Problem 10 - Fundamentals of Physics

Question

The magnitude \(J\) of the current density in a certain lab wire with a circular cross section of radius \(R=2.00 \mathrm{~mm}\) is given by \(J=\left(3.00 \times 10^{8}\right) r^{2},\) with \(J\) in amperes per square meter and radial distance \(r\) in meters. What is the current through the outer section bounded by \(r=0.900 R\) and \(r=R ?\)

Step-by-Step Solution

Answer

The current is approximately \(8.147 \text{ mA}\).

Step 1: Understand the Problem

We need to find the current flowing through the section of the wire from \(r = 0.900R\) to \(r = R\), where \(R = 2.00\, \text{mm}\). The formula for the current density \(J\) is given, which depends on the radial position \(r\). The current can be calculated by integrating the current density over the area of the section.

Step 2: Set Up the Integral

The current \(I\) through an infinitesimally small annular strip at radius \(r\) and thickness \(dr\) can be expressed as \(dI = J \, dA\), where \(dA = 2 \pi r \, dr\) is the area of the strip. Substitute the expression for \(J\):\[dI = (3.00 \times 10^{8}) r^{2} \, 2 \pi r \, dr = 6 \pi \times 10^8 \, r^3 \, dr\]

Step 3: Define the Limits of Integration

The problem requires us to find the current from \(r = 0.900R\) to \(r = R\). Convert these to meters:- \(0.900R = 0.900 \times 2.00 \times 10^{-3}\, \text{m} = 1.80 \times 10^{-3}\, \text{m}\)- \(R = 2.00 \times 10^{-3} \text{ m}\)Set the limits of integration from \(1.80 \times 10^{-3}\, \text{m}\) to \(2.00 \times 10^{-3}\, \text{m}\).

Step 4: Integrate to Find the Current

Evaluate the integral:\[I = \int_{1.80 \times 10^{-3}}^{2.00 \times 10^{-3}} 6 \pi \times 10^8 \, r^3 \, dr\]Perform the integration:\[I = 6 \pi \times 10^8 \left[ \frac{r^4}{4} \right]_{1.80 \times 10^{-3}}^{2.00 \times 10^{-3}}\]Calculate the result:\[I = 6 \pi \times 10^8 \left( \frac{(2.00 \times 10^{-3})^4}{4} - \frac{(1.80 \times 10^{-3})^4}{4} \right)\]

Step 5: Solve the Integral Expression

Compute each term:- \(\frac{(2.00 \times 10^{-3})^4}{4} = \frac{16 \times 10^{-12}}{4} = 4 \times 10^{-12}\)- \(\frac{(1.80 \times 10^{-3})^4}{4} = \frac{10.4976 \times 10^{-12}}{4} = 2.6244 \times 10^{-12}\)Plug these back into the formula:\[I = 6 \pi \times 10^8 ((4 - 2.6244) \times 10^{-12})\]\[I = 6 \pi \times 10^8 \times 1.3756 \times 10^{-12}\]\[I \approx 2.5935 \times 10^{-3} \pi \text{ Amperes}\]Calculate the numerical value using \(\pi \approx 3.14159\), \[I \approx 8.147 \text{ mA}\]

Step 6: Conclusion

The calculated current through the section of the wire bounded by \(r = 0.900R\) to \(r=R\) is approximately 8.147 \(\text{mA}\), which can be rounded as needed for presentation.

Key Concepts

Integral Calculus

Integral calculus is a fundamental tool in mathematics that helps us understand change and accumulation. When applied to problems like the one we have, it allows us to calculate not just discrete values, but quantities spread out over a range. In this problem, integral calculus is used to calculate the total current from the current density over a specified area.

This is achieved by setting up an integral that incorporates the current density function and the geometric dimensions of the wire. We use the expression \(dI = J \, dA\), where \(dI\) is the differential element of current, \(J\) is the current density, and \(dA\) is the differential area element.

• The current density \(J\) was given as a function of radial distance \(r\), \(J = (3.00 \times 10^{8}) r^{2}\).
• The area element \(dA\) for a circular strip is \(2 \pi r \, dr\), derived from the geometry of the circle.
• Integrating \(dI = J \, dA\) over the desired limits gives us the total current \(I\).

Through integration, you can take the radial-dependent current density and find the accumulated current over a specified area, demonstrating the power of integral calculus in physics.

Radial Coordinates

Radial coordinates are essential for problems involving circular or spherical symmetry. This coordinate system uses a radius and angle (or angles) to define a point's position, rather than just rectangular coordinates. In this exercise, radial distance is crucial because we're dealing with a circular cross-section of a wire.

• We identify positions within the circle using a single radial distance \(r\), where \(r\) can vary from 0 (center of the wire) to \(R\) (edge of the wire).
• This radial coordinate simplifies our problem since the current density \(J\) is a function of \(r\), \(J = 3.00 \times 10^{8} r^2\).
• Moreover, calculating the current involves integrating with respect to this radial distance \(r\).

Understanding radial coordinates helps visualize the problem effectively. You can see the current density spreading outwards from the center, enabling precise integration calculations within specific bounds.

Current Calculation

Calculating current involves knowing both the density of current across a section and the area over which it spreads. In this exercise, the current density varies with radius \(r\), and we needed to calculate the total current within a specific section of the wire.

• The current \(I\) is determined by integrating the current density \(J\) over the specified area of the wire.
• The given problem sets the region of interest between \(r = 0.900R\) and \(r = R\), with \(R\) being the wire's radius (\(2.00\, \text{mm}\)).
• By converting these radial limits to meters, the integration bounds become \(1.80 \times 10^{-3}\, \text{m}\) to \(2.00 \times 10^{-3}\, \text{m}\).

The integration \[ I = \int_{1.80 \times 10^{-3}}^{2.00 \times 10^{-3}} 6 \pi \times 10^8 \, r^3 \, dr \] proceeds by calculating the definite integral of the current density expression over this radial distance.

The result, \(I \approx 8.147 \, \text{mA}\), showcases how modifying the bounds and performing the integration provides a precise measure of current in the desired section.