Problem 42 - University Physics with Modern Physics

Question

A voltage, \(V\), is applied across a cylindrical conductor of radius \(r\), length \(L\), and resistance \(R\). As a result, a current, \(i\), is flowing through the conductor, which gives rise to a magnetic field, \(B\). The conductor is placed along the \(y\) -axis, and the current is flowing in the positive \(y\) -direction. Assume that the electric field is uniform throughout the conductor. a) Find the magnitude and the direction of the Poynting vector at the surface of the conductor. b) Show that \(\int \vec{S} \cdot d \vec{A}=i^{2} R\)

Step-by-Step Solution

Answer

#Short Answer# The Poynting vector at the surface of the conductor has a magnitude of \(S = \frac{i^2}{2\pi r^3 \sigma}\) and is directed in the positive \(y\)-direction. By evaluating the surface integral of the Poynting vector over the surface of the cylindrical conductor, we find the total power \(\int \vec{S} \cdot d \vec{A}=i^{2} R\), which confirms the expected result.

Step 1: Determine the electric field

Assuming the electric field is uniform throughout the conductor, we can use Ohm's Law to find the electric field \(\vec{E}\). Ohm's Law is given by \(\vec{J} = \sigma \vec{E}\), where \(\sigma\) is the conductivity and \(\vec{J}\) is the current density. The current density is related to the total current by the equation \(i = \int \vec{J} \cdot d \vec{A}\). Since the problem assumes a uniform current density, let \(J\) be the magnitude of the current density. Then, \(i = JA\), where \(A = \pi r^2\) is the cross-sectional area of the conductor. So, \(J = \frac{i}{\pi r^2}\). From Ohm's Law, \(\vec{E} = \frac{\vec{J}}{\sigma}\) and its magnitude \(E = \frac{J}{\sigma}\).

Step 2: Determine the magnetic field

According to Ampere's Circuital Law, the magnetic field is given by \(\oint \vec{B} \cdot d \vec{l} = \mu_0 i_{enc}\). Since the problem statement assumes a uniform magnetic field, let \(B\) be its magnitude. When we encircle the conductor (circle with radius \(r\)), we have \(2\pi rB = \mu_0 i\), from which we can find the magnetic field's magnitude: \(B = \frac{\mu_0 i}{2\pi r}\).

Step 3: Determine the Poynting vector's magnitude and direction

Now that we have the magnitudes of the electric and magnetic fields, we can compute the Poynting vector's magnitude. From the expression \(\vec{S}= \frac{1}{\mu_{0}}\vec{E} \times \vec{B}\), we can find \(S = \frac{1}{\mu_{0}} EB\). Substitute the expressions for \(E\) and \(B\) from the previous steps to obtain \(S = \frac{1}{\mu_0} \cdot \frac{i}{\pi r^2 \sigma} \cdot \frac{\mu_0 i}{2\pi r}\). After simplifying, we have \(S = \frac{i^2}{2\pi r^3 \sigma}\). The direction of the Poynting vector can be found using the right-hand rule for cross products. When the fingers of the right hand wrap from the positive electric field direction to the positive magnetic field direction, the thumb points in the direction of the Poynting vector. In this case, the electric field is in the \(x\)-direction, and the magnetic field is in the \(z\)-direction, so the Poynting vector is in the positive \(y\)-direction. #b) Surface Integral of the Poynting Vector#

Step 1: Define the surface integral

We need to evaluate the integral \(\int \vec{S} \cdot d \vec{A}\) over the cylindrical surface of the conductor. We will do this by integrating the \(y\)-component of the Poynting vector along the conductor's surface.

Step 2: Calculate the surface integral

Since the Poynting vector is directed in the positive \(y\)-direction, the integral simplifies to \(\int S \cdot dA\). Substitute the expression for \(S\) we found previously: \(\int \frac{i^2}{2\pi r^3 \sigma} \cdot dA = \frac{i^2}{2\pi r^3 \sigma} \int dA\). To compute the surface integral, we need to integrate over the side surface area of the conductor: \(A_s = 2\pi rL\). Therefore, the integral becomes \(\frac{i^2}{2\pi r^3 \sigma} (2\pi rL)\).

Step 3: Prove the result

Simplify the integral: \(\frac{i^2}{2\pi r^3 \sigma} (2\pi rL) = i^2\frac{L}{r^2 \sigma}\). Now, recall that the resistance \(R = \frac{L}{\sigma A}\), where \(A = \pi r^2\). Plug in this expression for \(R\) and simplify further: \(i^2\frac{L}{r^2 \sigma} = i^2\frac{R}{A}=i^2 R\). Thus, we have shown that \(\int \vec{S} \cdot d \vec{A}=i^{2} R\).

Key Concepts

Electric Fields in Conductors

When an electric field is present within a conductor, free electrons move, creating an electric current. This current, in turn, implies the presence of an electric field, which in a conductor under steady-state conditions (like our cylindrical conductor with a constant current) is often uniform. In the given exercise, we utilize Ohm's Law to relate the constant electric field to the current density within the conductor. This concept is crucial because it helps us understand how an electric field can lead to a distribution of current, ensuring that charges move in a manner that conserves energy within the system.

Magnetic Field

A magnetic field arises due to the motion of electric charges, such as the flow of current in our cylindrical conductor. It is a vector field, which means it has a direction as well as a magnitude. Using Ampere's Law, we can calculate the magnetic field around the conductor, as is done in the exercise, to find a relationship between the current and the magnetic field. Here, the field forms concentric circles around the path of the current, with the direction given by the right-hand thumb rule.

Ampere's Law

Ampere's Law is a fundamental principle of magnetism which relates the magnetic field around a closed loop to the current passing through a surface bounded by the loop. It is represented by the equation \(\oint \vec{B} \cdot d \vec{l} = \mu_0 i_{enc}\) as seen in the step-by-step solution. This principle is crucial for calculating the magnetic field produced by a known current within the cylindrical conductor.

Ohm's Law

Ohm's Law states the relationship between electric current, resistance, and voltage. In mathematical terms, it is given by \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. In the context of the electric field, Ohm's Law translates to \(\vec{J} = \sigma \vec{E}\), with \(\vec{J}\) being the current density and \(\sigma\) the conductivity. This law is fundamental when working with electric circuits and conductors, as it allows us to deduce an electric field from known current and conductivity values.

Surface Integral

A surface integral is a mathematical way to add up all the values of a vector field over a surface. It's like covering the surface with a net and then summing up the contributions from each of its little patches. This concept is applied in the problem when we calculate the total power crossing the surface of the conductor. In particular, by integrating the Poynting vector over the entire surface of the conductor, we establish a link between the electromagnetic energy flow and the power dissipated in the conductor.

Current Density

Current density is a measure of the flow of electric charge per unit area, pointed in the direction charge flows, typically denoted by \(\vec{J}\). It indicates how dense the current is in a given area and is essential when trying to understand the distribution of current flow within a conductor. A higher current density means more charge is passing through a specific area and thus could potentially generate a stronger electric or magnetic field. In the exercise, current density is crucial to linking Ohm's Law to the Poynting vector.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces a third vector perpendicular to both of the original vectors, with a magnitude that is the product of their magnitudes and the sine of the angle between them. It is symbolically represented as \(\vec{A} \times \vec{B}\). In electromagnetic theory, the Poynting vector is found using the cross product between the electric field vector and the magnetic field vector, which gives us a new vector that represents the direction and magnitude of the electromagnetic power flow. In the context of our exercise, this concept helps us visualize and calculate how energy propagates around the conductor.