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The microwaves in a certain microwave oven have a wavelength of 12.2 cm. (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0 cm longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Short Answer

Expert verified
(a) 30.5 cm. (b) 2.46 GHz. (c) 2.11 GHz.

Step by step solution

01

Understand the Standing Wave Condition

In a standing wave pattern, antinodes are points of maximum amplitude. For a wave to have 5 antinodal planes across the breadth of the oven, we know that an antinodal plane occurs at every half-wavelength. Therefore, for 5 antinodal planes, the oven's width must contain 2.5 wavelengths. Let \( \lambda \) be the wavelength. The total width of the oven \( W \) is given by \( \frac{5}{2} \lambda \).
02

Calculate the Oven's Width

Given that the wavelength \( \lambda \) is 12.2 cm: \[W = \frac{5}{2} \times 12.2 \text{ cm} = 30.5 \text{ cm}\]Thus, the oven must be 30.5 cm wide.
03

Determine the Frequency of the Microwaves

The frequency \( f \) of the microwaves can be determined using the speed of light \( c \) and the relationship \( c = \lambda f \). For microwaves, \( c = 3.00 \times 10^8 \text{ m/s} \). First, convert the wavelength to meters:\[\lambda = 12.2 \text{ cm} = 0.122 \text{ m}\]Then solve for \( f \):\[f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{0.122 \text{ m}} \approx 2.46 \times 10^9 \text{ Hz}\] The frequency is approximately \( 2.46 \times 10^9 \text{ Hz} \).
04

Consider the Manufacturing Error

If the oven is made 5.0 cm longer, the new width \( W' \) is: \[W' = 30.5 \text{ cm} + 5.0 \text{ cm} = 35.5 \text{ cm}\]
05

Recalculate the Required Microwave Frequency

To still have 5 antinodal planes across the new width, equate \( W' \) to \( \frac{5}{2} \lambda' \) and solve for \( \lambda' \):\[\frac{5}{2} \lambda' = 35.5 \text{ cm} \Rightarrow \lambda' = \frac{35.5}{2.5} = 14.2 \text{ cm} = 0.142 \text{ m}\]Now, calculate the new frequency \( f' \):\[f' = \frac{c}{\lambda'} = \frac{3.00 \times 10^8 \text{ m/s}}{0.142 \text{ m}} \approx 2.11 \times 10^9 \text{ Hz}\]The new frequency is approximately \( 2.11 \times 10^9 \text{ Hz} \).

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Key Concepts

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

Standing Wave
In wave physics, standing waves are a fascinating concept, often seen when a wave reflects back and interferes with itself, creating a pattern. Think of a standing wave as a wave that oscillates in place rather than traveling from one point to another. This happens due to reflections that reinforce the wave's amplitude at certain points, called antinodes, and cancel out at nodes (points of no displacement).
This phenomenon is crucial for understanding how microwaves in an oven can pattern themselves to create maximum points of energy concentration (antinodal planes). In the case of microwaves, these standing waves ensure that food heats evenly when the antinodal planes distribute the energy efficiently through the oven.
Microwave Frequency
Frequency is vital in wave physics as it tells us how many oscillations happen over a time unit. For microwaves, frequency determines the energy these waves transmit. The frequency is inversely proportional to the wavelength, meaning as the wavelength decreases, the frequency increases, leading to greater energy transfer.
  • The frequency of electromagnetic waves like microwaves is linked with their ability to heat substances since higher frequency waves can impart more energy onto particles.
  • Using the known speed of light and wavelength, frequency can be found through the equation: \[ f = \frac{c}{\lambda} \] where \( c \) is the speed of light.
In microwave ovens, knowing this frequency helps to optimize the heating effect by ensuring energy focuses on the antinodal planes.
Wavelength Calculation
Wavelength is another core concept of wave physics, indicating the distance over which the wave's shape repeats. For standing wave patterns in microwave ovens, calculating the wavelength is key to ensuring waves fit within the oven's dimensions.
In practical scenarios, such as designing a microwave oven to heat efficiently, we must consider how waves of a certain wavelength interact with the appliance's size. For instance, if you want five antinodal planes, you calculate the necessary wavelength to fill the oven correctly:
  • Knowing the number of desired antinodal planes and that antinodes occur every half wavelength, we can determine how to fit them within a specific dimension by setting up an equation involving wavelength.
  • By rearranging and solving for the wavelength using given or measured dimensions, we ensure optimal wave distribution within the space.
This is how engineers tailor microwaves to maximize energy transfer where needed.
Antinodal Planes
Antinodal planes in a standing wave pattern are crucial because they represent points of maximum amplitude, where the wave achieves its greatest displacement, creating areas of high energy.
In the context of a microwave oven, these planes are vital for efficient heating.
  • An antinodal plane occurs half a wavelength apart in a standing wave.
  • To ensure adequate heating, microwaves are designed to create a specific number of these planes, arranging wave interference in such a way that these high-energy points are distributed evenly across the food placed inside.
  • The ability to calculate and control where these planes form is key to a microwave's cooking efficiency, ensuring that food does not end up with hot and cold spots.
Understanding how to manipulate and calculate these antinodal planes allows appliances to harness wave physics effectively, optimizing the cooking process.

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Most popular questions from this chapter

The company where you work has obtained and stored five lasers in a supply room. You have been asked to determine the intensity of the electromagnetic radiation produced by each laser. The lasers are marked with specifications, but unfortunately different information is given for each laser: Laser A: power = 2.6 W; diameter of cylindrical beam = 2.6 mm Laser B: amplitude of electric field = 480 V/m Laser C: amplitude of magnetic field = 8.7 \(\times\) 10\(^{-6}\) T Laser D: diameter of cylindrical beam = 1.8 mm; force on totally reflecting surface = 6.0 \(\times\) 10\(^{-8}\) N Laser E: average energy density in beam = 3.0 \(\times\) 10-7 J/m\(^3\) Calculate the intensity for each laser, and rank the lasers in order of increasing intensity. Assume that the laser beams have uniform intensity distributions over their cross sections.

Consider each of the following electric- and magneticfield orientations. In each case, what is the direction of propagation of the wave? (a) \(\vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{\jmath}\); (b) \(\vec{E} = E\hat{\jmath}\), \(\vec{B} = B\hat{\imath}\); (c) \(\vec{E} = -E\hat{k}\) , \(\vec{B} = -B\hat{\imath}\); (d) \(vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{k}\).

The sun emits energy in the form of electromagnetic waves at a rate of 3.9 \(\times\) 10\(^{26}\) W. This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r = R = 6.96 \times 10^5\) km) and at \(r = R/\)2, in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about 1.0 \(\times\) 10\(^4\) Pa; at \(r = R/\)2, the gas pressure is calculated from solar models to be about 4.7 \(\times\) 10$^{13} Pa. Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate \(E_{max}\) and \(B_{max}\) for the 700-nm light at a distance of 5.00 m from the source.

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\). Verify this by substituting E_y(x, t) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (\(Hint\): The field does work to move charges within the conductor. The current of these moving charges causes \(i^2R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of 1/\(e\) in a distance \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); permeability \(\mu = \mu_0\)). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

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