A straight thick long wire of uniform cross-section of radius `5cm` is carrying a steady current `5A`. Calculate the magnetic field at a point `2cm` from the axis of wire, well inside the wire.

Explanation

To calculate the magnetic field at a point 2 cm from the axis of a thick wire carrying a steady current, we can use Ampère's Law. Here’s a step-by-step solution:

### Step 1: Identify the parameters
- Radius of the wire, \( R = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Current through the wire, \( I = 5 \, \text{A} \)
- Distance from the axis of the wire to the point of interest, \( r = 2 \, \text{cm} = 0.02 \, \text{m} \)

### Step 2: Use Ampère's Law
Ampère's Law states that the line integral of the magnetic field \( \mathbf{B} \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the total current \( I_{\text{enc}} \) enclosed by the loop:

\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} \]

### Step 3: Determine the enclosed current
Since we are calculating the magnetic field inside the wire, we need to find the current enclosed by a circular path of radius \( r \) (2 cm) within the wire. The current density \( J \) can be calculated as:

\[ J = \frac{I}{A} \]

where \( A \) is the cross-sectional area of the wire. The area \( A \) of the wire is given by:

\[ A = \pi R^2 = \pi (0.05)^2 = \pi (0.0025) \, \text{m}^2 \]

Calculating \( A \):

\[ A \approx 0.00785 \, \text{m}^2 \]

Now, calculate the current density \( J \):

\[ J = \frac{5 \, \text{A}}{0.00785 \, \text{m}^2} \approx 636.62 \, \text{A/m}^2 \]

Now, we can find the current enclosed by the circular path of radius \( r = 0.02 \, \text{m} \):

\[ A_{\text{enc}} = \pi r^2 = \pi (0.02)^2 = \pi (0.0004) \, \text{m}^2 \approx 0.00125664 \, \text{m}^2 \]

The enclosed current \( I_{\text{enc}} \) is:

\[ I_{\text{enc}} = J \cdot A_{\text{enc}} = 636.62 \, \text{A/m}^2 \cdot 0.00125664 \, \text{m}^2 \approx 0.799 \, \text{A} \]

### Step 4: Apply Ampère's Law
Now, substituting \( I_{\text{enc}} \) into Ampère's Law:

\[ B \cdot (2 \pi r) = \mu_0 I_{\text{enc}} \]

Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

Substituting the values:

\[ B \cdot (2 \pi (0.02)) = (4\pi \times 10^{-7}) \cdot 0.799 \]

### Step 5: Solve for \( B \)
We can simplify this equation:

\[ B \cdot (0.1256 \pi) = (4\pi \times 10^{-7}) \cdot 0.799 \]

Dividing both sides by \( 0.1256 \pi \):

\[ B = \frac{(4 \times 10^{-7} \cdot 0.799)}{0.1256} \]

Calculating \( B \):

\[ B \approx \frac{3.196 \times 10^{-7}}{0.1256} \approx 2.54 \times 10^{-6} \, \text{T} \]

### Final Answer
The magnetic field at a point 2 cm from the axis of the wire is approximately \( 2.54 \, \mu T \).

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