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Chapter 31: Problem 63

What is the distance between successive heating antinodes in a microwave oven's cavity? A microwave oven typically operates at a frequency of \(2.4 \mathrm{GHz}\).

Short Answer

Expert verified
Answer: The distance between successive heating antinodes in a microwave oven's cavity is approximately 6.25 cm.

Step by step solution

01

Determine the Wavelength

To find the wavelength (位), we can use the equation: $$ \lambda = \frac{c}{f} $$ where c is the speed of light (\(3.0 \times 10^{8} \mathrm{m/s}\)) and f is the given frequency (\(2.4 \times 10^9 \mathrm{Hz}\) or \(2.4 \mathrm{GHz}\)).
02

Calculate the Wavelength

Now, we can plug the values of c and f into the equation from Step 1 to find the wavelength: $$ \lambda = \frac{3.0 \times 10^8 \mathrm{m/s}}{2.4 \times 10^9 \mathrm{Hz}} = \frac{3.0}{2.4} \times 10^{-1} \mathrm{m} = 1.25 \times 10^{-1} \mathrm{m} $$ So the wavelength of the microwaves is \(0.125 \mathrm{m}\) or \(12.5 \mathrm{cm}\).
03

Find the Distance Between Successive Antinodes

Since there are two antinodes within one complete wavelength and the distance between two successive antinodes is half of the wavelength, we can find the distance between the antinodes as follows: $$ D = \frac{\lambda}{2} = \frac{1.25 \times 10^{-1} \mathrm{m}}{2} = 0.625 \times 10^{-1} \mathrm{m} = 6.25 \mathrm{cm} $$ Therefore, the distance between successive heating antinodes in a microwave oven's cavity is approximately \(6.25 \mathrm{cm}\).

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Key Concepts

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

Microwave Frequency
Microwaves are a type of electromagnetic radiation with frequencies ranging from about 300 MHz to 300 GHz. The frequency is the number of times the wave oscillates in one second. Therefore, a microwave with a frequency of 2.4 GHz oscillates 2.4 billion times per second. This high frequency is perfect for heating food quickly and evenly in microwave ovens because it excites water molecules, causing them to heat up. A microwave oven, therefore, typically operates at a frequency of around 2.4 GHz, making it very efficient for domestic cooking tasks.
Wavelength Calculation
Wavelength refers to the distance between two corresponding points on consecutive waves, like from crest to crest. To determine the wavelength of a wave, you can use the formula:
  • \( \lambda = \frac{c}{f} \)
where \( \lambda \) is the wavelength, \( c \) is the speed of light, and \( f \) is the frequency. By substituting the speed of light \( (3.0 \times 10^{8} \mathrm{m/s}) \) and the microwave frequency \( (2.4 \times 10^9 \mathrm{Hz}) \), we can find the wavelength of a microwave to be approximately 0.125 meters or 12.5 centimeters. The wavelength is crucial to understand how the microwave energy is distributed and absorbed by the substance being heated. In this context, shorter wavelengths can also result in more detailed wave patterns.
Speed of Light
The speed of light \((c)\) is a constant that plays a significant role in calculating wave properties. It is approximately \(3.0 \times 10^{8} \mathrm{m/s} \) and represents how fast light travels in a vacuum. This speed allows for the conversion between frequency and wavelength, and it effectively links the two key parameters of wave physics. Since the speed of light is finite, it determines how far and how quickly light and other electromagnetic waves like microwaves travel. This constant is essential when designing devices such as microwave ovens that depend on precise calculations to ensure efficient energy transfer.

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Most popular questions from this chapter

31.8 According to Gauss's Law for Magnetic Fields, all magnetic field lines form a complete loop. Therefore, the direction of the magnetic field \(\vec{B}\) points from _________ pole to ________ pole outside of an ordinary bar magnet and from ____ pole to pole _______ inside the magnet. a) north, south, north, south b) north, south, south, north c) south, north, south, north d) south, north, north, south

At the surface of the Earth, the Sun delivers an estimated \(1.00 \mathrm{~kW} / \mathrm{m}^{2}\) of energy. Suppose sunlight hits a \(10.0 \mathrm{~m}\) by \(30.0 \mathrm{~m}\) roof at an angle of \(90.0^{\circ}\) a) Estimate the total power incident on the roof. b) Find the radiation pressure on the roof.

What is the radiation pressure due to Betelgeuse (which has a luminosity, or power output, 10,000 times that of the Sun) at a distance equal to that of Uranus's orbit from it?

Electric and magnetic fields in many materials can be analyzed using the same relationships as for fields in vacuum, only substituting relative values of the permittivity and the permeability, \(\epsilon=\kappa \epsilon_{0}\) and \(\mu=\kappa_{\mathrm{m}} \mu_{0},\) for their vacuum values, where \(\kappa\) is the dielectric constant and \(\kappa_{\mathrm{m}}\) the relative permeability of the material. Calculate the ratio of the speed of electromagnetic waves in vacuum to their speed in such a material.

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 \(\int \vec{S} \cdot d \vec{A}=i^{2} R\)
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