \).

Substituting the values:
\[ L = (4\pi \times 10^{-7}) \frac{(1000)^2 (0.0050265)}{0.04\pi^2} \]
Calculating the values:
\[ L = (4\pi \times 10^{-7}) \frac{1000000 \times 0.0050265}{0.04\pi^2} \]
\[ L = (4 \times 10^{-7}) \frac{5026.5}{0.04\pi} \]
\[ L = (4 \times 10^{-7}) \frac{5026.5}{0.12566} \approx (4 \times 10^{-7}) \times 40000 \approx 16 \times 10^{-3} \text{ H} = 0.016 \text{ H} \]

### Step 3: Calculate the self-induced emf \( \mathcal{E} \).
The self-induced emf can be calculated using the formula:
\[ \mathcal{E} = L \frac{di}{dt} \]
Given that \( \frac{di}{dt} = 4.00 \, \text{A/s} \):
\[ \mathcal{E} = 0.016 \times 4 = 0.064 \text{ V} = 64 \text{ mV} \]

### Final Answer:
The self-induced emf in the solenoid is \( 64 \, \text{mV} \).

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