Current density in a cylindrical wire of radius R is given as `J={(J_(0)((x)/(R )-1)" for "0lex lt (R )/(2)),(J_(0)(x)/(R )" for "(R )/(2)le x le R):}`.
If the current flowing in the wire is `(k)/(12)piJ_(0)R^(2)`, find the value of k.

Answer

To find the value of \( k \) in the given problem, we will follow these steps:

### Step 1: Understand the Current Density Function
The current density \( J \) is given as:
\[ J = \begin{cases} J_0 \left( \frac{x}{R} - 1 \right) & \text{for } 0 \leq x < \frac{R}{2} \\ J_0 \left( \frac{x}{R} \right) & \text{for } \frac{R}{2} \leq x \leq R \end{cases} \]

### Step 2: Calculate the Current from Current Density
The total current \( I \) flowing through the wire can be calculated by integrating the current density over the cross-sectional area of the wire. The current \( I \) can be expressed as:
\[ I = \int_0^R J \, dA \]
where \( dA = 2\pi x \, dx \) is the differential area element.

### Step 3: Split the Integral
We will split the integral into two parts based on the piecewise definition of \( J \):
\[ I = \int_0^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2\pi x) \, dx + \int_{R/2}^{R} J_0 \left( \frac{x}{R} \right) (2\pi x) \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral**:
\[ I_1 = \int_0^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2\pi x) \, dx \]
This simplifies to:
\[ I_1 = 2\pi J_0 \int_0^{R/2} \left( \frac{x^2}{R} - x \right) \, dx \]
Evaluating this integral:
\[ I_1 = 2\pi J_0 \left[ \frac{x^3}{3R} - \frac{x^2}{2} \right]_0^{R/2} = 2\pi J_0 \left( \frac{(R/2)^3}{3R} - \frac{(R/2)^2}{2} \right) \]
\[ = 2\pi J_0 \left( \frac{R^2}{24} - \frac{R^2}{8} \right) = 2\pi J_0 \left( \frac{R^2}{24} - \frac{3R^2}{24} \right) = 2\pi J_0 \left( -\frac{2R^2}{24} \right) = -\frac{\pi J_0 R^2}{12} \]

2. **Second Integral**:
\[ I_2 = \int_{R/2}^{R} J_0 \left( \frac{x}{R} \right) (2\pi x) \, dx \]
This simplifies to:
\[ I_2 = 2\pi J_0 \int_{R/2}^{R} \frac{x^2}{R} \, dx \]
Evaluating this integral:
\[ I_2 = 2\pi J_0 \left[ \frac{x^3}{3R} \right]_{R/2}^{R} = 2\pi J_0 \left( \frac{R^3}{3R} - \frac{(R/2)^3}{3R} \right) \]
\[ = 2\pi J_0 \left( \frac{R^2}{3} - \frac{R^2}{24} \right) = 2\pi J_0 \left( \frac{8R^2}{24} - \frac{R^2}{24} \right) = 2\pi J_0 \left( \frac{7R^2}{24} \right) = \frac{14\pi J_0 R^2}{24} = \frac{7\pi J_0 R^2}{12} \]

### Step 5: Combine the Results
Now, combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = -\frac{\pi J_0 R^2}{12} + \frac{7\pi J_0 R^2}{12} = \frac{6\pi J_0 R^2}{12} = \frac{\pi J_0 R^2}{2} \]

### Step 6: Set Equal to Given Current
We are given that:
\[ I = \frac{k}{12} \pi J_0 R^2 \]
Setting the two expressions for current equal gives:
\[ \frac{\pi J_0 R^2}{2} = \frac{k}{12} \pi J_0 R^2 \]
Dividing both sides by \( \pi J_0 R^2 \) (assuming \( J_0 \neq 0 \) and \( R \neq 0 \)):
\[ \frac{1}{2} = \frac{k}{12} \]

### Step 7: Solve for \( k \)
Multiplying both sides by 12:
\[ k = 6 \]

### Final Answer
Thus, the value of \( k \) is \( 6 \).