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 given solenoid, we can follow these steps:

### Step 1: Understand the parameters given
- Turn density (n) = 10 turns/cm = 1000 turns/m (since 1 cm = 0.01 m)
- Volume (V) = \(10^{-3} \, m^3\)
- Current (I) = 0.5 A
- Relative permeability (\(\mu_r\)) = 1000

### Step 2: Calculate the magnetic field strength (H)
The magnetic field strength (H) can be calculated using the formula:
\[ H = n \cdot I \]
Substituting the values:
\[ H = 1000 \, \text{turns/m} \times 0.5 \, \text{A} = 500 \, \text{A/m} \]

### Step 3: Calculate the magnetic permeability (\(\mu\))
The magnetic permeability (\(\mu\)) of the material can be calculated using:
\[ \mu = \mu_0 \cdot \mu_r \]
Where \(\mu_0\) (the permeability of free space) is approximately \(4\pi \times 10^{-7} \, T \cdot m/A\). Thus,
\[ \mu = (4\pi \times 10^{-7}) \cdot 1000 \approx 4\pi \times 10^{-4} \, T \cdot m/A \]

### Step 4: Calculate the magnetic flux density (B)
The magnetic flux density (B) can be calculated using:
\[ B = \mu \cdot H \]
Substituting the values:
\[ B = (4\pi \times 10^{-4}) \cdot 500 \approx 2\pi \times 10^{-1} \, T \approx 0.628 \, T \]

### Step 5: Calculate the intensity of magnetization (I)
The intensity of magnetization (I) can be calculated using:
\[ I = \frac{B}{\mu_0} - H \]
Substituting the values:
\[ I = \frac{0.628}{4\pi \times 10^{-7}} - 500 \]
Calculating \(\frac{0.628}{4\pi \times 10^{-7}} \approx 50000\), thus:
\[ I \approx 50000 - 500 = 49500 \, A/m \]

### Step 6: Calculate the magnetic moment (M)
The magnetic moment (M) can be calculated using:
\[ M = I \cdot V \]
Substituting the values:
\[ M = 49500 \cdot 10^{-3} \approx 49.5 \, A \cdot m^2 \]

### Step 7: Final approximation
Since we need to express the answer in scientific notation:
\[ M \approx 5 \times 10^1 \, A \cdot m^2 \]

### Conclusion
The magnetic moment of the solenoid is approximately \(5 \times 10^1 \, A \cdot m^2\).