An iron rod is placed parallel to magnetic field of intensity `2000 Am^-1`. The magnetic flux through the rod is `6xx 10^-1` Wb and its cross-sectional area is `3 cm^2`. The magnetic permeability of the rod in Wb `A^-1 m^-1` is

Options

A. `10^(-1)`
B. `10^(-2)`
C. `10^(-3)`
D. `10^(-4)`

Answer

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

### Step 1: Identify the given values
- Magnetic field intensity, \( H = 2000 \, \text{A/m} \)
- Magnetic flux, \( \Phi = 6 \times 10^{-1} \, \text{Wb} \)
- Cross-sectional area, \( A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2 \)

### Step 2: Use the relationship between magnetic flux, magnetic field (B), and area
The magnetic flux \( \Phi \) is given by the formula:
\[ \Phi = B \cdot A \]
From this, we can express \( B \):
\[ B = \frac{\Phi}{A} \]

### Step 3: Calculate the magnetic field \( B \)
Substituting the values of \( \Phi \) and \( A \):
\[ B = \frac{6 \times 10^{-1} \, \text{Wb}}{3 \times 10^{-4} \, \text{m}^2} \]

### Step 4: Perform the calculation for \( B \)
\[ B = \frac{6 \times 10^{-1}}{3 \times 10^{-4}} = 2 \times 10^{3} \, \text{T} \]

### Step 5: Use the relationship between magnetic induction (B), magnetic field intensity (H), and permeability (μ)
The relationship is given by:
\[ B = \mu \cdot H \]
From this, we can express \( \mu \):
\[ \mu = \frac{B}{H} \]

### Step 6: Substitute the values of \( B \) and \( H \) to find \( \mu \)
Substituting the values we have:
\[ \mu = \frac{2 \times 10^{3} \, \text{T}}{2000 \, \text{A/m}} \]

### Step 7: Perform the calculation for \( \mu \)
\[ \mu = \frac{2 \times 10^{3}}{2 \times 10^{3}} = 1 \, \text{Wb/A/m} \]

### Final Answer
The magnetic permeability of the rod is:
\[ \mu = 1 \, \text{Wb/A/m} \]

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