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

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

### Step 2: Applying Ampere's Circuital Law
According to Ampere's Circuital Law, the line integral of the magnetic field \( B \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the total current \( I \) enclosed by that loop.

\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]

### Step 3: Analyzing Different Regions
We need to consider different regions based on the value of \( x \):

1. **For \( x < r \)**:
- In this region, there is no current enclosed by the Amperian loop since the current is only flowing through the tube. Therefore, the magnetic field \( B \) is zero.
\[ B = 0 \quad \text{for } x < r \]

2. **For \( x = r \)**:
- At this point, we can use the formula derived from Ampere's Law:
\[ B = \frac{\mu_0 I}{2 \pi r} \]

3. **For \( x > r \)**:
- In this case, the Amperian loop encloses the entire current \( I \). Thus, the magnetic field can be calculated as:
\[ B = \frac{\mu_0 I}{2 \pi x} \]

### Step 4: Summarizing the Results
- For \( 0 < x < r \): \( B = 0 \)
- For \( x = r \): \( B = \frac{\mu_0 I}{2 \pi r} \)
- For \( x > r \): \( B = \frac{\mu_0 I}{2 \pi x} \)

### Conclusion
The magnetic field \( B \) is zero for distances less than the radius of the tube. It is non-zero and inversely proportional to \( x \) for distances greater than the radius.