The current-density magnitude in a certain circular wire is `J= (2.75 xx 10^(10) A//m^(4))r^(2)`, where r is the radial distance out to the wire's radius of 3.00 mm. The potential applied to the wire (end to end) is 80.0 V. How much energy is converted to thermal energy in 1.00 h?

Answer

To solve the problem, we need to follow these steps:

### Step 1: Understand the Current Density
The current density \( J \) is given by:
\[ J = 2.75 \times 10^{10} \, \text{A/m}^4 \cdot r^2 \]
where \( r \) is the radial distance from the center of the wire, and the radius of the wire is \( 3.00 \, \text{mm} = 3.00 \times 10^{-3} \, \text{m} \).

### Step 2: Calculate the Total Current \( I \)
To find the total current \( I \) flowing through the wire, we need to integrate the current density over the cross-sectional area of the wire.

The differential area element \( dA \) in cylindrical coordinates is:
\[ dA = 2\pi r \, dr \]

Thus, the differential current \( dI \) can be expressed as:
\[ dI = J \cdot dA = J \cdot (2\pi r \, dr) \]

Substituting for \( J \):
\[ dI = (2.75 \times 10^{10} \, r^2) \cdot (2\pi r \, dr) = 2\pi \cdot 2.75 \times 10^{10} \cdot r^3 \, dr \]

Now, integrate \( dI \) from \( r = 0 \) to \( r = 3 \times 10^{-3} \):
\[ I = \int_0^{3 \times 10^{-3}} 2\pi \cdot 2.75 \times 10^{10} \cdot r^3 \, dr \]

### Step 3: Perform the Integration
Calculating the integral:
\[ I = 2\pi \cdot 2.75 \times 10^{10} \cdot \left[ \frac{r^4}{4} \right]_0^{3 \times 10^{-3}} \]
Substituting the limits:
\[ I = 2\pi \cdot 2.75 \times 10^{10} \cdot \left[ \frac{(3 \times 10^{-3})^4}{4} - 0 \right] \]
Calculating \( (3 \times 10^{-3})^4 \):
\[ (3 \times 10^{-3})^4 = 81 \times 10^{-12} = 8.1 \times 10^{-11} \]
Now substituting back:
\[ I = 2\pi \cdot 2.75 \times 10^{10} \cdot \frac{8.1 \times 10^{-11}}{4} \]
Calculating:
\[ I = 2\pi \cdot 2.75 \times 10^{10} \cdot 2.025 \times 10^{-11} \]
\[ I \approx 3.5 \, \text{A} \]

### Step 4: Calculate the Energy Converted to Thermal Energy
The energy converted to thermal energy \( E \) is given by the formula:
\[ E = V \cdot I \cdot t \]
where \( V = 80.0 \, \text{V} \), \( I = 3.5 \, \text{A} \), and \( t = 1 \, \text{h} = 3600 \, \text{s} \).

Substituting the values:
\[ E = 80.0 \cdot 3.5 \cdot 3600 \]
Calculating:
\[ E = 80.0 \cdot 3.5 = 280 \, \text{W} \]
\[ E = 280 \cdot 3600 = 1.008 \times 10^6 \, \text{J} \]

### Final Answer
The energy converted to thermal energy in 1 hour is approximately:
\[ E \approx 1.1 \times 10^6 \, \text{J} \]