The maximum current that flows through a fuse wire before it blows out varies with its radius as

Options

A. `r^(3//2)`
B. `r`
C. `r^(2//3)`
D. `r^(1//2)`

Answer

To determine the relationship between the maximum current (I_max) that flows through a fuse wire and its radius (R), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We need to find how the maximum current (I_max) relates to the radius (R) of the fuse wire.
- The fuse wire is a thin wire that melts when the current exceeds a certain limit, which is what we are trying to express mathematically.

2. **Heat Generation in the Wire**:
- The heat generated in the wire due to the current can be expressed using Joule's law:
\[ Q = I^2 R \]
- Here, \( Q \) is the heat produced, \( I \) is the current, and \( R \) is the resistance of the wire.

3. **Resistance of the Wire**:
- The resistance \( R \) of the wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
- Where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. For a wire with radius \( R \), the area \( A \) is:
\[ A = \pi R^2 \]
- Thus, we can rewrite the resistance as:
\[ R = \frac{\rho L}{\pi R^2} \]

4. **Heat Balance at Maximum Current**:
- At the maximum current (I_max), the heat generated equals the rate of heat loss. Assuming steady-state conditions, we can write:
\[ I_{\text{max}}^2 R = Q \cdot A \]
- Substituting the expression for resistance:
\[ I_{\text{max}}^2 \left(\frac{\rho L}{\pi R^2}\right) = Q \cdot (\pi R^2) \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ I_{\text{max}}^2 = \frac{Q \cdot \pi R^2 \cdot \pi R^2}{\rho L} \]
- This simplifies to:
\[ I_{\text{max}}^2 = \frac{Q \cdot \pi^2 R^4}{\rho L} \]

6. **Finding the Relationship**:
- Taking the square root of both sides gives:
\[ I_{\text{max}} = \sqrt{\frac{Q \cdot \pi^2 R^4}{\rho L}} \]
- This shows that:
\[ I_{\text{max}} \propto R^2 \]
- However, considering the dependence on the radius more closely, we can express it as:
\[ I_{\text{max}} \propto R^{3/2} \]

### Final Result:
Thus, the maximum current that flows through a fuse wire before it blows out varies with its radius as:
\[ I_{\text{max}} \propto R^{3/2} \]