A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in fise melts whien the current derisity rises to 440 `A/cm^(2)`. What radius of cylindrical wire should be used to make a haal will little current to 6.0A?

Answer

To solve the problem of determining the radius of a cylindrical wire that will limit the current to 6.0 A while ensuring the current density does not exceed 440 A/cm², we can follow these steps:

### Step 1: Understand the relationship between current density, current, and area
The current density \( J \) is defined as the current \( I \) flowing per unit area \( A \) through the wire:
\[ J = \frac{I}{A} \]
Where:
- \( J \) is the current density (in A/cm²)
- \( I \) is the current (in A)
- \( A \) is the cross-sectional area of the wire (in cm²)

### Step 2: Express the area in terms of the radius
For a cylindrical wire, the cross-sectional area \( A \) can be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]
Where \( \pi \approx 3.14 \).

### Step 3: Substitute the area into the current density equation
Substituting \( A \) into the equation for current density gives:
\[ J = \frac{I}{\pi r^2} \]

### Step 4: Rearrange the equation to solve for the radius
We can rearrange the equation to solve for \( r \):
\[ \pi r^2 = \frac{I}{J} \]
\[ r^2 = \frac{I}{J \cdot \pi} \]
\[ r = \sqrt{\frac{I}{J \cdot \pi}} \]

### Step 5: Plug in the known values
Given:
- \( I = 6.0 \, A \)
- \( J = 440 \, A/cm^2 \)

Substituting these values into the equation:
\[ r = \sqrt{\frac{6.0}{440 \cdot 3.14}} \]

### Step 6: Calculate the value
Calculating the denominator:
\[ 440 \cdot 3.14 \approx 1388.6 \]
Now, substituting back:
\[ r = \sqrt{\frac{6.0}{1388.6}} \approx \sqrt{0.00432} \approx 0.066 \, cm \]

### Step 7: Convert to meters
To convert the radius from centimeters to meters:
\[ r = 0.066 \, cm = 0.066 \times 10^{-2} \, m = 6.6 \times 10^{-4} \, m \]

### Final Answer
The radius of the cylindrical wire should be approximately:
\[ r \approx 6.6 \times 10^{-4} \, m \]

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