91Ó°ÊÓ

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\) (a) What are the magnitude and direction of the electric-field vector \(E\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\vec { \boldsymbol { B } }\) at the same point? \(( \mathrm { c } )\) What are the magnitude and direction of the Poynting vector \(\vec { \boldsymbol { S } }\) at the same point? (The direction of \(\vec { \boldsymbol { S } }\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (Hint: Integrate \(\vec { S }\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

Step by step solution

01

Calculate the Electric Field Inside the Conductor

To find the magnitude of the electric field \(E\) at the surface of the conductor (at distance \(a\) from the axis), use Ohm's Law for a conductor, \(V = I R, \where \V = E \cdot l\), the potential difference across the length \(l\) \(\text{of the conductor} \), and \(R = \frac{\rho l}{A}\). Substitute the given expressions and solve for \(E\). \\[E = \frac{\rho I}{\pi a^2}\] where a is the cross-sectional radius, and \\(\rho\) is the resistivity.

02

Determine the Magnetic Field Just Inside the Conductor

Use Ampere's Law in its integral form, \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}\), where \(I_{\text{enc}} = I\) for the entire current inside the path. The magnetic field \(\vec{B}\) at a distance \(a\) perpendicular to the flow of the current forms concentric circles,\[B = \frac{\mu_0 I}{2\pi a} \]where \(\mu_0\) is the permeability of free space.

03

Find the Poynting Vector at the Surface of the Conductor

The Poynting vector \(\vec{S}\) is given by \(\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})\). The magnitude \((S)\) of this cross product is \[S = \frac{1}{\mu_0} E B \= \frac{1}{\mu_0} \left(\frac{\rho I}{\pi a^2}\right)\left(\frac{\mu_0 I}{2\pi a}\right) \= \frac{\rho I^2}{2\pi^2 a^3} \\]and the direction will be radially inward, perpendicular to both \(E\) and \(B\).

04

Calculate Energy Flow into the Volume

The rate of energy flow into the cylindrical volume of length \(l\) is given by the surface integral of the Poynting vector \(\vec{S}\) over the cylindrical surface area. With \((S)\) calculated already: \\[\Phi = \int S \, dA = S (2 \pi a l)= \left(\frac{\rho I^2}{2\pi^2 a^3}\right) (2 \pi a l) = \frac{\rho I^2 l}{\pi a^2}\\]This matches the power dissipated \(P = I^2 R = I^2 (\frac{\rho l}{\pi a^2})\). The energy dissipated as heat is effectively fed by energy entering through the sides of the conductor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field

The electric field inside a conductor is determined by the relationship between the applied voltage, the current flowing through the conductor, and its geometrical and material properties. We use Ohm's Law, given by the formula\[ V = I R, \]where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. The resistance \( R \) can also be expressed as \( R = \frac{\rho l}{A} \), where \( \rho \) is the material's resistivity, \( l \) is the length of the conductor, and \( A \) is its cross-sectional area.
By substituting and rearranging these expressions, we find the electric field \( E \) at the surface just inside the conductor as \[ E = \frac{V}{l} = \frac{\rho I}{\pi a^2}, \]where \( a \) is the cross-sectional radius of the wire.
This electric field is directed along the axis of the conductor, indicating the direction in which positive charges would move under the influence of the field.

Magnetic Field

Inside a conductor carrying current, a magnetic field arises due to the flow of charge. Using Ampere's Law, we quantify this field. The law relates the magnetic field around a closed loop to the current passing through the loop:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}. \]For a cylindrical conductor, this loop is a circle of radius \( a \) centered around the conductor's axis.
The magnetic field \( \vec{B} \) forms concentric circles around the wire and is given by:
\[ B = \frac{\mu_0 I}{2\pi a}, \]
where \( \mu_0 \) is the permeability of free space. It is important to note that the direction of \( B \) is perpendicular to both the electric field and the direction of current flow, illustrating the right-hand rule application for magnetic fields.

Poynting Vector

The Poynting vector \( \vec{S} \) is a key concept in understanding electromagnetic energy flow. It represents the directional energy flux (or power per unit area) of an electromagnetic field. The vector is defined as:
\[ \vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}). \]
This expression allows us to calculate the electromagnetic energy flow around and into the conductor. For the given cylindrical conductor, the magnitude of the Poynting vector at the surface is:
\[ S = \frac{\rho I^2}{2\pi^2 a^3}. \]
The direction of \( \vec{S} \) is determined by the cross product \( \vec{E} \times \vec{B} \), which results in a vector that is perpendicular to both the electric and magnetic fields, pointing radially inward. This shows the pathway through which electromagnetic energy moves into the conductor.

Energy Flow in Conductors

Energy flow in conductors can be visualized using the Poynting vector to understand how electromagnetic energy is distributed and transformed within a conductor. The energy flowing into the conductor can be determined by integrating the Poynting vector over its surface area.
The expression for energy flux \( \Phi \) through a cylindrical conductor of length \( l \) is:
\[ \Phi = \int S \cdot dA = \frac{\rho I^2 l}{\pi a^2}. \]
This result matches the known power dissipated due to resistance \( P = I^2 R \), where \( R = \frac{\rho l}{\pi a^2} \). Remarkably, this implies that the energy dissipated as heat inside the wire can be thought of as entering through its cylindrical sides. Thus, resistive heating, a conversion of incoming electromagnetic energy into thermal energy, highlights the intricate interplay between electromagnetic and thermal processes in conductive materials.