Problem 52 - Fundamentals of Physics

Question

The current-density magnitude in a certain circular wire is \(J=\left(2.75 \times 10^{10} \mathrm{~A} / \mathrm{m}^{4}\right) r^{2},\) where \(r\) is the radial distance out to the wire's radius of \(3.00 \mathrm{~mm} .\) The potential applied to the wire (end to end) is \(60.0 \mathrm{~V}\). How much energy is converted to thermal energy in \(1.00 \mathrm{~h} ?\)

Step-by-Step Solution

Answer

The energy converted is approximately 237,384 Joules.

Step 1: Understanding Current Density

Current density is given by the equation:\[J = (2.75 \times 10^{10} \mathrm{~A/m^4}) \cdot r^2\]The variable \( r \) represents the radial distance from the center of the wire.

Step 2: Area and Total Current

We can express the differential current \( dI \) across a differential area \( dA \) using:\[dI = J \cdot dA = J \, 2\pi r\, dr\]Substituting in the expression for \( J \),\[dI = (2.75 \times 10^{10} \mathrm{~A/m^4}) \cdot r^2 \cdot 2\pi r \, dr = (5.5 \times 10^{10} \pi) r^3 \, dr\]

Step 3: Integrating for Total Current

Integrate \( dI \) from \( r = 0 \) to \( r = 3.00\mathrm{~mm} = 3.00 \times 10^{-3} \mathrm{~m} \) to find the total current \( I \):\[I = \int_{0}^{3.00 \times 10^{-3}} (5.5 \times 10^{10} \pi) r^3 \, dr\]\[\I = (5.5 \times 10^{10} \pi) \left[ \frac{r^4}{4} \right]_0^{3.00 \times 10^{-3}}\] \[I = (5.5 \times 10^{10} \pi) \left( \frac{(3.00 \times 10^{-3})^4}{4} \right)\]\[=(5.5 \times 10^{10} \pi) \left( \frac{81 \times 10^{-12}}{4} \right)\]\[=(5.5 \times 10^{10} \pi) \times 20.25 \times 10^{-12} \]\[= 349.65 \times 10^{-2} \pi \, \mathrm{A}\]\[\approx 1.099 \, \mathrm{A} \]

Step 4: Calculating Power

Calculate the power \( P \) dissipated using the formula:\[P = V \cdot I\]where \( V = 60.0 \, \mathrm{V} \). Substituting the values, we have:\[P = 60 \, \mathrm{V} \times 1.099 \, \mathrm{A} \]\[P \approx 65.94 \, \mathrm{W}\]

Step 5: Converting Energy over Time

To find the energy converted to thermal energy over 1 hour, use the formula:\[E = P \cdot t \]where \( t = 3600 \, \mathrm{seconds} \, (1 \, \mathrm{hour}) \). Substituting in, we get:\[E = 65.94 \, \mathrm{W} \times 3600 \, \mathrm{s}\]\[E \approx 237384 \, \mathrm{J}\]

Step 6: Final Answer

The energy converted to thermal energy in 1 hour is approximately 237,384 Joules.

Key Concepts

Current Density

In the realm of electricity, current density is a crucial concept. It measures how much electric current is flowing through a specific area of a conductor. Think of it as the amount of current per unit area. The formula for current density \( J \) in this exercise is given by \( J = (2.75 \times 10^{10} \mathrm{~A/m^4}) \cdot r^2 \). Here, \( r \) is the radial distance from the wire's center. This formula tells us that the current density increases as the distance from the center increases, due to its dependence on \( r^2 \).
Understanding current density helps in analyzing the flow of electrons through conductors, and it's a fundamental parameter in assessing the behavior of electrical circuits.
This eventually feeds into knowing how efficient a wire or a conductor really is when used in electrical components and systems.

Power Dissipation

Power dissipation is an important concept when considering electrical circuits and energy conversion. It basically tells us how much electrical energy is being converted into heat within a circuit. In this exercise, we've calculated the power dissipation using the formula \( P = V \cdot I \), where \( V \) is the voltage across the wire, and \( I \) is the current flowing through it.
The computed power dissipation indicates how efficiently energy is being used or lost in the form of thermal energy. The significance of power dissipation comes to the forefront when we need to manage heat in technological devices, as excessive power dissipation may lead to overheating and damage.
Understanding power dissipation helps designers and engineers create systems that do not just function efficiently, but also last longer by avoiding issues related to heat damage.

Integration in Physics

Integration is a powerful mathematical tool used extensively in physics to solve problems related to continuous quantities. In this exercise, integration is used to find the total current flowing through the wire. We started with a differential current \( dI \) over a differential area \( dA \), then used integration to add up these small currents over the entire radius of the wire.
The integration from \( r = 0 \) to the final radius \( r = 3.00 \times 10^{-3} \, \mathrm{m} \) allows us to compute the total current \( I \). This method is crucial not only in electricity and magnetism but also in various other fields such as fluid dynamics and thermodynamics.
By using integration, we can calculate quantities that would otherwise be difficult or impossible to measure directly, providing a deeper understanding and accurate solutions to complex problems.

Electric Potential

Electric potential is a central concept in understanding how electric fields work. It refers to the potential energy per unit charge in an electric field, simply put, it's what drives the current through a wire when a potential difference is applied. In this problem, the electric potential difference applied across the wire is \( 60.0 \, \mathrm{V} \).
By applying this potential difference, an electric field is created, prompting charges to flow, producing current. This potential energy is then used to do work, or in this case, is converted into thermal energy, leading to power dissipation.
Understanding electric potential is vital for anyone studying circuit design and electrical engineering because it aids in knowing how much work can be done by electric charges and in predicting the flow of current through different materials and components.