An iron rod of `0.2cm^2` cross-Sectional area is subjected to a magnetising field of `1200A//m`. The susceptibility of iron is 599. The permeability will be

Options

A. `7.9xx10^5T m//A`
B. `8.0xx10^22T m//A`
C. `7.5xx10^-4T m//A`
D. `8xx10^-5T m//A`

Answer

To find the permeability of the iron rod given the susceptibility, we can use the formula for permeability in terms of the permeability of free space and the susceptibility of the material.

### Step-by-Step Solution:

1. **Identify the given values:**
- Cross-sectional area of the rod: \( A = 0.2 \, \text{cm}^2 \) (not needed for permeability calculation)
- Magnetizing field strength: \( H = 1200 \, \text{A/m} \) (not needed for permeability calculation)
- Susceptibility of iron: \( \chi = 599 \)

2. **Recall the formula for permeability:**
The permeability \( \mu \) of a material is given by the formula:
\[ \mu = \mu_0 (1 + \chi) \]
where \( \mu_0 \) is the permeability of free space.

3. **Substitute the value of \( \mu_0 \):**
The permeability of free space \( \mu_0 \) is given by:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \]

4. **Calculate \( 1 + \chi \):**
\[ 1 + \chi = 1 + 599 = 600 \]

5. **Substitute \( \mu_0 \) and \( 1 + \chi \) into the permeability formula:**
\[ \mu = 4\pi \times 10^{-7} \times 600 \]

6. **Calculate \( \mu \):**
First, calculate \( 4\pi \):
\[ 4\pi \approx 4 \times 3.14 = 12.56 \]
Now, substitute this back into the equation:
\[ \mu = 12.56 \times 10^{-7} \times 600 \]
\[ \mu = 12.56 \times 600 \times 10^{-7} \]
\[ \mu = 7536 \times 10^{-7} \, \text{H/m} \]
\[ \mu = 7.536 \times 10^{-4} \, \text{H/m} \]

7. **Final Answer:**
The permeability of the iron rod is:
\[ \mu = 7.536 \times 10^{-4} \, \text{H/m} \]