There is one tightly wound long solenoid carrying current 2 A. Solenoid have 100 turns per cm. `H_1` is the magnetic intensity and `B_1` is the magnetic field at the centre of solenoid. Now an iron core is inserted inside the solenoid. Intensity of magnetization in the core is `4 xx 10^6` A/m. New values of magnetic intensity and magnetic field at the centre becomes `H_2` and `B_2`
`Value of `H_1` is

Options

A. `2 xx 10^4 A//m`
B. `3 xx 10^4 A//m`
C. `2 xx 10^6 A//m`
D. `3xx10^6 A//m`

Explanation

To solve the problem, we need to calculate the magnetic intensity \( H_1 \) at the center of a long solenoid carrying a current. Here are the steps to find \( H_1 \):

### Step 1: Identify the given values
- Current \( I = 2 \, \text{A} \)
- Number of turns per centimeter \( n = 100 \, \text{turns/cm} \)

### Step 2: Convert the number of turns to turns per meter
Since we need to work in SI units, we convert the number of turns from per centimeter to per meter:
\[ n = 100 \, \text{turns/cm} = 100 \times 100 = 10,000 \, \text{turns/m} \]

### Step 3: Use the formula for magnetic intensity
The magnetic intensity \( H \) in a solenoid is given by the formula:
\[ H = n \cdot I \]
Substituting the values we have:
\[ H_1 = 10,000 \, \text{turns/m} \times 2 \, \text{A} \]

### Step 4: Calculate \( H_1 \)
Now, we can calculate \( H_1 \):
\[ H_1 = 10,000 \times 2 = 20,000 \, \text{A/m} \]
Thus, we have:
\[ H_1 = 2 \times 10^4 \, \text{A/m} \]

### Final Answer
The value of \( H_1 \) is:
\[ H_1 = 2 \times 10^4 \, \text{A/m} \]

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