An iron rod of volume `10^(-4)m^(3)` and relative permeability 1000 is placed inside a long solenoid wound with `5 "turns"//"cm"`. If a current of `0.5 A` is passed through the solenoid, then the magnetic moment of the rod is

Options

A. `10 AM^(2)`
B. `15 AM^(2)`
C. `20 AM^(2)`
D. `25 AM^(2)`

Explanation

To find the magnetic moment of the iron rod placed inside a solenoid, we can follow these steps:

### Step 1: Understand the Parameters
- Volume of the iron rod, \( V = 10^{-4} \, m^3 \)
- Relative permeability of the iron rod, \( \mu_r = 1000 \)
- Number of turns per centimeter in the solenoid, \( n = 5 \, \text{turns/cm} = 500 \, \text{turns/m} \)
- Current through the solenoid, \( I = 0.5 \, A \)

### Step 2: Calculate the Magnetic Field Intensity (H)
The magnetic field intensity \( H \) in a solenoid is given by the formula:
\[ H = n \cdot I \]
Substituting the values:
\[ H = 500 \, \text{turns/m} \cdot 0.5 \, A = 250 \, A/m \]

### Step 3: Calculate the Magnetic Field (B)
The magnetic field \( B \) inside the solenoid can be calculated using the formula:
\[ B = \mu_0 \cdot \mu_r \cdot H \]
Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, T \cdot m/A \). Thus:
\[ B = (4\pi \times 10^{-7}) \cdot 1000 \cdot 250 \]
Calculating this gives:
\[ B = 4\pi \times 10^{-4} \, T \cdot 250 \approx 3.14 \times 10^{-4} \cdot 250 \approx 0.246 \, T \]

### Step 4: Calculate the Magnetic Moment (M)
The magnetic moment \( M \) of the rod can be calculated using the formula:
\[ M = \chi \cdot V \cdot B \]
Where \( \chi \) (the magnetic susceptibility) is given by:
\[ \chi = \mu_r - 1 = 1000 - 1 = 999 \]
Now substituting the values:
\[ M = 999 \cdot 10^{-4} \cdot 0.246 \]
Calculating this gives:
\[ M \approx 999 \cdot 10^{-4} \cdot 0.246 \approx 0.246 \, A \cdot m^2 \]

### Step 5: Final Result
Thus, the magnetic moment of the iron rod is approximately:
\[ M \approx 0.246 \, A \cdot m^2 \]