# 4.3 The electromagnetic spectrum  (Page 16/33)

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A certain 50.0-Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 13.0 kV/m. (a) What is the wavelength of this very low frequency electromagnetic wave? (b) What is its maximum magnetic field strength?

(a) $6.00×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m}$

(b) $4\text{.}\text{33}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$

During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? (c) What is the wavelength of the electromagnetic wave?

(a) The ideal size (most efficient) for a broadcast antenna with one end on the ground is one-fourth the wavelength ( $\lambda /4$ ) of the electromagnetic radiation being sent out. If a new radio station has such an antenna that is 50.0 m high, what frequency does it broadcast most efficiently? Is this in the AM or FM band? (b) Discuss the analogy of the fundamental resonant mode of an air column closed at one end to the resonance of currents on an antenna that is one-fourth their wavelength.

(a) 1.50 × 10 6 Hz, AM band
(b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation.

(a) What is the wavelength of 100-MHz radio waves used in an MRI unit? (b) If the frequencies are swept over a $±1\text{.}\text{00}\text{}$ range centered on 100 MHz, what is the range of wavelengths broadcast?

(a) What is the frequency of the 193-nm ultraviolet radiation used in laser eye surgery? (b) Assuming the accuracy with which this EM radiation can ablate the cornea is directly proportional to wavelength, how much more accurate can this UV be than the shortest visible wavelength of light?

(a) $1\text{.}\text{55}×{\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\text{Hz}$

(b) The shortest wavelength of visible light is 380 nm, so that

$\begin{array}{}\frac{{\lambda }_{\text{visible}}}{{\lambda }_{\text{UV}}}\\ =\frac{\text{380 nm}}{\text{193 nm}}\\ =1\text{.}\text{97}\text{.}\end{array}$

In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate!

TV-reception antennas for VHF are constructed with cross wires supported at their centers, as shown in [link] . The ideal length for the cross wires is one-half the wavelength to be received, with the more expensive antennas having one for each channel. Suppose you measure the lengths of the wires for particular channels and find them to be 1.94 and 0.753 m long, respectively. What are the frequencies for these channels?

Conversations with astronauts on lunar walks had an echo that was used to estimate the distance to the Moon. The sound spoken by the person on Earth was transformed into a radio signal sent to the Moon, and transformed back into sound on a speaker inside the astronaut’s space suit. This sound was picked up by the microphone in the space suit (intended for the astronaut’s voice) and sent back to Earth as a radio echo of sorts. If the round-trip time was 2.60 s, what was the approximate distance to the Moon, neglecting any delays in the electronics?

$3\text{.}\text{90}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}$

Lunar astronauts placed a reflector on the Moon’s surface, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. (a) To what accuracy in meters can the distance to the Moon be determined, if this time can be measured to 0.100 ns? (b) What percent accuracy is this, given the average distance to the Moon is $3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}$ ?

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away is the planet Venus if the echo time is 1000 s? (b) What is the echo time for a car 75.0 m from a Highway Police radar unit? (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 10.0 m?

(a) $1\text{.}\text{50}×{\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{m}$

(b) $0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}\mu s$

(c) 66.7 ns

Integrated Concepts

(a) Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm. (b) Compare this with the ratio of highest to lowest frequencies the ear can hear.

Integrated Concepts

(a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely in the infrared) from $1\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ of the Earth’s surface at night. Assume the emissivity is 0.90, the temperature of the Earth is $\text{15º}\text{C}$ , and that of outer space is 2.7 K. (b) Compare the intensity of this radiation with that coming to the Earth from the Sun during the day, which averages about $\text{800}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , only half of which is absorbed. (c) What is the maximum magnetic field strength in the outgoing radiation, assuming it is a continuous wave?

(a) $-3\text{.}5×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$

(b) 88%

(c) $1\text{.}7\phantom{\rule{0.25em}{0ex}}\mu T$

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
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salma
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Sherica
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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I'm interested in Nanotube
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