A long straight solid conductor of radius `5 (~cm)` carries a current of 2 A waich is uniformly distributed over its circular cross-section. If the magnetic field induction at a distunce of 3 cm from the axis of the conductor is `y times 10^(-6) T`, then find the value of `y`.

Answer

To solve the problem of finding the magnetic field induction at a distance of 3 cm from the axis of a long straight solid conductor carrying a current of 2 A, we can use Ampère's Law and the formula for the magnetic field inside a cylindrical conductor.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Radius of the conductor, \( R = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Current, \( I = 2 \, \text{A} \)
- Distance from the axis where we want to find the magnetic field, \( r = 3 \, \text{cm} = 0.03 \, \text{m} \)

2. **Check the position of the point:**
Since \( r = 3 \, \text{cm} \) is less than \( R = 5 \, \text{cm} \), we are inside the conductor.

3. **Use the formula for the magnetic field inside a cylindrical conductor:**
The magnetic field \( B \) at a distance \( r \) from the axis of a long straight conductor with uniform current distribution is given by:
\[ B = \frac{\mu_0 I}{2 \pi R^2} r \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space.

4. **Substitute the values into the formula:**
\[ B = \frac{(4\pi \times 10^{-7}) \times 2}{2 \pi (0.05)^2} \times 0.03 \]

5. **Simplify the expression:**
- The \( 2\pi \) cancels out:
\[ B = \frac{(4 \times 10^{-7}) \times 2}{(0.05)^2} \times 0.03 \]
- Calculate \( (0.05)^2 = 0.0025 \):
\[ B = \frac{8 \times 10^{-7}}{0.0025} \times 0.03 \]
- Calculate \( \frac{8 \times 10^{-7}}{0.0025} = 3.2 \times 10^{-4} \):
\[ B = 3.2 \times 10^{-4} \times 0.03 = 9.6 \times 10^{-6} \, \text{T} \]

6. **Express the magnetic field in terms of \( y \):**
We have \( B = 9.6 \times 10^{-6} \, \text{T} \), which can be expressed as \( y \times 10^{-6} \, \text{T} \). Thus, \( y = 9.6 \).

### Final Answer:
The value of \( y \) is \( 9.6 \).