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.

Explanation

To solve the problem step by step, we will calculate the permeability and susceptibility of the iron bar using the given data.

### Step 1: Understand the given data
We have the following values:
- Magnetizing field intensity (H) = 1500 A/m
- Magnetic flux (Φ) = 2.4 × 10^-5 Wb
- Cross-sectional area (A) = 0.5 cm² = 0.5 × 10^-4 m² (conversion from cm² to m²)

### Step 2: Calculate the magnetic field (B)
The magnetic field (B) can be calculated using the formula:
\[ B = \frac{\Phi}{A} \]

Substituting the values:
\[ B = \frac{2.4 \times 10^{-5} \text{ Wb}}{0.5 \times 10^{-4} \text{ m}^2} \]

Calculating:
\[ B = \frac{2.4 \times 10^{-5}}{0.5 \times 10^{-4}} = 4.8 \times 10^{-1} \text{ T} \]

### Step 3: Calculate the permeability (μ)
The permeability (μ) can be calculated using the formula:
\[ \mu = \frac{B}{H} \]

Substituting the values:
\[ \mu = \frac{4.8 \times 10^{-1} \text{ T}}{1500 \text{ A/m}} \]

Calculating:
\[ \mu = 3.2 \times 10^{-4} \text{ H/m} \]

### Step 4: Calculate the relative permeability (μ_r)
The relative permeability (μ_r) is calculated using:
\[ \mu_r = \frac{\mu}{\mu_0} \]
where \( \mu_0 = 4\pi \times 10^{-7} \text{ H/m} \).

Substituting the values:
\[ \mu_r = \frac{3.2 \times 10^{-4}}{4\pi \times 10^{-7}} \]

Calculating:
\[ \mu_r \approx \frac{3.2 \times 10^{-4}}{1.2566 \times 10^{-6}} \approx 254 \]

### Step 5: Calculate the magnetic susceptibility (χ)
The magnetic susceptibility (χ) can be calculated using the relationship:
\[ \chi = \mu_r - 1 \]

Substituting the value of μ_r:
\[ \chi = 254 - 1 = 253 \]

### Final Answers
- Permeability (μ) = \( 3.2 \times 10^{-4} \text{ H/m} \)
- Susceptibility (χ) = \( 253 \)