A magnetising field of 5000 A/m produces a magnetic flux of `5xx10^(-5)` weber in an iron rod. If the area of cross section of the rod is `0.5cm^(2)`, then the permeability of the rod will be

Options

A. `1xx10^(-3)`
B. `2xx10^(-4)`
C. `3xx10^(-5)`
D. `4xx10^(-6)`

Explanation

To find the permeability of the iron rod, we can follow these steps:

### Step 1: Write down the given values
- Magnetizing field intensity (H) = 5000 A/m
- Magnetic flux (Φ) = \(5 \times 10^{-5}\) Weber
- Area of cross-section (A) = 0.5 cm² = \(0.5 \times 10^{-4}\) m² (conversion from cm² to m²)

### Step 2: Use the formula for magnetic flux
The magnetic flux (Φ) through a surface is given by the equation:
\[ \Phi = B \cdot A \]
Where B is the magnetic flux density. Rearranging this gives:
\[ B = \frac{\Phi}{A} \]

### Step 3: Calculate B
Substituting the values of Φ and A into the equation:
\[ B = \frac{5 \times 10^{-5}}{0.5 \times 10^{-4}} \]

### Step 4: Simplify the calculation
Calculating the above expression:
\[ B = \frac{5 \times 10^{-5}}{0.5 \times 10^{-4}} = \frac{5}{0.5} \times 10^{-5 + 4} = 10 \times 10^{-1} = 1 \, \text{T} \]

### Step 5: Use the relationship between B, H, and μ
The magnetic flux density (B) is related to the magnetizing field (H) and the permeability (μ) by the equation:
\[ B = \mu \cdot H \]
Rearranging this gives:
\[ \mu = \frac{B}{H} \]

### Step 6: Calculate μ
Substituting the values of B and H into the equation:
\[ \mu = \frac{1}{5000} \]

### Step 7: Simplify the calculation
Calculating the above expression:
\[ \mu = 2 \times 10^{-4} \, \text{H/m} \]

### Final Answer
The permeability of the iron rod is:
\[ \mu = 2 \times 10^{-4} \, \text{H/m} \]

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