A rod of cross sectional area `10cm^(2)` is placed with its length parallel to a magnetic field of intensity 1000 A/M the flux through the rod is `10^(4)` webers. Then the permeability of material of rod is

Options

A. `10^(4)wb//Am`
B. `10^(3)wb//Am`
C. `10^(2)wb//Am`
D. 10 wb/Am

Explanation

To find the permeability of the material of the rod, we will follow these steps:

### Step 1: Identify the given values
- Cross-sectional area of the rod, \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \)
- Magnetic field intensity, \( H = 1000 \, \text{A/m} \)
- Magnetic flux through the rod, \( \Phi = 10^4 \, \text{Wb} \)

### Step 2: Use the formula for magnetic flux
The magnetic flux \( \Phi \) through a surface is given by the formula:
\[ \Phi = B \cdot A \]
where \( B \) is the magnetic flux density.

### Step 3: Relate magnetic flux density \( B \) to magnetic field intensity \( H \)
The relationship between magnetic flux density \( B \), magnetic field intensity \( H \), and permeability \( \mu \) is given by:
\[ B = \mu H \]

### Step 4: Substitute \( B \) in the flux formula
From the equation of magnetic flux:
\[ \Phi = B \cdot A = (\mu H) \cdot A \]

### Step 5: Rearrange to find permeability \( \mu \)
Rearranging the equation gives:
\[ \mu = \frac{\Phi}{H \cdot A} \]

### Step 6: Substitute the known values
Now substitute the values we have:
\[ \mu = \frac{10^4 \, \text{Wb}}{1000 \, \text{A/m} \cdot 10 \times 10^{-4} \, \text{m}^2} \]

### Step 7: Calculate the denominator
Calculate the denominator:
\[ 1000 \, \text{A/m} \cdot 10 \times 10^{-4} \, \text{m}^2 = 1000 \cdot 10^{-3} = 1 \, \text{Wb} \]

### Step 8: Calculate permeability \( \mu \)
Now substituting back:
\[ \mu = \frac{10^4}{1} = 10^4 \, \text{Wb/A·m} \]

### Step 9: Final answer
Thus, the permeability of the material of the rod is:
\[ \mu = 10^4 \, \text{Wb/A·m} \]