A magnetising field of `2xx10^(3)Am^(-1)` produces a magnetic flux density of `8piT` in an iron rod. The relative permeability of the rod will be

Options

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

Explanation

To find the relative permeability of the iron rod, we can use the formula for relative permeability (\( \mu_r \)) given by:

\[ \mu_r = \frac{B}{\mu_0 H} \]

where:
- \( B \) is the magnetic flux density,
- \( \mu_0 \) is the permeability of free space, and
- \( H \) is the magnetizing field intensity.

### Step 1: Identify the given values
- The magnetizing field intensity \( H = 2 \times 10^3 \, \text{A/m} \)
- The magnetic flux density \( B = 8\pi \, \text{T} \)
- The permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

### Step 2: Substitute the values into the formula
We can now substitute the values into the formula for relative permeability:

\[ \mu_r = \frac{8\pi}{(4\pi \times 10^{-7})(2 \times 10^3)} \]

### Step 3: Simplify the equation
First, we can cancel \( \pi \) from the numerator and denominator:

\[ \mu_r = \frac{8}{4 \times 10^{-7} \times 2 \times 10^3} \]

### Step 4: Calculate the denominator
Now, calculate the denominator:

\[ 4 \times 2 = 8 \]
\[ 10^{-7} \times 10^3 = 10^{-4} \]
Thus, the denominator becomes:

\[ 8 \times 10^{-4} \]

### Step 5: Substitute back into the equation
Now substitute back into the equation for \( \mu_r \):

\[ \mu_r = \frac{8}{8 \times 10^{-4}} = \frac{8}{8} \times 10^4 = 1 \times 10^4 \]

### Final Result
Therefore, the relative permeability \( \mu_r \) of the rod is:

\[ \mu_r = 10^4 \]