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Chapter 20: Problem 44

Microwave ovens can have cold spots and hot spots due to standing electromagnetic waves, analogous to standing wave nodes and antinodes in strings (v Fig. 20.32). (a) The longer the distance between the cold spots, (1) the higher the frequency of the waves, (2) the lower the frequency of the waves, (3) the frequency of the waves is independent of this distance. Why? (b) In your microwave the cold spots (nodes) occur approximately every \(5.0 \mathrm{~cm},\) but your neighbor's microwave produces them at every \(6.0 \mathrm{~cm}\). Which microwave operates at a higher frequency and by how much?

Short Answer

Expert verified
(a) The lower the frequency, as wavelength and frequency are inversely related. (b) Your microwave operates at a higher frequency by 500 MHz.

Step by step solution

01

Understanding Standing Waves

In a microwave oven, the electromagnetic waves create a standing wave pattern with nodes (cold spots) and antinodes (hot spots). These waves behave similarly to standing waves on a string, where the distance between nodes is half the wavelength of the wave. The frequency of a wave is related to its wavelength by the equation: \( c = \lambda f \), where \( c \) is the speed of light.
02

Relating Node Distance with Frequency

The distance between cold spots (nodes) is half the wavelength \( \lambda / 2 \). Thus, if the distance between nodes increases, the wavelength \( \lambda \) increases. Since frequency \( f = c/\lambda \), an increase in wavelength results in a decrease in frequency. Thus, the statement (2) is correct: the longer the distance between the cold spots, the lower the frequency of the waves.
03

Calculate Wavelength of Your Microwave

For your microwave, the distance between cold spots is \(5.0 \, \text{cm}\). This is half the wavelength of the microwave: \(\frac{\lambda_1}{2} = 0.05 \, \text{m}\), so \(\lambda_1 = 0.1 \, \text{m}\).
04

Calculate Frequency of Your Microwave

Using the relation \( f = \frac{c}{\lambda} \), where \( c \approx 3 \times 10^8 \, \text{m/s} \), the frequency of your microwave is \( f_1 = \frac{3 \times 10^8}{0.1} = 3 \times 10^9 \, \text{Hz} \).
05

Calculate Wavelength of Neighbor's Microwave

For the neighbor's microwave, the nodes occur every \(6.0 \, \text{cm}\). So, \(\frac{\lambda_2}{2} = 0.06 \, \text{m}\), thus \(\lambda_2 = 0.12 \, \text{m}\).
06

Calculate Frequency of Neighbor's Microwave

Using the same wave equation, \( f_2 = \frac{c}{\lambda_2} = \frac{3 \times 10^8}{0.12} \approx 2.5 \times 10^9 \, \text{Hz} \).
07

Determining Which Microwave Has Higher Frequency

Comparing frequencies, your microwave has a frequency of \(3 \times 10^9 \, \text{Hz} \) and the neighbor's is \(2.5 \times 10^9 \, \text{Hz} \). Thus, your microwave operates at a higher frequency. The difference in frequency is \(3 \times 10^9 - 2.5 \times 10^9 = 0.5 \times 10^9 = 500 \, \text{MHz} \).

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are a type of wave that can travel through both empty space and matter. These waves carry energy and are made up of electric and magnetic fields oscillating perpendicular to one another. The speed of electromagnetic waves in a vacuum is the speed of light, denoted as \(c\), and is approximately \(3 \times 10^8\) meters per second.
  • They encompass a broad spectrum, including radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.
  • The wavelength and frequency vary across the electromagnetic spectrum, but all electromagnetic waves travel at the same speed in a vacuum.
In microwave ovens, electromagnetic waves in the microwave range are used to heat food. These microwaves generate standing wave patterns inside the oven, helping to cook food efficiently.
Microwave Ovens
Microwave ovens are a common household appliance that uses microwaves, a type of electromagnetic wave, to heat and cook food. These appliances operate by emitting waves at frequencies typically around 2.45 GHz.
  • Microwaves penetrate food and agitate water molecules, causing them to vibrate and generate heat.
  • Inside the oven, electromagnetic waves create standing patterns of high and low energy, leading to areas that cook faster (antinodes) and slower (nodes).
The standing pattern inside a microwave is why food might heat unevenly, leading to hot and cold spots.
Frequency and Wavelength
Frequency and wavelength are fundamental properties of waves. Frequency, denoted as \(f\), is the number of wave cycles that pass a point per second, measured in hertz (Hz). Wavelength, denoted as \(\lambda\), is the distance between successive crests or troughs of a wave.

