A magnetic material of volume `30cm^(3)` is placed in a magnetic field of intensity `5` oversted. The manetic moment produced due it is `6 amp-m^(2)`. The value of magnetic induction will be.

Options

A. `0.2517` Tesla
B. `0.025` Tesla
C. `0.0025` Tesla
D. `25` Tesla

Explanation

To find the magnetic induction \( B \) produced in a magnetic material, we can use the following formula:

\[ B = \mu_0 (H + I) \]

where:
- \( B \) is the magnetic induction,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( H \) is the magnetic field intensity,
- \( I \) is the intensity of magnetization.

### Step-by-Step Solution:

**Step 1: Convert the volume from cm³ to m³.**
\[ \text{Volume} = 30 \, \text{cm}^3 = 30 \times 10^{-6} \, \text{m}^3 = 3.0 \times 10^{-5} \, \text{m}^3 \]

**Step 2: Calculate the intensity of magnetization \( I \).**
The intensity of magnetization \( I \) is given by:
\[ I = \frac{\text{Magnetic Moment}}{\text{Volume}} = \frac{6 \, \text{A m}^2}{3.0 \times 10^{-5} \, \text{m}^3} = 2 \times 10^{5} \, \text{A/m} \]

**Step 3: Convert the magnetic field intensity \( H \) from Oersted to SI units.**
Given \( H = 5 \, \text{Oersted} \):
\[ H = 5 \times \frac{10^3}{4\pi} \approx 397.887 \, \text{A/m} \]

**Step 4: Calculate the magnetic induction \( B \).**
Using the formula:
\[ B = \mu_0 (H + I) \]
Substituting the values:
\[ B = (4\pi \times 10^{-7}) \left(397.887 + 2 \times 10^{5}\right) \]
Calculating:
\[ B \approx (4\pi \times 10^{-7}) \left(200397.887\right) \]
\[ B \approx 0.2517 \, \text{T} \]

### Final Answer:
The value of magnetic induction \( B \) is approximately \( 0.2517 \, \text{T} \).

---