Solution:

To calculate the magnetic field

$B$

at a point

$P$

within the outer shell of a coaxial cable (

$b < r < c$

), carrying equal and opposite currents

$i$

on the inner and outer conductors, we use Ampère's law and superposition principles.

---

1. Current Distribution in the Outer Shell

The outer shell carries current

$-i$

, distributed uniformly across the cross-sectional area of the shell between radii

$b$

and

$c$

.

The current density

$J$

in the shell is:

 

$$J = \frac{-i}{\pi (c^2 - b^2)}.$$

 

The current enclosed within a radius

$r$

(

$b < r < c$

) in the outer shell is the current

$-i$

contributed by the region from

$b$

to

$r$

:

 

$$I_{\text{enc,shell}} = J \cdot \text{area of shell portion} = J \cdot \pi (r^2 - b^2).$$

 

Substituting

$J$

:

 

$$I_{\text{enc,shell}} = \frac{-i}{\pi (c^2 - b^2)} \cdot \pi (r^2 - b^2) = \frac{-i (r^2 - b^2)}{c^2 - b^2}.$$

 

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2. Net Enclosed Current at Radius $r$

 

At any point

$P$

within the shell (

$b < r < c$

), the net current enclosed by a loop of radius

$r$

is:

 

$$I_{\text{enc}} = \text{current from the inner wire} + \text{current from the outer shell}.$$

 

The inner wire contributes

$+i$

, and the shell contributes

$I_{\text{enc,shell}}$

:

 

$$I_{\text{enc}} = i + \frac{-i (r^2 - b^2)}{c^2 - b^2}.$$

 

Simplify:

 

$$I_{\text{enc}} = i \left(1 - \frac{r^2 - b^2}{c^2 - b^2}\right).$$

 

Factorize:

 

$$I_{\text{enc}} = i \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

 

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3. Magnetic Field at Radius $r$

 

Using Ampère's law, the magnetic field

$B$

at radius

$r$

is:

 

$$B \cdot 2\pi r = \mu_0 I_{\text{enc}}.$$

 

Substitute

$I_{\text{enc}}$

:

 

$$B \cdot 2\pi r = \mu_0 \cdot i \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

 

Solve for

$B$

:

 

$$B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

 

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Final Answer:

 

$$\boxed{B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}}$$