A long straight wire of radius R carries a current i. The magnetic field inside the wire at distance`r(r lt R)`, from its centre is expressed as:

Options

A. `((mu_(0)i)/(piR^(2))).r`
B. `((2mu_(0)i)/(piR^(2))).r`
C. `(mu_(0)i)/(2piR^(2)).r`
D. `(mu_(0)i)/(2piR).r`

Explanation

To find the magnetic field inside a long straight wire of radius \( R \) carrying a current \( i \) at a distance \( r \) from its center (where \( r < R \)), we can use Ampere's Circuital Law. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
We have a long straight wire with a uniform current \( i \). We need to find the magnetic field \( B \) at a distance \( r \) from the center of the wire, where \( r < R \).

### Step 2: Apply Ampere's Circuital Law
According to Ampere's Circuital Law:
\[ \oint B \cdot dl = \mu_0 I_{in} \]
where \( I_{in} \) is the current enclosed by the Amperian loop of radius \( r \).

### Step 3: Define the Amperian Loop
We consider a circular Amperian loop of radius \( r \) inside the wire. The magnetic field \( B \) is constant along this loop and is tangent to the loop.

### Step 4: Calculate the Enclosed Current
The current density \( J \) in the wire is given by:
\[ J = \frac{i}{\pi R^2} \]
The enclosed current \( I_{in} \) for the Amperian loop of radius \( r \) is:
\[ I_{in} = J \cdot \text{Area of the loop} = J \cdot \pi r^2 = \frac{i}{\pi R^2} \cdot \pi r^2 = \frac{i r^2}{R^2} \]

### Step 5: Substitute into Ampere's Law
Now substituting \( I_{in} \) into Ampere's Circuital Law:
\[ B \cdot (2 \pi r) = \mu_0 \cdot \frac{i r^2}{R^2} \]

### Step 6: Solve for \( B \)
Rearranging the equation to solve for \( B \):
\[ B = \frac{\mu_0 i r}{2 \pi R^2} \]

### Final Result
Thus, the magnetic field inside the wire at a distance \( r \) from its center is given by:
\[ B = \frac{\mu_0 i r}{2 \pi R^2} \]