Problem 39 - University Physics with Modern Physics

Question

Displacement Current in a Wire. A long, straight, copper wire with a circular cross-scctional area of 2.1 \(\mathrm{mm}^{2}\) carries a current of 16 \(\mathrm{A}\) . The resistivity of the material is \(20 \times 10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the uniform electric field in the material? (b) If the cur- rent is changing at the rate of 4000 \(\mathrm{A} / \mathrm{s}\) , at what rate is the electric field in the material changing? (c) What is the displacement current density in the material in part (b)? (Hint: Since \(K\) for copper is very close to \(1,\) use \(\epsilon=\epsilon_{0} . )\) (d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 \(\mathrm{cm}\) from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of \(B\) . Is the contribution from the displacement current significant?

Step-by-Step Solution

Answer

(a) The electric field is \(1.52 \times 10^{-3} \text{ V/m}\). (b) The rate of change of electric field is \(0.038 \text{ V/m/s}\). (c) Displacement current density is \(3.36 \times 10^{-13} \text{ A/m}^2\). (d) The magnetic field is dominated by conduction current; displacement current contribution is negligible.

Step 1: Calculate the Current Density

The current density \( J \) can be calculated as the current \( I \) divided by the cross-sectional area \( A \). First, convert the area from \( \text{mm}^2 \) to \( \text{m}^2 \): \[ A = 2.1 \times 10^{-6} \text{ m}^2 \]. Then, \[ J = \frac{I}{A} = \frac{16 \text{ A}}{2.1 \times 10^{-6} \text{ m}^2} \].

Step 2: Determine the Electric Field

Using Ohm's Law in the form \( E = \rho J \), where \( E \) is the electric field, \( \rho \) is the resistivity \( 20 \times 10^{-8} \Omega \cdot \text{m} \), and \( J \) is the current density calculated previously: \[ E = (20 \times 10^{-8} \Omega \cdot \text{m}) \times J \].

Step 3: Calculate the Rate of Change of Electric Field

The rate of change of the electric field \( \frac{dE}{dt} \) is given by \( \frac{dE}{dt} = \frac{d}{dt}(\rho J) = \rho \frac{dJ}{dt} \). Since \( \frac{dI}{dt} = 4000 \text{ A/s} \), \[ \frac{dJ}{dt} = \frac{1}{A} \frac{dI}{dt} = \frac{4000}{2.1 \times 10^{-6}} \text{ A/m}^2/\text{s} \].

Step 4: Find Displacement Current Density

The displacement current density \( J_d \) is given by \( \epsilon_0 \frac{dE}{dt} \), with \( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \). So, \[ J_d = \epsilon_0 \times \frac{dE}{dt} \].

Step 5: Calculate Magnetic Field with Both Currents

Use Ampere's law, which states \( B \times 2\pi r = \mu_0 (I + I_d) \), where \( I_d = J_d \times A \), to find \( B \) at 6 cm from the center: \[ B = \frac{\mu_0 (I + I_d)}{2\pi r}\]. Here, \( r = 0.06 \text{ m} \) and \( \mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A} \).

Step 6: Determine Significance of Displacement Current

Calculate the percentage of the magnetic field due to the displacement current compared to the total magnetic field by dividing the contribution from \( I_d \) by the total \( I + I_d \). If the percentage is very small, the displacement current's contribution is not significant.

Key Concepts

Current Density

Current density is an essential concept when studying electric currents in conductors. It refers to the amount of electric current flowing per unit area through a given cross-section of a conductor. Think of it like the average flow of electric charge in an area, much like how traffic flow is measured on a road.
To calculate current density (\( J \)), divide the total current (\( I \)) by the cross-sectional area (\( A \)):

• Convert area from square millimeters (\( ext{mm}^2 \)) to square meters (\( ext{m}^2 \)). For instance, given \( A = 2.1 imes 10^{-6} ext{ m}^2 \).
• Use the formula: \( J = \frac{I}{A} \).

A high current density can be indicative of more significant energy transfer, often leading to increased heat due to resistive losses.

Electric Field

The electric field inside a conductor influences how charged particles move. It's essentially the force per unit charge that a charged particle would experience within the field.
To find the electric field within a conducting material, Ohm's Law can be adapted into the form:

• \( E = \rho J \), where \( E \) is the electric field, \( \rho \) is the material's resistivity, and \( J \) is the current density.
• The resistivity (\( \rho \)) for copper, as given, is \( 20 imes 10^{-8} ext{ Ω⋅m} \).

In essence, this relationship shows that electric fields arise in materials due to the inherent resistance they offer to moving charges.

Rate of Change of Electric Field

When the electric field inside a conductor changes with time, we describe this as the rate of change of the electric field. It signifies how rapidly the field is varying, often due to changes in the current.
The formula for calculating the rate of change (\( \frac{dE}{dt} \)) is:

• \( \frac{dE}{dt} = \rho \frac{dJ}{dt} \), where \( \frac{dJ}{dt} \) is the rate of change of the current density.
• This rate can be determined using the changing current over time: \( \frac{dI}{dt} \).

Understanding this concept helps predict the behavior of charges in time varying electrical conditions.

Magnetic Field

The magnetic field is a vector field surrounding electric currents and magnetic moments, influencing forces on nearby moving charges. In our context, when current flows through a wire, it produces a magnetic field around it.
To calculate the magnetic field (\( B \)) resulting from both conduction and displacement currents, Ampere's Law is used:

• Given: \( B \times 2\pi r = \mu_0 (I + I_d) \).
• Solve for \( B \) considering displacement current density \( J_d \).

This equation provides insight into how both traditional and time-varying currents contribute to magnetic field generation.

Ohm's Law

Ohm's Law is fundamental in understanding the linear relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through most conductors is directly proportional to the voltage across it.
In a formula form, it can be expressed as:

• \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
• For continuous media like wires, the version \( E = \rho J \) is used—which relates electric field, resistivity, and current density.

Ohm's Law helps in understanding how electrical energy is translated to mechanical or thermal energy in systems.

Ampere's Law

Ampere's Law is crucial for relating integrated magnetic field around a closed loop to the current passing through the loop. It's often used alongside Maxwell's equations to understand electromagnetic phenomena.
In specific scenarios like ours:

• \( B \times 2\pi r = \mu_0 (I + I_d) \)
• Where \( \mu_0 \) is the permeability of free space, \( I_d \) is the displacement current, and \( I \) is the conduction current.

By including both types of current, this law provides a comprehensive understanding of the resulting magnetic fields, highlighting elements neglected under simpler models.