A cylindrical wire of radius R has current density varying with distance r form its axis as `J(x)=J_0(1-(r^2)/(R^2))` . The total current through the wire is

Options

A. `(piJ_0R^2)/2`
B. `(2piJ_0R^2)/3`
C. `(4piJ_0R^2)/3`
D. none of these

Answer

To find the total current through a cylindrical wire with a given current density that varies with distance from the axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Density Function**:
The current density \( J(r) \) is given by:
\[ J(r) = J_0 \left( 1 - \frac{r^2}{R^2} \right) \]
where \( J_0 \) is a constant, \( r \) is the distance from the axis of the cylinder, and \( R \) is the radius of the cylinder.

2. **Determine the Differential Area**:
For a cylindrical shell at a distance \( r \) from the axis with thickness \( dr \), the differential area \( dA \) can be expressed as:
\[ dA = 2\pi r \, dr \]

3. **Calculate the Differential Current**:
The differential current \( dI \) through this differential area can be calculated using the formula:
\[ dI = J(r) \cdot dA \]
Substituting the expressions for \( J(r) \) and \( dA \):
\[ dI = J_0 \left( 1 - \frac{r^2}{R^2} \right) \cdot (2\pi r \, dr) \]

4. **Integrate to Find Total Current**:
To find the total current \( I \), integrate \( dI \) from \( r = 0 \) to \( r = R \):
\[ I = \int_0^R dI = \int_0^R J_0 \left( 1 - \frac{r^2}{R^2} \right) (2\pi r \, dr) \]
This can be separated into two integrals:
\[ I = 2\pi J_0 \int_0^R \left( r - \frac{r^3}{R^2} \right) dr \]

5. **Evaluate the Integrals**:
Calculate the two integrals:
- The first integral:
\[ \int_0^R r \, dr = \left[ \frac{r^2}{2} \right]_0^R = \frac{R^2}{2} \]
- The second integral:
\[ \int_0^R r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4} \]

Substitute these results back into the expression for \( I \):
\[ I = 2\pi J_0 \left( \frac{R^2}{2} - \frac{1}{R^2} \cdot \frac{R^4}{4} \right) \]
Simplifying this gives:
\[ I = 2\pi J_0 \left( \frac{R^2}{2} - \frac{R^2}{4} \right) = 2\pi J_0 \left( \frac{2R^2}{4} - \frac{R^2}{4} \right) = 2\pi J_0 \cdot \frac{R^2}{4} = \frac{\pi J_0 R^2}{2} \]

6. **Final Result**:
The total current through the wire is:
\[ I = \frac{\pi J_0 R^2}{2} \]