The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are `1 Wbm^(-2)` and `150Am^(-1)` respectively. The relative permeability of iron is : `(mu_(0)=4pixx10^(-7)"henry"//m)`

Options

A. `(10^(5))/(6 pi)`
B. `(10^(5))/(pi)`
C. `(10^(5))/(3 pi)`
D. `(10^(5))/(5 pi)`

Explanation

To find the relative permeability of iron (\( \mu_r \)), we can use the relationship between the magnetic induction (\( B \)), the magnetic field intensity (\( H \)), and the permeability of free space (\( \mu_0 \)).

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic induction, \( B = 1 \, \text{Wb/m}^2 \)
- Magnetic field intensity, \( H = 150 \, \text{A/m} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)

2. **Use the formula for magnetic induction:**
The relationship between \( B \), \( \mu_r \), \( \mu_0 \), and \( H \) is given by:
\[ B = \mu_r \mu_0 H \]

3. **Rearranging the formula to find \( \mu_r \):**
We can rearrange the formula to solve for the relative permeability \( \mu_r \):
\[ \mu_r = \frac{B}{\mu_0 H} \]

4. **Substituting the known values:**
Substitute the values of \( B \), \( \mu_0 \), and \( H \) into the equation:
\[ \mu_r = \frac{1 \, \text{Wb/m}^2}{(4\pi \times 10^{-7} \, \text{H/m})(150 \, \text{A/m})} \]

5. **Calculating the denominator:**
First, calculate \( \mu_0 H \):
\[ \mu_0 H = (4\pi \times 10^{-7})(150) = 600\pi \times 10^{-7} \, \text{H/m} \]

6. **Substituting back into the equation:**
Now substitute back into the equation for \( \mu_r \):
\[ \mu_r = \frac{1}{600\pi \times 10^{-7}} \]

7. **Simplifying the expression:**
To simplify:
\[ \mu_r = \frac{10^7}{600\pi} \]

8. **Final calculation:**
Now, calculate \( \frac{10^7}{600\pi} \):
\[ \mu_r \approx \frac{10^7}{1884.96} \approx 5295.5 \]

### Conclusion:
The relative permeability of iron is approximately \( \mu_r \approx 5295.5 \).