Find B. Given n = 1000 turns/m `mu_r = 500` `mu_0 = 4 pi x 10^-7 Tm/A` I = 10 A

Options

A. `2 pi Tesla`
B. `3 pi Tesla`
C. `5 pi Tesla`
D. `7 pi Tesla`

Answer

To find the magnetic field \( B \) inside a solenoid, we can use the formula:

\[ B = \mu_0 \mu_r n I \]

Where:
- \( \mu_0 \) is the permeability of free space, given as \( 4\pi \times 10^{-7} \, \text{Tm/A} \)
- \( \mu_r \) is the relative permeability of the medium, given as \( 500 \)
- \( n \) is the number of turns per unit length, given as \( 1000 \, \text{turns/m} \)
- \( I \) is the current flowing through the solenoid, given as \( 10 \, \text{A} \)

### Step-by-Step Solution:

1. **Identify the given values:**
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \)
- \( \mu_r = 500 \)
- \( n = 1000 \, \text{turns/m} \)
- \( I = 10 \, \text{A} \)

2. **Substitute the values into the formula:**
\[ B = (4\pi \times 10^{-7}) \times 500 \times 1000 \times 10 \]

3. **Calculate \( \mu_0 \mu_r n I \):**
- First, calculate \( \mu_0 \mu_r \):
\[ \mu_0 \mu_r = (4\pi \times 10^{-7}) \times 500 = 2000\pi \times 10^{-7} \]
- Now, multiply by \( n \) and \( I \):
\[ B = (2000\pi \times 10^{-7}) \times 1000 \times 10 \]

4. **Simplify the expression:**
- Combine the numbers:
\[ B = 2000 \times 1000 \times 10 \times \pi \times 10^{-7} = 20000000\pi \times 10^{-7} \]
- This can be rewritten as:
\[ B = 20\pi \times 10^{-1} = 2\pi \, \text{T} \]

5. **Final result:**
\[ B \approx 2\pi \, \text{T} \]

### Conclusion:
The magnetic field \( B \) inside the solenoid is approximately \( 2\pi \, \text{T} \).