An iron rod of `0*2cm^2` cross-sectional area is subjected to a magnetising field of `1200Am^-1`. The suscaptibility of iron is `599`. Find the permeability and the magnetic flux produced.

Options

A. `0.904Wb`
B. `1.81xx10^(-5)Wb`
C. `0.904xx10^(-5)Wb`
D. `5.43xx10^(-5)Wb`

Explanation

To solve the problem step by step, we will find the permeability and the magnetic flux produced in the iron rod.

### Step 1: Convert the cross-sectional area from cm² to m²
The given cross-sectional area \( A \) is \( 0.2 \, \text{cm}^2 \).

\[ A = 0.2 \, \text{cm}^2 = 0.2 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-5} \, \text{m}^2 \]

**Hint:** Remember to convert cm² to m² by multiplying by \( 10^{-4} \).

### Step 2: Use the formula for permeability
The permeability \( \mu \) can be calculated using the formula:

\[ \mu = \mu_0 (1 + \chi) \]

Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (the permeability of free space)
- \( \chi = 599 \) (the susceptibility of iron)

Substituting the values:

\[ \mu = 4\pi \times 10^{-7} \times (1 + 599) \]

Calculating \( 1 + 599 = 600 \):

\[ \mu = 4\pi \times 10^{-7} \times 600 \]

Calculating \( 4\pi \approx 12.566 \):

\[ \mu \approx 12.566 \times 10^{-7} \times 600 \]

Calculating:

\[ \mu \approx 7.539 \times 10^{-4} \, \text{H/m} = 7539 \times 10^{-7} \, \text{H/m} \]

### Step 3: Calculate the magnetic field \( B \)
The magnetic field \( B \) can be calculated using the formula:

\[ B = \mu H \]

Where:
- \( H = 1200 \, \text{A/m} \)

Substituting the values:

\[ B = (7.539 \times 10^{-7} \, \text{H/m}) \times (1200 \, \text{A/m}) \]

Calculating:

\[ B \approx 0.90468 \, \text{T} \approx 0.905 \, \text{Wb/m}^2 \]

### Final Answers
- The permeability \( \mu \) is approximately \( 7.539 \times 10^{-4} \, \text{H/m} \) or \( 7539 \times 10^{-7} \, \text{H/m} \).
- The magnetic flux density \( B \) is approximately \( 0.905 \, \text{Wb/m}^2 \).