A cylinder wire of radius R is carrying uniformly distributed current I over its cross-section. If a circular loop of radius r is taken as amperian loop, then the variation value of `oint vec(B)* vec(dl)` over this loop with radius 'r' of loop will be best represented by

Options

A.
B.
C.
D.

Answer

To solve the problem, we will analyze the magnetic field around a cylindrical wire carrying a uniformly distributed current using Ampere's Circuital Law.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a cylindrical wire of radius \( R \) carrying a current \( I \) uniformly distributed across its cross-section. We want to analyze the magnetic field around this wire using an Amperian loop of radius \( r \).

2. **Applying Ampere's Circuital Law**:
According to Ampere's Circuital Law, the line integral of the magnetic field \( \vec{B} \) around a closed loop is equal to \( \mu_0 \) times the current enclosed by that loop:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}} \]

3. **Determining the Current Enclosed**:
- If \( r < R \) (the loop is inside the wire), we need to find the current enclosed by the loop. The current density \( J \) is given by:
\[ J = \frac{I}{\pi R^2} \]
- The area of the circular loop of radius \( r \) is \( \pi r^2 \). Thus, the current enclosed \( I_{\text{enclosed}} \) is:
\[ I_{\text{enclosed}} = J \times \text{Area} = \left(\frac{I}{\pi R^2}\right) \times \pi r^2 = \frac{I r^2}{R^2} \]

4. **Calculating the Line Integral**:
- For \( r < R \):
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 \left(\frac{I r^2}{R^2}\right) \]
- The left side can be simplified since \( \vec{B} \) is constant along the circular path of radius \( r \):
\[ B(2\pi r) = \mu_0 \left(\frac{I r^2}{R^2}\right) \]
- Therefore, we can express \( B \):
\[ B = \frac{\mu_0 I r}{2 \pi R^2} \]

5. **For \( r \geq R \)**:
- The entire current \( I \) is enclosed, so:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I \]
- Thus, for \( r \geq R \):
\[ B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2 \pi r} \]

6. **Graphing the Results**:
- For \( r < R \), \( B \) varies linearly with \( r \) (specifically, \( B \propto r \)).
- For \( r \geq R \), \( B \) decreases with \( r \) (specifically, \( B \propto \frac{1}{r} \)).
- The graph of \( \oint \vec{B} \cdot d\vec{l} \) versus \( r \) will show a parabolic increase for \( r < R \) and a hyperbolic decrease for \( r \geq R \).

### Conclusion:
The variation of \( \oint \vec{B} \cdot d\vec{l} \) over the loop will be best represented by a curve that increases parabolically for \( r < R \) and then becomes constant for \( r \geq R \).