Current density in a cylindrical wire of radius R is gives as
`{{:(J_(0)((X)/(R)-1)"for "0lexlt(R)/(2),),(J_(0)(X)/(R)" ""for"(R)/(2)leXleR,):}` . The current flowing in the wire is

Options

A. `(7)/(24)piJ_(0)R^(2)`
B. `(1)/(6)piJ_(0)R^(2)`
C. `(7)/(12)piJ_(0)R^(2)`
D. `(5)/(12)piJ_(0)R^(2)`

Answer

To find the current flowing in the cylindrical wire with the given current density, we can follow these steps:

### Step 1: Understand the Current Density Function
The current density \( J \) is given in two regions:
1. For \( 0 \leq x < \frac{R}{2} \):
\[ J = J_0 \left( \frac{x}{R} - 1 \right) \]
2. For \( \frac{R}{2} \leq x \leq R \):
\[ J = J_0 \left( \frac{x}{R} \right) \]

### Step 2: Define the Differential Current Element
The differential current \( dI \) flowing through a differential area \( dA \) is given by:
\[ dI = J \cdot dA \]
For a cylindrical wire, the differential area \( dA \) at a distance \( x \) from the center is:
\[ dA = 2 \pi x \, dx \]

### Step 3: Set Up the Integral for Current
The total current \( I \) can be found by integrating \( dI \) over the entire cross-sectional area of the wire:
\[ I = \int_0^R J \cdot dA = \int_0^R J \cdot (2 \pi x \, dx) \]

### Step 4: Split the Integral into Two Parts
We will split the integral into two parts based on the regions defined for \( J \):
\[ I = \int_0^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2 \pi x) \, dx + \int_{R/2}^{R} J_0 \left( \frac{x}{R} \right) (2 \pi x) \, dx \]

### Step 5: Calculate the First Integral
For the first integral:
\[ I_1 = \int_0^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2 \pi x) \, dx \]
This simplifies to:
\[ I_1 = 2 \pi J_0 \int_0^{R/2} \left( \frac{x^2}{R} - x \right) \, dx \]
Calculating the integral:
\[ = 2 \pi J_0 \left[ \frac{x^3}{3R} - \frac{x^2}{2} \right]_0^{R/2} = 2 \pi J_0 \left[ \frac{(R/2)^3}{3R} - \frac{(R/2)^2}{2} \right] = 2 \pi J_0 \left[ \frac{R^2}{24} - \frac{R^2}{8} \right] = 2 \pi J_0 \left[ \frac{R^2}{24} - \frac{3R^2}{24} \right] = 2 \pi J_0 \left[ -\frac{2R^2}{24} \right] = -\frac{\pi J_0 R^2}{12} \]

### Step 6: Calculate the Second Integral
For the second integral:
\[ I_2 = \int_{R/2}^{R} J_0 \left( \frac{x}{R} \right) (2 \pi x) \, dx \]
This simplifies to:
\[ I_2 = 2 \pi J_0 \int_{R/2}^{R} \frac{x^2}{R} \, dx = \frac{2 \pi J_0}{R} \left[ \frac{x^3}{3} \right]_{R/2}^{R} = \frac{2 \pi J_0}{R} \left[ \frac{R^3}{3} - \frac{(R/2)^3}{3} \right] = \frac{2 \pi J_0}{R} \left[ \frac{R^3}{3} - \frac{R^3}{24} \right] = \frac{2 \pi J_0}{R} \left[ \frac{8R^3}{24} - \frac{R^3}{24} \right] = \frac{2 \pi J_0}{R} \cdot \frac{7R^3}{24} = \frac{14 \pi J_0 R^2}{24} = \frac{7 \pi J_0 R^2}{12} \]

### Step 7: Combine the Results
Now, we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = -\frac{\pi J_0 R^2}{12} + \frac{7 \pi J_0 R^2}{12} = \frac{6 \pi J_0 R^2}{12} = \frac{\pi J_0 R^2}{2} \]

### Final Answer
The total current flowing in the wire is:
\[ I = \frac{\pi J_0 R^2}{2} \]