A rod of ferromegnetic materical with dimensions `10cmxx0.5cmxx2cm` is placed in a magnetising field of intensity `2xx10^(5)A//m`. The magnetic moment produced due it is `6` `amop-m^(2)`. The value of magnetic induction will be------`10^(-2)T`.

Options

A. `100.48`
B. `200.28`
C. `50.24`
D. `300.48`

Explanation

To solve the problem, we need to find the magnetic induction (B) produced in a ferromagnetic rod placed in a magnetizing field. We will use the relationship between magnetic induction, magnetizing field intensity (H), and intensity of magnetization (I).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Dimensions of the rod: \(10 \, \text{cm} \times 0.5 \, \text{cm} \times 2 \, \text{cm}\)
- Magnetizing field intensity \(H = 2 \times 10^5 \, \text{A/m}\)
- Magnetic moment \(m = 6 \, \text{Am}^2\)

2. **Calculate the Volume of the Rod:**
\[ \text{Volume} = \text{length} \times \text{width} \times \text{height} = 10 \, \text{cm} \times 0.5 \, \text{cm} \times 2 \, \text{cm} \]
Converting to meters:
\[ = 0.1 \, \text{m} \times 0.005 \, \text{m} \times 0.02 \, \text{m} = 0.00001 \, \text{m}^3 = 10^{-5} \, \text{m}^3 \]

3. **Calculate the Intensity of Magnetization (I):**
\[ I = \frac{m}{\text{Volume}} = \frac{6 \, \text{Am}^2}{10^{-5} \, \text{m}^3} = 6 \times 10^5 \, \text{A/m}^2 \]

4. **Use the Formula for Magnetic Induction (B):**
For ferromagnetic materials, the magnetic induction \(B\) is given by:
\[ B = \mu_0 (H + I) \]
where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\).

5. **Calculate \(B\):**
\[ B = 4\pi \times 10^{-7} \times (2 \times 10^5 + 6 \times 10^5) \]
\[ = 4\pi \times 10^{-7} \times 8 \times 10^5 \]
\[ = 32\pi \times 10^{-2} \, \text{T} \]

6. **Approximate \(\pi\):**
Using \(\pi \approx 3.14\):
\[ B \approx 32 \times 3.14 \times 10^{-2} \approx 100.48 \times 10^{-2} \, \text{T} \]

7. **Final Result:**
Thus, the value of magnetic induction \(B\) is approximately:
\[ B \approx 100.48 \times 10^{-2} \, \text{T} \approx 1.0048 \, \text{T} \]

### Summary:
The magnetic induction produced in the ferromagnetic rod is approximately \(1.0048 \, \text{T}\).