Consider current I passing through a resistor of radius r, length L, and resistance R. Show that the flux of the Poynting vector (i.e., the integral of S → ⋅ d A → \overrightarrow{S}\cdot d\overrightarrow{A} ) over the surface of the resistor is I 2 R. Then give an interpretation of this result.

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A cylindrical wire of radius A length B and resistance R carries a current I if the power per unit area carried by an electric and magnetic field at the surface of the wire is I squared multiplied by R divided by two multiplied by pi multiplied by a multiplied by B what should be the flux of the pointing vector over the surface of the wire? So that's our end goal. So trying to figure out what the flux of the pointing vector over the surface of the wire will be for this case. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. A is negative. I multiplied by RB is I divided by R C is I multiplied by R squared and D is negative I squared multiplied by R. OK. So first off, let us recall that a pointing vector is the power per unit area carried by electric and magnetic fields. Therefore, a pointing vector s at the surface of a cylindrical wire is equal to I squared multiplied by R all divided by two, multiplied by pi multiplied by A multiplied by B. And I should take a minute here to emphasize that A, in this case is lower case, A because in a second, we're gonna introduce the surface area which is capital A. So let's talk about that. So now we also know that the surface area capital A of a cylindrical wire can be written as so capital A, the surface area of the cylindrical wire can be is equal to two multiplied by pi multiplied by lower case A multiplied by lowercase B. Thus the total power flowing out is the integral of the pointing vector over the surface of the resistor, which can be written as the integral of SD A capital A is equal to the integral faces of SD capital A plus the integral of the integral wall of SD ad capital A. OK. So the pointing vector S at the surface of the cylindrical wire is directed into the resistor for the faces, the direction S and D A are perpendicular to each other. So the integral faces of SD A will equal zero. So let's write that down. So the integral faces of S D A will equal zero for the wall, the direction of S and D are opposite to each other. So the integral which is an elongated S, the integral wall of SD A is equal to negative S multiplied by capital A which can simplify to the integral of SD A equals zero minus S A or S multiplied by A. So we can therefore go on to write that the integral of SD A is equal to negative I squared multiplied by R all divided by two multiplied by pi multiplied by lower case A multiplied by lower case B multiplied by two pi or two multiplied by pi multiplied by lowercase A multiplied by lowercase B which is equal to negative I squared multiplied by R. And that is our final answer. Hooray, we did it. So the correct answer has to be the letter D which is negative I squared multiplied by R. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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32. Electromagnetic Waves

Intensity of EM Waves

Physics

32. Electromagnetic Waves

Intensity of EM Waves

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Problem 66c

Textbook Question

Consider current I passing through a resistor of radius r, length L, and resistance R. Show that the flux of the Poynting vector (i.e., the integral of $\vec{S} \cdot d \vec{A}$ ) over the surface of the resistor is I²R. Then give an interpretation of this result.

Verified step by step guidance

Step 1: Recall the Poynting vector, $ \overrightarrow{S} $, which represents the power per unit area flowing through a surface due to electromagnetic fields. It is given by $ \overrightarrow{S} = \frac{1}{\mu_0} \overrightarrow{E} \times \overrightarrow{B} $, where $ \overrightarrow{E} $ is the electric field and $ \overrightarrow{B} $ is the magnetic field.

Step 2: For a resistor, the electric field $ \overrightarrow{E} $ inside the resistor is related to the current density $ \overrightarrow{J} $ by Ohm's law: $ \overrightarrow{E} = \rho \overrightarrow{J} $, where $ \rho $ is the resistivity of the material. The current density $ \overrightarrow{J} $ is given by $ \overrightarrow{J} = \frac{I}{A} $, where $ A = \pi r^2 $ is the cross-sectional area of the resistor.

Step 3: The magnetic field $ \overrightarrow{B} $ around the resistor can be determined using Ampère's law: $ \oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I $. For a cylindrical resistor, the magnetic field at a distance $ r $ from the axis is $ B = \frac{\mu_0 I}{2 \pi r} $.

Step 4: The Poynting vector $ \overrightarrow{S} $ at the surface of the resistor is then calculated as $ \overrightarrow{S} = \frac{1}{\mu_0} \overrightarrow{E} \times \overrightarrow{B} $. Substituting $ \overrightarrow{E} $ and $ \overrightarrow{B} $, and noting that the cross product is perpendicular to the surface, the magnitude of $ \overrightarrow{S} $ is $ S = \frac{E B}{\mu_0} $.

Step 5: To find the total power flow, integrate $ \overrightarrow{S} \cdot d\overrightarrow{A} $ over the cylindrical surface of the resistor. This yields $ P = \int \overrightarrow{S} \cdot d\overrightarrow{A} = I^2 R $, which matches the power dissipated in the resistor due to Joule heating. This result shows that the electromagnetic energy flow into the resistor is entirely converted into thermal energy, consistent with energy conservation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Poynting Vector

The Poynting vector, denoted as extbf{S} = rac{1}{ extmu_0} extbf{E} imes extbf{B}, represents the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field. It combines the electric field extbf{E} and the magnetic field extbf{B} to describe how electromagnetic energy flows through space. In the context of a resistor, the Poynting vector helps quantify the energy being dissipated as heat due to the current flowing through the resistor.

Resistance and Ohm's Law

Resistance (R) is a measure of the opposition to the flow of electric current in a conductor. According to Ohm's Law, the relationship between voltage (V), current (I), and resistance is given by V = I * R. This relationship is crucial for understanding how energy is dissipated in a resistor, as the power (P) dissipated can be expressed as P = I^2 * R, which directly relates to the Poynting vector's flux over the surface of the resistor.

Surface Integral

A surface integral is a mathematical operation that calculates the integral of a function over a surface in three-dimensional space. In this context, the surface integral of the Poynting vector over the surface of the resistor quantifies the total electromagnetic energy flow through that surface. This integral is essential for deriving the relationship between the Poynting vector and the power dissipated in the resistor, leading to the conclusion that the flux equals I^2R.

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