A solenoid has core of a material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of the solenoid is 500per meter. The magnetization of the material is nearly

Options

A. `2.25xx10^(3)Am^(-1)`
B. `2.25xx10^(5)Am^(-1)`
C. `2.0xx10^(3)Am^(-1)`
D. `2.0xx10^(5)Am^(-1)`

Explanation

To solve the problem of finding the magnetization of a solenoid with a core material of relative permeability 500, we can follow these steps:

### Step 1: Identify the Given Data
- Relative permeability (\( \mu_r \)) = 500
- Current (\( I \)) = 1 A
- Number of turns per meter (\( n \)) = 500 turns/m

### Step 2: Calculate the Magnetic Intensity (\( H \))
The magnetic intensity \( H \) can be calculated using the formula:
\[ H = n \cdot I \]
Substituting the values:
\[ H = 500 \, \text{turns/m} \times 1 \, \text{A} = 500 \, \text{A/m} \]

### Step 3: Relate Relative Permeability to Magnetic Susceptibility
The relationship between relative permeability (\( \mu_r \)) and magnetic susceptibility (\( \chi \)) is given by:
\[ \mu_r = 1 + \chi \]
Rearranging this equation to find \( \chi \):
\[ \chi = \mu_r - 1 \]
Substituting the value of \( \mu_r \):
\[ \chi = 500 - 1 = 499 \]

### Step 4: Calculate the Magnetization (\( M \))
The magnetization \( M \) is given by the formula:
\[ M = \chi \cdot H \]
Substituting the values of \( \chi \) and \( H \):
\[ M = 499 \cdot 500 \, \text{A/m} \]
Calculating this:
\[ M = 249500 \, \text{A/m} = 2.495 \times 10^5 \, \text{A/m} \]

### Step 5: Final Answer
Thus, the magnetization of the material is approximately:
\[ M \approx 2.5 \times 10^5 \, \text{A/m} \]