current enclosed is (1+2+3-1-4)= 1A

using ampere circuital law we get the answer.

To solve for the magnetic field at a distance of

$\frac{5R}{4}$

from the axis of a hollow cylindrical wire carrying current

$I$

, with inner radius

$R$

and outer radius

$2R$

, we use Ampère's Law. The key steps are:

1. Magnetic Field Inside a Hollow Cylinder

For

$R < r < 2R$

(points within the shell of the cylinder):

• Current density
  $J$ 

  is uniform, given by: 

  $$J = \frac{I}{\pi \left((2R)^2 - R^2\right)} = \frac{I}{3\pi R^2}.$$ 

• The current enclosed within a radius
  $r$ 

  (where $R < r < 2R$ 

  ) is: 

  $$I_{\text{enc}} = J \cdot \text{area enclosed} = J \cdot \pi (r^2 - R^2).$$Substituting

  $J$:

   

  $$I_{\text{enc}} = \frac{I}{3\pi R^2} \cdot \pi (r^2 - R^2) = \frac{I}{3R^2}(r^2 - R^2).$$ 

• Using Ampère's Law: 

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

  $$B = \frac{\mu_0}{2\pi r} \cdot \frac{I}{3R^2}(r^2 - R^2).$$ 

2. Magnetic Field at $r = \frac{5R}{4}$

Since

$\frac{5R}{4}$

lies within the shell (

$R < r < 2R$

), substitute

$r = \frac{5R}{4}$

into the above equation:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2}\left(\left(\frac{5R}{4}\right)^2 - R^2\right).$$

Simplify:

• 
  $r^2 = \left(\frac{5R}{4}\right)^2 = \frac{25R^2}{16}$ 

  ,

• 
  $r^2 - R^2 = \frac{25R^2}{16} - R^2 = \frac{25R^2}{16} - \frac{16R^2}{16} = \frac{9R^2}{16}$ 

  .

Thus:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2} \cdot \frac{9R^2}{16}.$$

Simplify further:

$$B = \frac{\mu_0}{2\pi} \cdot \frac{4}{5R} \cdot \frac{9I}{48} = \frac{\mu_0 I}{2\pi} \cdot \frac{36}{240R}.$$

Final simplification:

$$B = \frac{3}{40} \frac{\mu_0 I}{\pi R}.$$

Final Answer:

$$\boxed{B = \frac{3}{40} \frac{\mu_0 I}{\pi R}}$$