Information

What is the maximum electric potential that can occur in the human body?


Neurons can generate a voltage, if there is an action potential. Also, membranes where an ion concentration gradient is present, generate some membrane potential. What I wish to know is, which cellular element in the human body creates the greatest potential differences? And how high can they be so high? Are there also biochemical reactions, where significantly high electric/electrostatic voltages are important? How strong are these?


To the best of my knowledge, the potential difference across the hair cells in the inner ear (cochlea) (Fig. 1) is the highest in the (human) body. It is about 120 mV, mainly due to the exceptionally high positive potential of the scala media as provided by the stria vascularis, referred to as the endocochlear potential. This potential is mainly built up by the high amount of K+ pumped into the scala media by ion pumps located in the marginal cells and basal cells (Salt et al, 1987) in the stria vascularis. Adding the potential difference of -40 mV of the hair cell it yields a total difference of 120 mV (Fig. 2).

References
- Morrill & He, J Otology (2017); 12: 151-64
- Salt et al., Laryngoscope (1987); 97(8 Pt 1): 984-91


Fig. 1. The cochlea. source: Morrill & He (2017)


Fig. 2. The potential difference across the hair cells is 120 mV, due to the endocochlear potential of +80 mV in the scala media. source: University of Minnesota Dulluth


The inner membrane of the mitochondria has a huge membrane potential generated by the pumping of protons as part of the respiratory chain.

In rat cortical neurons this value is around -140 mV and can variate between -108 and -158 depending on the metabolic activity of the cell.

Ref: J Physiol. 2012 Jun 15; 590(Pt 12): 2845-2871.


Volcanic Lightning

Stephen R. McNutt , Ronald J. Thomas , in The Encyclopedia of Volcanoes (Second Edition) , 2015

4.1 Background

Electrical phenomena are widespread and important, even though people may not be aware of them. For example, the electric field in normal undisturbed atmosphere is approximately 100 V/m (vertically). It can be substantially higher in and near a thunderstorm. Electrical breakdown occurs when the electric field exceeds the dielectric breakdown. A familiar small-scale example is walking on a carpet. An electric charge develops on the person, and bringing a finger near a metal object such as a doorknob causes a spark to jump across the air gap. The relevant parameters are the charge magnitude (moderate in this case), the distance (short), and the permittivity of air (nearly constant). Basic physics theory shows that the maximum electric field at the surface of a uniformly charged spherical volume of radius R is Rρ/3e0, where ρ is the charge density and e0 is the permittivity of free space. For ordinary thunderstorm lightning, the charges are on the order of tens of Coulombs and the distances a few kilometers. The permittivity again is constant, while the breakdown decreases slowly with increasing altitude. Hence, the lengths of lightning flashes in ordinary thunderstorms are many kilometers. Lightning flashes propagate at speeds of about 10 5 m/s (much slower than the speed of light, which is 10 8 m/s), so there is a systematic relation between the duration of a flash and its length. Vent discharges occur with very short durations and hence short-scale lengths. