An iron specimen has relative permeability of `600` when placed in uniform magnetic field of intensity `110 amp//m`. Then the magnetic flux density inside is……tesls.

Options

A. `18.29xx10^(-3)`
B. `8.29xx10^(-2)`
C. `66xx10^(3)`
D. `7.536xx10^(-4)`

Explanation

To find the magnetic flux density \( B \) inside the iron specimen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Relative permeability \( \mu_r = 600 \)
- Magnetic field intensity \( H = 110 \, \text{A/m} \)

2. **Understand the Relationship**:
- The magnetic flux density \( B \) is related to the magnetic field intensity \( H \) and the permeability \( \mu \) of the medium by the equation:
\[ B = \mu H \]

3. **Calculate the Permeability of the Medium**:
- The permeability \( \mu \) of the medium can be expressed in terms of the relative permeability \( \mu_r \) and the permeability of free space \( \mu_0 \):
\[ \mu = \mu_r \mu_0 \]
- The permeability of free space \( \mu_0 \) is given by:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \]

4. **Substitute Values**:
- Now substitute the values into the equation for \( \mu \):
\[ \mu = 600 \times (4\pi \times 10^{-7}) \, \text{T m/A} \]

5. **Calculate \( \mu \)**:
- Calculate \( \mu \):
\[ \mu = 600 \times (4 \times 3.14 \times 10^{-7}) = 600 \times 1.256 \times 10^{-6} \approx 7.536 \times 10^{-4} \, \text{T m/A} \]

6. **Calculate Magnetic Flux Density \( B \)**:
- Now substitute \( \mu \) and \( H \) into the equation for \( B \):
\[ B = \mu H = (7.536 \times 10^{-4}) \times 110 \]
- Calculate \( B \):
\[ B \approx 8.2896 \times 10^{-2} \, \text{T} \approx 0.0829 \, \text{T} \]

7. **Final Answer**:
- The magnetic flux density inside the iron specimen is approximately:
\[ B \approx 0.0829 \, \text{T} \]