An iron ring of mean circumferential length 30 cm and cross-section 1 `cm^(2)` is wound uniformly with 300 turns of wire . When a current of `0.032` A flows in the windings , flux in the ring `2 xx 10^(-6) `Wb . Find the flux density in the ring , magnetising field intensity and relative permeability of iron.

Answer

To solve the problem step by step, we will find the flux density, magnetizing field intensity, and relative permeability of the iron ring.

### Step 1: Calculate the Flux Density (B)
The flux density \( B \) is defined as the magnetic flux \( \Phi \) per unit area \( A \).

Given:
- Flux \( \Phi = 2 \times 10^{-6} \, \text{Wb} \)
- Cross-sectional area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)

Using the formula:
\[ B = \frac{\Phi}{A} \]
Substituting the values:
\[ B = \frac{2 \times 10^{-6}}{1 \times 10^{-4}} = 2 \times 10^{-2} \, \text{T} \]

### Step 2: Calculate the Magnetizing Field Intensity (H)
The magnetizing field intensity \( H \) can be calculated using the formula:
\[ H = N \cdot I \]
where \( N \) is the number of turns per unit length and \( I \) is the current.

Given:
- Total number of turns \( N = 300 \)
- Mean circumferential length \( L = 30 \, \text{cm} = 0.3 \, \text{m} \)
- Current \( I = 0.032 \, \text{A} \)

First, calculate the number of turns per unit length:
\[ N_{\text{per unit length}} = \frac{N}{L} = \frac{300}{0.3} = 1000 \, \text{turns/m} \]

Now, substituting into the formula for \( H \):
\[ H = 1000 \times 0.032 = 32 \, \text{A/m} \]

### Step 3: Calculate the Relative Permeability (\( \mu_r \))
The relative permeability \( \mu_r \) can be calculated using the relationship:
\[ B = \mu \cdot H \]
where \( \mu \) is the permeability of the material. The relative permeability is defined as:
\[ \mu_r = \frac{\mu}{\mu_0} \]
where \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \).

From the earlier equation, we can express \( \mu \) as:
\[ \mu = \frac{B}{H} \]
Substituting the values we found:
\[ \mu = \frac{2 \times 10^{-2}}{32} = \frac{2}{32} \times 10^{-2} = \frac{1}{16} \times 10^{-2} \, \text{H/m} \]

Now, substituting this into the relative permeability formula:
\[ \mu_r = \frac{\mu}{\mu_0} = \frac{\frac{1}{16} \times 10^{-2}}{4\pi \times 10^{-7}} \]
Calculating \( \mu_r \):
\[ \mu_r = \frac{1 \times 10^{-2}}{16 \times 4\pi \times 10^{-7}} = \frac{1}{64\pi} \times 10^5 \]
Approximating \( \pi \approx 3.14 \):
\[ \mu_r \approx \frac{1}{64 \times 3.14} \times 10^5 \approx \frac{1}{200} \times 10^5 \approx 500 \]

### Summary of Results
- Flux Density \( B = 2 \times 10^{-2} \, \text{T} \)
- Magnetizing Field Intensity \( H = 32 \, \text{A/m} \)
- Relative Permeability \( \mu_r \approx 500 \)