The current density varies radical distance r as `J=ar^(2)`, in a cylindrical wire of radius R. The current passing through the wire between radical distance `R//3 and R//2` is,

Options

A. `(65piaR^(4))/(2592)`
B. `(25piaR^(4))/(72)`
C. `(65pia^(2)R^(3))/(2938)`
D. `(81pia^(2)R^(4))/(144)`

Explanation

To solve the problem of finding the current passing through a cylindrical wire between the radial distances \( \frac{R}{3} \) and \( \frac{R}{2} \) where the current density \( J \) varies with the radial distance \( r \) as \( J = A r^2 \), we will follow these steps:

### Step 1: Understand the Current Density
The current density \( J \) is given by the equation:
\[ J = A r^2 \]
where \( A \) is a constant and \( r \) is the radial distance from the center of the cylindrical wire.

### Step 2: Express Current in Terms of Current Density
The current \( I \) flowing through a differential area \( dA \) can be expressed as:
\[ dI = J \cdot dA \]
For a cylindrical wire, the differential area \( dA \) at a distance \( r \) is given by the circumference of the circle multiplied by an infinitesimal thickness \( dr \):
\[ dA = 2\pi r \, dr \]
Thus, we can write:
\[ dI = J \cdot dA = A r^2 \cdot (2\pi r \, dr) = 2\pi A r^3 \, dr \]

### Step 3: Integrate to Find Total Current
To find the total current \( I \) flowing through the wire between the limits \( r = \frac{R}{3} \) and \( r = \frac{R}{2} \), we need to integrate \( dI \):
\[ I = \int_{R/3}^{R/2} 2\pi A r^3 \, dr \]

### Step 4: Perform the Integration
The integral can be computed as follows:
\[ I = 2\pi A \int_{R/3}^{R/2} r^3 \, dr \]
The integral of \( r^3 \) is:
\[ \int r^3 \, dr = \frac{r^4}{4} \]
Thus, we have:
\[ I = 2\pi A \left[ \frac{r^4}{4} \right]_{R/3}^{R/2} \]
Substituting the limits:
\[ I = 2\pi A \left( \frac{(R/2)^4}{4} - \frac{(R/3)^4}{4} \right) \]
Calculating the terms:
\[ I = 2\pi A \left( \frac{R^4}{64} - \frac{R^4}{324} \right) \cdot \frac{1}{4} \]
\[ I = \frac{\pi A R^4}{8} \left( \frac{1}{64} - \frac{1}{324} \right) \]

### Step 5: Simplify the Expression
To simplify \( \frac{1}{64} - \frac{1}{324} \):
Finding a common denominator (which is \( 64 \times 324 = 20736 \)):
\[ \frac{1}{64} = \frac{324}{20736}, \quad \frac{1}{324} = \frac{64}{20736} \]
Thus:
\[ \frac{1}{64} - \frac{1}{324} = \frac{324 - 64}{20736} = \frac{260}{20736} \]
Now substituting back:
\[ I = \frac{\pi A R^4}{8} \cdot \frac{260}{20736} \]
\[ I = \frac{260 \pi A R^4}{165888} \]

### Final Result
Thus, the current passing through the wire between the radial distances \( \frac{R}{3} \) and \( \frac{R}{2} \) is:
\[ I = \frac{65 \pi A R^4}{41472} \]