Consider a co axial cable which consists of an inner wilre of radius `a` surrounded by an outer shell of inner and outer radii `b` and `c` respectively. The inner wire caries a current `I` and outer shell carries an equal and opposite current. The magnetic field at a distance `x` from the axis where `bltxltc` is

Options

A. `(mu_0I(c^2-b^2))/(2pix(c^2-a^2))`
B. `(mu_0I(c^2-x^2))/(2pix(c^2-a^2))`
C. `(mu_0I(c^2-x^2))/(2pix(c^2-b^2))`
D. zero

Explanation

To find the magnetic field at a distance \( x \) from the axis of a coaxial cable where \( b < x < c \), we can follow these steps:

### Step 1: Understand the Configuration
We have a coaxial cable with:
- An inner wire of radius \( a \) carrying a current \( I \).
- An outer shell with inner radius \( b \) and outer radius \( c \) carrying an equal and opposite current \( -I \).

### Step 2: Choose an Amperian Loop
To apply Ampere's Law, we consider an Amperian loop of radius \( x \) such that \( b < x < c \). The magnetic field will be uniform along this loop due to symmetry.

### Step 3: Calculate the Area of the Amperian Loop
The area \( A \) of the Amperian loop is given by:
\[ A = 2\pi x \cdot dx \]
However, since we are interested in the current enclosed, we will focus on the current density.

### Step 4: Determine the Current Density of the Outer Shell
The current density \( J \) in the outer shell can be calculated as:
\[ J = \frac{I}{\pi(c^2 - b^2)} \]

### Step 5: Calculate the Current Enclosed by the Amperian Loop
The current enclosed \( I_{\text{enc}} \) by the Amperian loop can be expressed as:
\[ I_{\text{enc}} = J \cdot A_{\text{loop}} = J \cdot \pi(x^2 - b^2) \]
Substituting the expression for \( J \):
\[ I_{\text{enc}} = \frac{I}{\pi(c^2 - b^2)} \cdot \pi(x^2 - b^2) = \frac{I(x^2 - b^2)}{c^2 - b^2} \]

### Step 6: Apply Ampere's Law
According to Ampere's Law:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} \]
For our Amperian loop:
\[ B \cdot 2\pi x = \mu_0 \left( I - \frac{I(x^2 - b^2)}{c^2 - b^2} \right) \]
This simplifies to:
\[ B \cdot 2\pi x = \mu_0 \left( I \left( 1 - \frac{x^2 - b^2}{c^2 - b^2} \right) \right) \]

### Step 7: Simplify the Expression
Rearranging gives:
\[ B = \frac{\mu_0 I}{2\pi x} \left( \frac{c^2 - x^2}{c^2 - b^2} \right) \]

### Final Expression
Thus, the magnetic field \( B \) at a distance \( x \) from the axis where \( b < x < c \) is:
\[ B = \frac{\mu_0 I}{2\pi x} \cdot \frac{c^2 - x^2}{c^2 - b^2} \]