An iron ring having 500 turns of wire and mean diameter of 12 cm carries a current of 0.3 A. The relative permeability of iron is 600. What is the magnetic flux density in the core ? What is the magnetisation field intensity ? What part of the density is due to the electronic loop currents in the core ?

Answer

To solve the problem step by step, we will calculate the magnetic flux density in the core, the magnetizing field intensity, and the part of the density due to the electronic loop currents in the core.

### Step 1: Calculate the Mean Length of the Iron Ring
The mean diameter of the iron ring is given as 12 cm. First, we convert this to meters:
\[ \text{Mean diameter} = 12 \, \text{cm} = 0.12 \, \text{m} \]
The mean length \( L \) of the ring can be calculated using the formula:
\[ L = \pi \times d = \pi \times 0.12 \, \text{m} \]
Calculating this gives:
\[ L \approx 0.37699 \, \text{m} \]

**Hint:** Use the formula for the circumference of a circle to find the mean length.

### Step 2: Calculate the Magnetizing Field Intensity \( H \)
The magnetizing field intensity \( H \) is given by the formula:
\[ H = \frac{N \cdot I}{L} \]
where \( N \) is the number of turns (500), \( I \) is the current (0.3 A), and \( L \) is the mean length calculated in the previous step.

Substituting the values:
\[ H = \frac{500 \times 0.3}{0.37699} \approx 397.9 \, \text{A/m} \]

**Hint:** Remember to substitute the values correctly into the formula for \( H \).

### Step 3: Calculate the Magnetic Flux Density \( B \)
The magnetic flux density \( B \) can be calculated using the formula:
\[ B = \mu \cdot H \]
where \( \mu \) is the permeability of the medium, which can be calculated as:
\[ \mu = \mu_0 \cdot \mu_r \]
Here, \( \mu_0 \) (the permeability of free space) is \( 4\pi \times 10^{-7} \, \text{H/m} \) and \( \mu_r \) (the relative permeability of iron) is 600.

Calculating \( \mu \):
\[ \mu = (4\pi \times 10^{-7}) \times 600 \]
Calculating this gives:
\[ \mu \approx 7.54 \times 10^{-4} \, \text{H/m} \]
Now substituting into the formula for \( B \):
\[ B = (7.54 \times 10^{-4}) \times 397.9 \approx 0.3 \, \text{T} \]

**Hint:** Make sure to calculate the permeability correctly before using it to find \( B \).

### Step 4: Calculate the Part of Density Due to Electronic Loop Currents
The part of the magnetic flux density \( B \) due to electronic loop currents can be found using:
\[ B_{\text{electronic}} = B - \mu_0 \cdot H \]
Calculating \( \mu_0 \cdot H \):
\[ \mu_0 \cdot H = (4\pi \times 10^{-7}) \times 397.9 \approx 4.99 \times 10^{-4} \, \text{T} \]
Now substituting into the equation for \( B_{\text{electronic}} \):
\[ B_{\text{electronic}} = 0.3 - 4.99 \times 10^{-4} \approx 0.2995 \, \text{T} \]

**Hint:** Remember to subtract the contribution of the free space permeability from the total magnetic flux density to find the electronic contribution.

### Final Answers
1. **Magnetic Flux Density \( B \)**: \( 0.3 \, \text{T} \)
2. **Magnetizing Field Intensity \( H \)**: \( 397.9 \, \text{A/m} \)
3. **Density due to Electronic Loop Currents**: \( 0.2995 \, \text{T} \)