There is an iron core solenoid of turn density 10 turns/cm and volume `10^(-3) m^3`. It carries a current of 0.5A and relative permeability of iron core is `mu_r = 1000`. The magnetic moment of this solenoid is approximately (A-m^2)

Options

A. `5 x 10^2`
B. `5 x 10^3`
C. `5 x 10^4`
D. `5 x 10^6`

Answer

To find the magnetic moment of the iron core solenoid, we can use the formula for the magnetic moment \( m \) of a solenoid, which is given by:

\[ m = n \cdot I \cdot A \]

where:
- \( n \) is the turn density (number of turns per unit length),
- \( I \) is the current flowing through the solenoid,
- \( A \) is the cross-sectional area of the solenoid.

### Step 1: Convert turn density to turns per meter
The turn density is given as 10 turns/cm. To convert this to turns per meter, we multiply by 100:

\[ n = 10 \, \text{turns/cm} \times 100 \, \text{cm/m} = 1000 \, \text{turns/m} \]

### Step 2: Calculate the cross-sectional area \( A \)
We know the volume \( V \) of the solenoid is \( 10^{-3} \, \text{m}^3 \). The volume of a solenoid can also be expressed as:

\[ V = A \cdot L \]

where \( L \) is the length of the solenoid. To find the area, we need to express \( L \) in terms of the volume and area:

\[ A = \frac{V}{L} \]

However, we do not have \( L \) directly. We will keep \( A \) in terms of \( L \) for now.

### Step 3: Substitute values into the magnetic moment formula
The magnetic moment can now be expressed as:

\[ m = n \cdot I \cdot A = n \cdot I \cdot \frac{V}{L} \]

Substituting the known values:

- \( n = 1000 \, \text{turns/m} \)
- \( I = 0.5 \, \text{A} \)
- \( V = 10^{-3} \, \text{m}^3 \)

We get:

\[ m = 1000 \cdot 0.5 \cdot \frac{10^{-3}}{L} \]

### Step 4: Calculate the magnetic moment
Since we do not have \( L \), we can express the magnetic moment in terms of \( L \):

\[ m = \frac{500 \cdot 10^{-3}}{L} = \frac{0.5}{L} \, \text{A-m}^2 \]

### Step 5: Determine the effective length
To find \( L \), we can use the relationship of \( V \) and \( A \). However, we can also assume that the length of the solenoid is such that it is a reasonable value for a solenoid. For practical purposes, we can assume \( L \) is around 1 meter (as a rough estimate):

\[ m \approx 0.5 \, \text{A-m}^2 \]

### Final Result
The magnetic moment of the solenoid is approximately:

\[ m \approx 0.5 \, \text{A-m}^2 \]