A cylinder conductor AB of non uniform area of cross-section carries a current of 5A. The radius of the conductor at one end A is 0.5 cm.The current density at the other end of the conductor is half of the value at A. The radius of the conductor at the end B is nearly

Options

A. 1.4 cm
B. 0.7 cm
C. 0.6cm
D. None of these

Explanation

To solve the problem, we need to find the radius of the conductor at the end B, given that the current density at B is half of that at A. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Current (I) = 5 A
- Radius at end A (r_A) = 0.5 cm
- Current density at end B (J_B) = 0.5 * J_A

2. **Formula for Current Density**:
- The current density (J) is defined as:
\[ J = \frac{I}{A} \]
- Where A is the cross-sectional area of the conductor. For a circular cross-section, the area A is given by:
\[ A = \pi r^2 \]

3. **Calculate Current Density at End A**:
- The area at end A (A_A) can be calculated as:
\[ A_A = \pi (r_A)^2 = \pi (0.5 \, \text{cm})^2 = \pi (0.25 \, \text{cm}^2) \]
- Thus, the current density at end A (J_A) is:
\[ J_A = \frac{I}{A_A} = \frac{5 \, \text{A}}{\pi (0.25 \, \text{cm}^2)} = \frac{20}{\pi} \, \text{A/cm}^2 \]

4. **Current Density at End B**:
- Since J_B = 0.5 * J_A, we have:
\[ J_B = 0.5 \times \frac{20}{\pi} = \frac{10}{\pi} \, \text{A/cm}^2 \]

5. **Area at End B**:
- The area at end B (A_B) can be expressed in terms of the radius at B (r_B):
\[ A_B = \pi (r_B)^2 \]
- Using the definition of current density at end B:
\[ J_B = \frac{I}{A_B} \implies \frac{10}{\pi} = \frac{5}{\pi (r_B)^2} \]

6. **Solving for r_B**:
- Rearranging the equation gives:
\[ 10 \cdot (r_B)^2 = 5 \implies (r_B)^2 = \frac{5}{10} = 0.5 \]
- Taking the square root:
\[ r_B = \sqrt{0.5} = \frac{1}{\sqrt{2}} \approx 0.707 \, \text{cm} \]

7. **Final Answer**:
- The radius at end B (r_B) is approximately 0.7 cm.

### Conclusion:
The radius of the conductor at the end B is nearly **0.7 cm**.