A magnetising field of `1500 A//m` produces a flux of `2*4xx10^-5` weber in a bar of iron of cross-sectional area `0*5cm^2`. Calculate the permeability and susceptibility of the iron bar used.

Options

A. 245
B. 250
C. 252
D. 255

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Magnetizing field \( H = 1500 \, \text{A/m} \)
- Magnetic flux \( \Phi = 2.4 \times 10^{-5} \, \text{Wb} \)
- Cross-sectional area \( A = 0.5 \, \text{cm}^2 = 0.5 \times 10^{-4} \, \text{m}^2 \)

### Step 2: Calculate the magnetic field \( B \)
Using the formula for magnetic field \( B \):
\[ B = \frac{\Phi}{A} \]
Substituting the values:
\[ B = \frac{2.4 \times 10^{-5} \, \text{Wb}}{0.5 \times 10^{-4} \, \text{m}^2} \]
Calculating this gives:
\[ B = \frac{2.4 \times 10^{-5}}{0.5 \times 10^{-4}} = 4.8 \times 10^{-1} \, \text{T} \]

### Step 3: Calculate the permeability \( \mu \)
Using the formula for permeability \( \mu \):
\[ \mu = \frac{B}{H} \]
Substituting the values:
\[ \mu = \frac{4.8 \times 10^{-1} \, \text{T}}{1500 \, \text{A/m}} \]
Calculating this gives:
\[ \mu = 3.2 \times 10^{-4} \, \text{H/m} \]

### Step 4: Calculate the permeability relative \( \mu_r \)
Using the formula for relative permeability:
\[ \mu_r = \frac{\mu}{\mu_0} \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (the permeability of free space).
Substituting the values:
\[ \mu_r = \frac{3.2 \times 10^{-4}}{4\pi \times 10^{-7}} \]
Calculating this gives:
\[ \mu_r \approx 255 \]

### Step 5: Calculate the susceptibility \( \chi_m \)
Using the formula for susceptibility:
\[ \chi_m = \mu_r - 1 \]
Substituting the value of \( \mu_r \):
\[ \chi_m = 255 - 1 = 254 \]

### Final Results
- Permeability \( \mu_r \approx 255 \)
- Susceptibility \( \chi_m = 254 \)