The magnitude of the current density in a certain lab wire with a circular cross section of radius R=2.50 mm is given by `J= (3.00 xx 10^(8))r^(2)`, with J in amperes per square meter and radial distance rin meters. What is the current through the outer section bounded by r= 0.900R and r=R?

Answer

To solve the problem of finding the current through the outer section of the wire bounded by \( r = 0.9R \) and \( r = R \), we will follow these steps:

### Step 1: Understand the Current Density
The current density \( J \) is given by the equation:
\[ J = 3.00 \times 10^8 r^2 \]
where \( J \) is in amperes per square meter and \( r \) is the radial distance in meters.

### Step 2: Define the Area Element
For a circular cross-section, the differential area \( dA \) at a distance \( r \) from the center is given by:
\[ dA = 2\pi r \, dr \]

### Step 3: Relate Current to Current Density
The differential current \( dI \) flowing through the area \( dA \) can be expressed as:
\[ dI = J \cdot dA \]
Substituting the expression for \( dA \):
\[ dI = J \cdot (2\pi r \, dr) \]

### Step 4: Substitute for Current Density
Now substitute \( J \) into the equation:
\[ dI = (3.00 \times 10^8 r^2) \cdot (2\pi r \, dr) = 6\pi \times 10^8 r^3 \, dr \]

### Step 5: Set Up the Integral for Total Current
To find the total current \( I \) through the section bounded by \( r = 0.9R \) and \( r = R \), we integrate \( dI \) from \( r = 0.9R \) to \( r = R \):
\[ I = \int_{0.9R}^{R} 6\pi \times 10^8 r^3 \, dr \]

### Step 6: Perform the Integration
Calculating the integral:
\[ I = 6\pi \times 10^8 \int_{0.9R}^{R} r^3 \, dr \]
The integral of \( r^3 \) is:
\[ \int r^3 \, dr = \frac{r^4}{4} \]
Thus,
\[ I = 6\pi \times 10^8 \left[ \frac{r^4}{4} \right]_{0.9R}^{R} \]
Calculating the limits:
\[ I = 6\pi \times 10^8 \left( \frac{R^4}{4} - \frac{(0.9R)^4}{4} \right) \]
\[ = 6\pi \times 10^8 \left( \frac{R^4}{4} - \frac{0.6561R^4}{4} \right) \]
\[ = 6\pi \times 10^8 \left( \frac{R^4}{4} (1 - 0.6561) \right) \]
\[ = 6\pi \times 10^8 \left( \frac{R^4}{4} \times 0.3439 \right) \]

### Step 7: Substitute the Value of R
Given \( R = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \):
\[ I = 6\pi \times 10^8 \left( \frac{(2.5 \times 10^{-3})^4}{4} \times 0.3439 \right) \]

### Step 8: Calculate the Final Current
Calculating \( (2.5 \times 10^{-3})^4 \):
\[ (2.5 \times 10^{-3})^4 = 3.90625 \times 10^{-11} \, \text{m}^4 \]
Now substituting this back into the equation:
\[ I = 6\pi \times 10^8 \left( \frac{3.90625 \times 10^{-11}}{4} \times 0.3439 \right) \]
\[ = 6\pi \times 10^8 \times 9.8475 \times 10^{-12} \]
\[ = 6 \times 3.14 \times 10^8 \times 9.8475 \times 10^{-12} \]
Calculating the numerical values:
\[ I \approx 1.8398 \times 10^{-3} \, \text{A} \]

### Final Answer
The current through the outer section bounded by \( r = 0.9R \) and \( r = R \) is approximately:
\[ I \approx 1.84 \, \text{mA} \]