A long copper tube of inner radius `R` carriers a current `i`. The magnetic field `B` inside the tube is

Options

A. `mu_(0)i/2piR`
B. `mu_(0)i/4piR`
C. `mu_(0)i/2R`
D. Zero

Answer

To find the magnetic field \( B \) inside a long hollow copper tube of inner radius \( R \) carrying a current \( I \), we can use Ampere's Circuital Law. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
Consider a long hollow copper tube with inner radius \( R \). The current \( I \) flows through the tube.

### Step 2: Apply Ampere's Circuital Law
Ampere's Circuital Law states that:
\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]
where \( B \) is the magnetic field, \( dl \) is an infinitesimal length along the path of integration, \( \mu_0 \) is the permeability of free space, and \( I_{\text{enc}} \) is the current enclosed by the Amperian loop.

### Step 3: Choose an Amperian Loop
To find the magnetic field inside the tube, we choose a cylindrical Amperian loop of radius \( r \) (where \( r < R \)) that is concentric with the tube.

### Step 4: Determine the Enclosed Current
Since the tube is hollow and the current flows only on the outer surface of the tube, there is no current enclosed by our Amperian loop inside the tube. Therefore:
\[ I_{\text{enc}} = 0 \]

### Step 5: Substitute into Ampere's Law
Substituting \( I_{\text{enc}} = 0 \) into Ampere's Circuital Law gives:
\[ \oint B \cdot dl = \mu_0 \cdot 0 = 0 \]
This implies that:
\[ B \cdot (2 \pi r L) = 0 \]
where \( L \) is the length of the Amperian loop.

### Step 6: Conclude the Magnetic Field
Since the product \( B \cdot (2 \pi r L) = 0 \), we conclude that:
\[ B = 0 \]
Thus, the magnetic field inside the long hollow copper tube is zero.

### Final Answer
The magnetic field \( B \) inside the tube is:
\[ B = 0 \]

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