A long wire in the shape of a cylindrical shell of inner and outer radii `R/2` and R respectively carries a steady current I in a direction parallel to its length. This current is uniformly distributed across the area of cross-section of the wire. The magnetic field intensity at a point at a distance `(3R)/4` from the axis of the wire is `m((mu_(0)I)/(pi R))`.. The value of m is _________.

Explanation

To solve the problem, we need to find the magnetic field intensity at a distance \( \frac{3R}{4} \) from the axis of a cylindrical shell carrying a steady current \( I \). The shell has inner radius \( \frac{R}{2} \) and outer radius \( R \).

### Step-by-Step Solution:

1. **Determine the Current Density (J)**:
The current density \( J \) in the cylindrical shell is given by the formula:
\[ J = \frac{I}{A} \]
where \( A \) is the cross-sectional area of the wire. The cross-sectional area \( A \) can be calculated as the difference between the area of the outer circle and the area of the inner circle:
\[ A = \pi R^2 - \pi \left(\frac{R}{2}\right)^2 = \pi R^2 - \pi \frac{R^2}{4} = \pi R^2 \left(1 - \frac{1}{4}\right) = \pi R^2 \cdot \frac{3}{4} = \frac{3\pi R^2}{4} \]
Thus, the current density becomes:
\[ J = \frac{I}{\frac{3\pi R^2}{4}} = \frac{4I}{3\pi R^2} \]

2. **Apply Ampere's Law**:
According to Ampere's Law, the magnetic field \( B \) around a long straight conductor is given by:
\[ B \cdot 2\pi r = \mu_0 \cdot I_{\text{enc}} \]
where \( I_{\text{enc}} \) is the current enclosed by the Amperian loop of radius \( r \). We need to find \( I_{\text{enc}} \) at \( r = \frac{3R}{4} \).

3. **Calculate the Enclosed Current**:
The distance \( \frac{3R}{4} \) is between the inner radius \( \frac{R}{2} \) and the outer radius \( R \). The current enclosed by the Amperian loop can be calculated using the current density:
\[ I_{\text{enc}} = J \cdot A_{\text{enc}} \]
where \( A_{\text{enc}} \) is the area of the annular region from \( \frac{R}{2} \) to \( \frac{3R}{4} \):
\[ A_{\text{enc}} = \pi \left(\left(\frac{3R}{4}\right)^2 - \left(\frac{R}{2}\right)^2\right) = \pi \left(\frac{9R^2}{16} - \frac{R^2}{4}\right) = \pi \left(\frac{9R^2}{16} - \frac{4R^2}{16}\right) = \pi \cdot \frac{5R^2}{16} \]
Therefore, the enclosed current is:
\[ I_{\text{enc}} = J \cdot A_{\text{enc}} = \frac{4I}{3\pi R^2} \cdot \left(\pi \cdot \frac{5R^2}{16}\right) = \frac{4I \cdot 5}{3 \cdot 16} = \frac{20I}{48} = \frac{5I}{12} \]

4. **Calculate the Magnetic Field (B)**:
Now substituting \( I_{\text{enc}} \) into Ampere's Law:
\[ B \cdot 2\pi \left(\frac{3R}{4}\right) = \mu_0 \cdot \frac{5I}{12} \]
Simplifying this gives:
\[ B = \frac{\mu_0 \cdot \frac{5I}{12}}{2\pi \cdot \frac{3R}{4}} = \frac{\mu_0 \cdot 5I}{12 \cdot 2\pi \cdot \frac{3R}{4}} = \frac{\mu_0 \cdot 5I \cdot 4}{12 \cdot 2\pi \cdot 3R} = \frac{20\mu_0 I}{72\pi R} = \frac{5\mu_0 I}{18\pi R} \]

5. **Compare with Given Magnetic Field**:
The problem states that the magnetic field intensity at that point is \( m \left(\frac{\mu_0 I}{\pi R}\right) \). Thus, we can equate:
\[ \frac{5\mu_0 I}{18\pi R} = m \left(\frac{\mu_0 I}{\pi R}\right) \]
From this, we find:
\[ m = \frac{5}{18} \]

### Final Answer:
The value of \( m \) is \( \frac{5}{18} \).