A solenoid has a core of a material with relative permeability of 500. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per meter, then magnetisation is

Options

A. `7.78xx10^(5)Am^(-1)`
B. `8.88xx10^(5)Am^(-1)`
C. `9.98xx10^(5)Am^(-1)`
D. `10.2xx10^(5)Am^(-1)`

Explanation

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

### Step 1: Identify the given values
- Relative permeability (μ_r) = 500
- Current (I) = 2 A
- Number of turns per meter (n) = 1000 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:
\[ H = 1000 \, \text{turns/m} \times 2 \, \text{A} = 2000 \, \text{A/m} \]

### Step 3: Calculate the magnetic susceptibility (χ)
The magnetic susceptibility (χ) is related to the relative permeability (μ_r) by the formula:
\[ μ_r = 1 + χ \]
Rearranging this gives:
\[ χ = μ_r - 1 \]
Substituting the value of μ_r:
\[ χ = 500 - 1 = 499 \]

### Step 4: Calculate the magnetization (M)
The magnetization (M) can be calculated using the formula:
\[ M = χ \cdot H \]
Substituting the values:
\[ M = 499 \cdot 2000 \, \text{A/m} \]
Calculating this gives:
\[ M = 998000 \, \text{A/m} = 9.98 \times 10^5 \, \text{A/m} \]

### Conclusion
The magnetization of the solenoid is:
\[ M = 9.98 \times 10^5 \, \text{A/m} \]

### Final Answer
The correct option is \( 9.98 \times 10^5 \, \text{A/m} \).

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