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 `B_1` is

Options

A. 25mT
B. 50 mT
C. `100 mu T`
D. `200 mT`

Explanation

To solve the problem, we need to calculate the magnetic field \( B_1 \) at the center of the solenoid before the iron core is inserted. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Magnetic Intensity \( H_1 \)

The magnetic intensity \( H \) in a solenoid is given by the formula:

\[ H = N \cdot I \]

where:
- \( N \) is the number of turns per unit length (in turns per meter),
- \( I \) is the current flowing through the solenoid.

Given:
- Current \( I = 2 \, \text{A} \)
- Number of turns \( N = 100 \, \text{turns/cm} = 100 \times 100 = 10^4 \, \text{turns/m} \)

Now substituting the values:

\[ H_1 = 10^4 \, \text{turns/m} \times 2 \, \text{A} = 2 \times 10^4 \, \text{A/m} \]

### Step 2: Calculate the Magnetic Field \( B_1 \)

The magnetic field \( B \) in a solenoid is related to the magnetic intensity \( H \) by the formula:

\[ B = \mu_0 H \]

where \( \mu_0 \) is the permeability of free space, given by:

\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \]

Now substituting \( H_1 \):

\[ B_1 = \mu_0 H_1 = (4\pi \times 10^{-7} \, \text{T m/A}) \times (2 \times 10^4 \, \text{A/m}) \]

Calculating this:

\[ B_1 = 8\pi \times 10^{-3} \, \text{T} \]

### Step 3: Numerical Calculation of \( B_1 \)

Using \( \pi \approx 3.14 \):

\[ B_1 \approx 8 \times 3.14 \times 10^{-3} \, \text{T} = 25.12 \times 10^{-3} \, \text{T} \approx 25 \, \text{mT} \]

### Final Answer

Thus, the value of \( B_1 \) is approximately:

\[ B_1 \approx 25 \, \text{mT} \]

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