The current density is a solid cylindrical wire a radius R, as a function of radial distance r is given by `J(r )=J_(0)(1-(r )/(R ))` . The total current in the radial regon r = 0 to `r=(R )/(4)` will be :

Options

A. `(5J_(0)pi R^(2))/(32)`
B. `(5J_(0)piR^(2))/(96)`
C. `(3J_(0)piR^(2))/(64)`
D. `(J_(0)piR^(3))/(128)`

Explanation

To solve the problem, we need to calculate the total current flowing through a cylindrical wire with a given current density function. The current density \( J(r) \) is defined as:

\[ J(r) = J_0 \left(1 - \frac{r}{R}\right) \]

where \( J_0 \) is a constant, \( r \) is the radial distance from the center of the cylinder, and \( R \) is the radius of the cylinder.

### Step-by-step Solution:

1. **Define the Elemental Area**:
The elemental area \( dA \) of a cylindrical shell at radius \( r \) with thickness \( dr \) is given by:
\[ dA = 2\pi r \, dr \]

2. **Define the Elemental Current**:
The elemental current \( dI \) through this area can be expressed as:
\[ dI = J(r) \, dA = J_0 \left(1 - \frac{r}{R}\right) \cdot (2\pi r \, dr) \]

3. **Set Up the Integral for Total Current**:
To find the total current \( I \) from \( r = 0 \) to \( r = \frac{R}{4} \), we integrate \( dI \):
\[ I = \int_0^{R/4} dI = \int_0^{R/4} J_0 \left(1 - \frac{r}{R}\right) (2\pi r \, dr) \]

4. **Simplify the Integral**:
Substitute \( J_0 \) and \( 2\pi \) outside the integral:
\[ I = 2\pi J_0 \int_0^{R/4} \left(1 - \frac{r}{R}\right) r \, dr \]

5. **Expand the Integrand**:
The integrand can be expanded:
\[ I = 2\pi J_0 \int_0^{R/4} \left(r - \frac{r^2}{R}\right) \, dr \]

6. **Calculate the Integral**:
Now, calculate the integral:
\[ I = 2\pi J_0 \left( \int_0^{R/4} r \, dr - \frac{1}{R} \int_0^{R/4} r^2 \, dr \right) \]
The integrals can be computed as:
\[ \int_0^{R/4} r \, dr = \left[\frac{r^2}{2}\right]_0^{R/4} = \frac{(R/4)^2}{2} = \frac{R^2}{32} \]
\[ \int_0^{R/4} r^2 \, dr = \left[\frac{r^3}{3}\right]_0^{R/4} = \frac{(R/4)^3}{3} = \frac{R^3}{192} \]

7. **Substitute Back into the Total Current Expression**:
Substitute these results back into the expression for \( I \):
\[ I = 2\pi J_0 \left( \frac{R^2}{32} - \frac{1}{R} \cdot \frac{R^3}{192} \right) \]
Simplifying this gives:
\[ I = 2\pi J_0 \left( \frac{R^2}{32} - \frac{R^2}{192} \right) \]

8. **Combine the Terms**:
To combine the fractions, find a common denominator (which is 192):
\[ \frac{R^2}{32} = \frac{6R^2}{192} \]
Thus,
\[ I = 2\pi J_0 \left( \frac{6R^2}{192} - \frac{R^2}{192} \right) = 2\pi J_0 \cdot \frac{5R^2}{192} \]

9. **Final Expression for Total Current**:
Therefore, the total current is:
\[ I = \frac{10\pi J_0 R^2}{192} = \frac{5\pi J_0 R^2}{96} \]

### Conclusion:
The total current in the radial region from \( r = 0 \) to \( r = \frac{R}{4} \) is:
\[ I = \frac{5\pi J_0 R^2}{96} \]