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 find the magnetization (M) of the core material in the solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between B, H, and M**:
The magnetic flux density (B) is related to the magnetic field strength (H) and the magnetization (M) of the material. The relationship is given by:
\[ B = \mu_0 (H + M) \]
where \(\mu_0\) is the permeability of free space.

2. **Express B in terms of H**:
We can express B in terms of the permeability (\(\mu\)) of the material:
\[ B = \mu H \]
where \(\mu = \mu_r \mu_0\) and \(\mu_r\) is the relative permeability of the material.

3. **Substituting B into the equation**:
By substituting the expression for B into the first equation, we have:
\[ \mu H = \mu_0 (H + M) \]

4. **Rearranging to find M**:
Rearranging the equation to isolate M gives:
\[ M = \mu H - \mu_0 H \]
This can be simplified to:
\[ M = (\mu - \mu_0) H \]

5. **Substituting \(\mu\)**:
Since \(\mu = \mu_r \mu_0\), we can substitute this into the equation:
\[ M = (\mu_r \mu_0 - \mu_0) H = \mu_0 (\mu_r - 1) H \]

6. **Finding H**:
The magnetic field strength (H) for a solenoid is given by:
\[ H = nI \]
where \(n\) is the number of turns per unit length and \(I\) is the current.

7. **Substituting values**:
Given:
- \(\mu_r = 500\)
- \(n = 500 \, \text{turns/m}\)
- \(I = 1 \, \text{A}\)

We can substitute these values into the equation for M:
\[ M = \mu_0 (500 - 1) (500 \times 1) \]

8. **Calculating M**:
The value of \(\mu_0\) is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\). Therefore:
\[ M = (4\pi \times 10^{-7}) \times 499 \times 500 \]
Calculating this gives:
\[ M \approx 2.25 \times 10^5 \, \text{A/m} \]

### Final Answer:
The magnetization of the material is approximately:
\[ M \approx 2.25 \times 10^5 \, \text{A/m} \]

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