A long cylindrical wire carrying current of `10 A` has radius of `5 mm` find its its magnetic field induction at a point `2 mm` from the centre of the wire

Options

A. `16xx10^(-4)T`
B. `2.4xx10^(-4)T`
C. `3.2xx10^(-4)T`
D. `0.8xx10^(-4)T`

Explanation

To find the magnetic field induction at a point 2 mm from the center of a long cylindrical wire carrying a current of 10 A, we can use Ampere's Law. The formula for the magnetic field (B) inside a long cylindrical wire is given by:

\[ B = \frac{\mu_0 I r}{2 \pi R^2} \]

where:
- \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( I \) is the current (10 A)
- \( r \) is the distance from the center of the wire (2 mm = 2 × 10^{-3} m)
- \( R \) is the radius of the wire (5 mm = 5 × 10^{-3} m)

### Step-by-Step Solution:

1. **Convert units**:
- The radius of the wire \( R = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- The distance from the center \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

2. **Substitute values into the formula**:
\[ B = \frac{\mu_0 I r}{2 \pi R^2} \]
Substitute \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \), \( I = 10 \, \text{A} \), \( r = 2 \times 10^{-3} \, \text{m} \), and \( R = 5 \times 10^{-3} \, \text{m} \):
\[ B = \frac{(4\pi \times 10^{-7}) \times 10 \times (2 \times 10^{-3})}{2 \pi (5 \times 10^{-3})^2} \]

3. **Simplify the equation**:
- The \( \pi \) terms cancel out:
\[ B = \frac{(4 \times 10^{-7}) \times 10 \times (2 \times 10^{-3})}{2 \times (5 \times 10^{-3})^2} \]
- Calculate \( (5 \times 10^{-3})^2 = 25 \times 10^{-6} \):
\[ B = \frac{(4 \times 10^{-7}) \times 10 \times (2 \times 10^{-3})}{2 \times 25 \times 10^{-6}} \]

4. **Calculate the numerator**:
\[ = 4 \times 10^{-7} \times 10 \times 2 = 80 \times 10^{-7} \]

5. **Calculate the denominator**:
\[ = 2 \times 25 \times 10^{-6} = 50 \times 10^{-6} \]

6. **Combine the results**:
\[ B = \frac{80 \times 10^{-7}}{50 \times 10^{-6}} = \frac{80}{50} \times 10^{-1} = 1.6 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnetic field induction at a point 2 mm from the center of the wire is:
\[ B = 1.6 \times 10^{-4} \, \text{T} \]