A `470` - turn solenoid having a length of 25 cm and a diameter of 10 cm carries a current of `0.29`A. Calculate the magnitude of the magnetic field `vecB` inside the solenoid.

Answer

To calculate the magnitude of the magnetic field \( \vec{B} \) inside a solenoid, we can use the formula:

\[ B = \mu_0 \frac{N I}{L} \]

where:
- \( B \) is the magnetic field inside the solenoid,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( N \) is the number of turns in the solenoid,
- \( I \) is the current flowing through the solenoid,
- \( L \) is the length of the solenoid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \( N = 470 \)
- Length of the solenoid, \( L = 25 \, \text{cm} = 0.25 \, \text{m} \) (convert cm to m)
- Current, \( I = 0.29 \, \text{A} \)

2. **Substitute the values into the formula:**
- Use the value of \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
- Substitute \( N \), \( I \), and \( L \) into the formula:

\[ B = (4\pi \times 10^{-7}) \frac{470 \times 0.29}{0.25} \]

3. **Calculate the numerator:**
- First, calculate \( 470 \times 0.29 \):

\[ 470 \times 0.29 = 136.3 \]

4. **Calculate the fraction:**
- Now, divide by the length \( L \):

\[ \frac{136.3}{0.25} = 545.2 \]

5. **Multiply by \( \mu_0 \):**
- Now multiply by \( 4\pi \times 10^{-7} \):

\[ B = (4\pi \times 10^{-7}) \times 545.2 \]

- Using \( \pi \approx 3.14 \):

\[ B \approx (4 \times 3.14 \times 10^{-7}) \times 545.2 \approx 6.9 \times 10^{-4} \, \text{T} \]

6. **Convert to milliTesla:**
- Since \( 1 \, \text{T} = 1000 \, \text{mT} \):

\[ B \approx 0.69 \, \text{mT} \]

### Final Answer:
The magnitude of the magnetic field \( \vec{B} \) inside the solenoid is approximately \( 0.69 \, \text{mT} \).

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