These two properties are inversely related through the wave speed equation:
  • \(c = \lambda f\), where \(c\) is the speed of light.
  • A longer wavelength means a lower frequency when the speed of the wave remains constant.
  • In the context of microwaves, altering the wavelength directly affects the frequency, thus changing how energy is distributed in the oven.
Nodes and Antinodes
Nodes and antinodes in standing waves are key concepts for understanding the heating patterns inside a microwave oven.
  • Nodes: These are points where the wave exhibits minimum displacement 鈥� in microwaves, they correspond to cold spots where little heating occurs.
  • Antinodes: Points of maximum displacement in the wave鈥攖hese are hot spots where the most substantial heating happens.
  • In the microwave, the spacing of nodes influences wavelength, affecting the frequency according to the equation \(f = c/\lambda\).
  • The relationship between nodes and antinodes helps explain why some areas of the food might not heat as evenly as others.
Understanding nodes and antinodes is crucial for optimizing cooking performance in microwaves.

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Most popular questions from this chapter

A student makes a simple ac generator by using a single square wire loop \(10 \mathrm{~cm}\) on a side. The loop is then rotated at a frequency of \(60 \mathrm{~Hz}\) in a magnetic field of \(0.015 \mathrm{~T}\). (a) What is the maximum emf output? (b) If she wanted to make the maximum emf output ten times larger by adding loops, how many should she use in total?

A boy is traveling due north at a constant speed while carrying a metal rod. The rod's length is oriented in the east-west direction and is parallel to the ground. (a) There will be no induced emf when the rod is (1) at the equator, (2) near the Earth's magnetic poles, (3) somewhere between the equator and the poles. Why? (b) Assume that the Earth's magnetic field is \(1.0 \times 10^{-4} \mathrm{~T}\) near the North Pole and \(1.0 \times 10^{-5} \mathrm{~T}\) near the equator. If the boy runs with a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) northward near each location, and the rod is \(1.5 \mathrm{~m}\) long, calculate the induced emf in the rod in each location.

The starter motor in an automobile has a resistance of \(0.40 \Omega\) in its armature windings. The motor operates on \(12 \mathrm{~V}\) and has a back emf of \(10 \mathrm{~V}\) when running at normal operating speed. How much current does the motor draw (a) when running at its operating speed, (b) when running at half its final rotational speed, and (c) when starting up?

The transformer on a utility pole steps the voltage down from \(10000 \mathrm{~V}\) to \(220 \mathrm{~V}\) for use in a college science building. During the day, the transformer delivers electric energy at the rate of \(10.0 \mathrm{~kW}\). (a) Assuming the transformer to be ideal, during that time, what are the primary and secondary currents in the transformer? (b) If the transformer is only \(90 \%\) efficient (but still delivers electric power at \(10.0 \mathrm{~kW}\) ), how does its inputcurrent compare to the ideal case? (c) At what rate is heat lost in the nonideal transformer? (d) If you wanted to keep the transformer \(\mathrm{cool}\) and to do this needed to dissipate half of the joule heating of part (c) using water cooling lines (the other half is taken care of by air cooling), what should be the rate of flow (in liters per minute) of water in the lines? Assume the input cool water is at \(68^{\circ} \mathrm{F}\) and the maximum allowable output water temperature is \(98^{\circ} \mathrm{F}\).

A circular loop (radius of \(20 \mathrm{~cm}\) ) is in a uniform magnetic field of \(0.15 \mathrm{~T}\). What angle(s) between the normal to the plane of the loop and the field would result in a flux with a magnitude of \(1.4 \times 10^{-2} \mathrm{~T} \cdot \mathrm{m}^{2} ?\)
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