Skip to main content
Ch 31: Electromagnetic Fields and Waves
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 31: Electromagnetic Fields and WavesProblema 21
Capitolo 31, Problema 21

A microwave beam with a wavelength of 1.5 cm has an intensity of 25 W/m2. What is the magnetic field amplitude?

Guida verificata passo dopo passo
1
Step 1: Recall the relationship between the intensity of an electromagnetic wave and the amplitudes of the electric and magnetic fields. The intensity (I) is given by the formula: I = \(\frac{1}{2}\) \(\epsilon\)_0 c E_0^2, where \(\epsilon\)_0 is the permittivity of free space, c is the speed of light, and E_0 is the electric field amplitude.
Step 2: Use the relationship between the electric field amplitude (E_0) and the magnetic field amplitude (B_0) in an electromagnetic wave: E_0 = c B_0. Rearrange this equation to express B_0 in terms of E_0: B_0 = \(\frac{E_0}{c}\).
Step 3: Solve for the electric field amplitude (E_0) using the intensity formula. Rearrange the formula to isolate E_0: E_0 = \(\sqrt{\frac{2I}{\epsilon_0 c}\)}. Substitute the given intensity (I = 25 \ \(\text{W/m}\)^2), the permittivity of free space (\(\epsilon\)_0 = 8.85 \(\times\) 10^{-12} \ \(\text{F/m}\)), and the speed of light (c = 3 \(\times\) 10^8 \ \(\text{m/s}\)) into this equation.
Step 4: Once E_0 is calculated, substitute it into the equation B_0 = \(\frac{E_0}{c}\) to find the magnetic field amplitude. Use the value of c = 3 \(\times\) 10^8 \ \(\text{m/s}\) for the speed of light.
Step 5: Perform the calculations step by step to determine the numerical value of B_0. Ensure that all units are consistent throughout the calculations.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Electromagnetic Waves

Electromagnetic waves are oscillations of electric and magnetic fields that propagate through space. They consist of a transverse wave where the electric field (E) and magnetic field (B) oscillate perpendicular to each other and the direction of wave propagation. Understanding the nature of these waves is crucial for analyzing phenomena like microwaves, which are a type of electromagnetic radiation.
Video consigliato:
Percorso guidato
05:22
Introduction to Electromagnetic (EM) Waves

Intensity of Electromagnetic Waves

The intensity of an electromagnetic wave is defined as the power per unit area carried by the wave, typically measured in watts per square meter (W/m²). It is related to the electric and magnetic field amplitudes through the equation I = (1/2) * ε₀ * c * E², where ε₀ is the permittivity of free space and c is the speed of light. This relationship is essential for calculating the magnetic field amplitude from the given intensity.
Video consigliato:
Percorso guidato
02:32
Wave Intensity

Magnetic Field Amplitude

The magnetic field amplitude (B₀) of an electromagnetic wave can be derived from the electric field amplitude (E₀) using the relationship B₀ = E₀/c, where c is the speed of light. This relationship highlights how the electric and magnetic fields are interconnected in electromagnetic waves. Knowing the intensity allows us to find E₀, and subsequently B₀, which is necessary for solving the problem presented.
Video consigliato:
Percorso guidato
05:30
Magnetic Fields and Magnetic Dipoles
Pratica correlata
Domanda del libro di testo

Unpolarized light with intensity 350 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 131 W/m2. What is the angle from vertical of the axis of the second polarizing filter?

186
views
Domanda del libro di testo

What is the force (magnitude and direction) on the proton in FIGURE P31.28? Give the direction as an angle cw or ccw from vertical.

292
views
Domanda del libro di testo

The electric field of an electromagnetic wave in a vacuum is Ey=(20.0 V/m)cos[(6.28×108)x−ωt]E_y = (20.0 \(\text{ V/m}\)) \(\cos\)[(6.28 \(\times\) 10^8)x - \(\omega\) t], where x is in m and t is in s. What are the wave's magnetic field amplitude?

206
views
Domanda del libro di testo

FIGURE EX31.23 shows a horizontally polarized radio wave of frequency 1.0×106 Hz traveling into the figure. The maximum electric field strength is 1000 V/m. What is the maximum magnetic field strength?

88
views
Domanda del libro di testo

A radio receiver can detect signals with electric field amplitudes as small as 300 μV/m. What is the intensity of the smallest detectable signal?

132
views
Domanda del libro di testo

The magnetic field of an electromagnetic wave in a vacuum is Bz=(3.0 μT)sin[(1.00×107)x−ωt]B_{z}=(3.0\(\text{ }\)\(\mu\) T)\(\sin\)[(1.00\(\times\)10^7)x-\(\omega\) t], where x is in m and t is in s. What are the wave's wavelength?

216
views