The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are `1 Wbm^(-2)` and `150Am^(-1)` respectively. The relative permeability of iron is : `(mu_(0)=4pixx10^(-7)"henry"//m)`

Options

A. 25
B. `10^(5)/(3 pi)`
C. `10^(5)/pi`
D. `10^(5)/(5 pi)`

Answer

To find the relative permeability of iron (\( \mu_r \)), we can follow these steps:

### Step 1: Write down the formula for permeability
The permeability (\( \mu \)) of a material is given by the formula:
\[ \mu = \frac{B}{H} \]
where \( B \) is the magnetic induction and \( H \) is the intensity of the magnetic field.

### Step 2: Substitute the known values
From the problem, we know:
- \( B = 1 \, \text{Wbm}^{-2} \)
- \( H = 150 \, \text{Am}^{-1} \)

Substituting these values into the formula:
\[ \mu = \frac{1}{150} \]

### Step 3: Relate permeability to relative permeability
The permeability of a material is also related to the permeability of free space (\( \mu_0 \)) and the relative permeability (\( \mu_r \)) by the equation:
\[ \mu = \mu_0 \cdot \mu_r \]
Thus, we can express \( \mu_r \) as:
\[ \mu_r = \frac{\mu}{\mu_0} \]

### Step 4: Substitute \( \mu_0 \) into the equation
The permeability of free space (\( \mu_0 \)) is given as:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \]
Now substituting \( \mu \) and \( \mu_0 \) into the equation for \( \mu_r \):
\[ \mu_r = \frac{\frac{1}{150}}{4\pi \times 10^{-7}} \]

### Step 5: Simplify the expression
To simplify:
\[ \mu_r = \frac{1}{150 \cdot 4\pi \times 10^{-7}} = \frac{1}{600\pi \times 10^{-7}} = \frac{10^{7}}{600\pi} \]

### Step 6: Final expression for relative permeability
Thus, the relative permeability of iron is:
\[ \mu_r = \frac{10^5}{6\pi} \]