A `25cm` long solenoid has radius `2cm` and `500` total number of turns. It carries as current of `15A`.If it is equivalent to a magnet of the same size and magnetization `bar(M)` (magnetic moment//volume) then `|bar(M)|` is:

Options

A. `3pi Am^(-1)`
B. `30000Am^(-1)`
C. `300Am^(-1)`
D. `30000pi Am^(-1)`

Explanation

To find the magnetization \(|\bar{M}|\) of the solenoid, we will follow these steps:

### Step 1: Identify the given values
- Length of the solenoid, \(L = 25 \, \text{cm} = 0.25 \, \text{m}\)
- Radius of the solenoid, \(r = 2 \, \text{cm} = 0.02 \, \text{m}\)
- Total number of turns, \(N = 500\)
- Current, \(I = 15 \, \text{A}\)

### Step 2: Calculate the area of the cross-section of the solenoid
The area \(A\) of the cross-section of the solenoid can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 0.00125664 \, \text{m}^2 \]

### Step 3: Calculate the magnetic moment \(m\)
The magnetic moment \(m\) of the solenoid can be calculated using the formula:
\[ m = N \cdot I \cdot A \]
Substituting the values:
\[ m = 500 \cdot 15 \cdot 0.00125664 \]
Calculating this gives:
\[ m = 500 \cdot 15 \cdot 0.00125664 \approx 9.4473 \, \text{A m}^2 \]

### Step 4: Calculate the volume \(V\) of the solenoid
The volume \(V\) of the solenoid can be calculated using the formula:
\[ V = A \cdot L \]
Substituting the values:
\[ V = 0.00125664 \cdot 0.25 = 0.00031416 \, \text{m}^3 \]

### Step 5: Calculate the magnetization \(|\bar{M}|\)
Magnetization \(|\bar{M}|\) is defined as the magnetic moment per unit volume:
\[ |\bar{M}| = \frac{m}{V} \]
Substituting the values:
\[ |\bar{M}| = \frac{9.4473}{0.00031416} \approx 30100.5 \, \text{A/m} \]

### Final Answer
Thus, the magnetization \(|\bar{M}|\) is approximately \(30100.5 \, \text{A/m}\).

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