Current flows through a straight, thin-walled tube of radius r. The magnetic field at a distance x from the axis of the tube has magnitude B.

Options

A. `B prop x`, for 0 lt x lt r
B. `B prop 1//x` , for 0 lt x lt r
C. B=0 , for `0 le x lt r`
D. B=0 , only for x=0

Answer

To solve the problem of determining the magnetic field \( B \) at a distance \( x \) from the axis of a straight, thin-walled tube carrying current, we can use Ampere's Circuital Law. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a straight, thin-walled tube of radius \( R \) carrying a current \( I \). We want to find the magnetic field at a distance \( x \) from the axis of the tube.

### Step 2: Identify the Regions
We need to consider two cases based on the distance \( x \):
1. When \( x < R \) (inside the tube)
2. When \( x > R \) (outside the tube)

### Step 3: Apply Ampere's Circuital Law
Ampere's Circuital Law states:
\[ \oint B \cdot dL = \mu_0 I_{\text{enc}} \]
where \( I_{\text{enc}} \) is the current enclosed by the Amperian loop.

### Step 4: Case 1: Inside the Tube (\( x < R \))
- For \( x < R \), the Amperian loop lies inside the tube.
- The current enclosed \( I_{\text{enc}} = 0 \) because all the current flows along the surface of the tube and does not pass through the area enclosed by the loop.
- Therefore, applying Ampere's Law:
\[ \oint B \cdot dL = 0 \implies B = 0 \quad \text{(inside the tube)} \]

### Step 5: Case 2: Outside the Tube (\( x > R \))
- For \( x > R \), the Amperian loop encloses the entire current \( I \).
- The current enclosed \( I_{\text{enc}} = I \).
- The magnetic field \( B \) is constant along the loop, and we can write:
\[ B \cdot (2\pi x) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi x} \quad \text{(outside the tube)} \]

### Conclusion
- For \( x < R \): \( B = 0 \)
- For \( x > R \): \( B = \frac{\mu_0 I}{2\pi x} \)