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 varying current density, we can follow these steps:

### Step 1: Understand the Given Current Density
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.

### Step 2: Define a Differential Current Element
To find the total current, we will consider a thin cylindrical shell of radius \( r \) and thickness \( dr \). The differential area \( dA \) of this shell is given by the circumference times the thickness:
\[ dA = 2\pi r \, dr \]

### Step 3: Express the Differential Current
The differential current \( dI \) flowing through this shell can be expressed as:
\[ dI = J(r) \cdot dA = J_0 \left( 1 - \frac{r^2}{R^2} \right) \cdot (2\pi r \, dr) \]

### Step 4: Integrate to Find Total Current
To find the total current \( I \), we integrate \( dI \) from \( r = 0 \) to \( r = R \):
\[ I = \int_0^R dI = \int_0^R J_0 \left( 1 - \frac{r^2}{R^2} \right) (2\pi r \, dr) \]

### Step 5: Simplify the Integral
We can factor out the constants:
\[ I = 2\pi J_0 \int_0^R \left( 1 - \frac{r^2}{R^2} \right) r \, dr \]
Now, we can split the integral:
\[ I = 2\pi J_0 \left( \int_0^R r \, dr - \frac{1}{R^2} \int_0^R r^3 \, dr \right) \]

### Step 6: Calculate the Integrals
1. The first integral:
\[ \int_0^R r \, dr = \left[ \frac{r^2}{2} \right]_0^R = \frac{R^2}{2} \]
2. The second integral:
\[ \int_0^R r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4} \]

### Step 7: Substitute Back into the Equation
Substituting these results back into the equation for \( I \):
\[ I = 2\pi J_0 \left( \frac{R^2}{2} - \frac{1}{R^2} \cdot \frac{R^4}{4} \right) \]
\[ I = 2\pi J_0 \left( \frac{R^2}{2} - \frac{R^2}{4} \right) \]
\[ I = 2\pi J_0 \left( \frac{2R^2}{4} - \frac{R^2}{4} \right) = 2\pi J_0 \cdot \frac{R^2}{4} \]

### Step 8: Final Result
Thus, the total current \( I \) through the wire is:
\[ I = \frac{\pi J_0 R^2}{2} \]

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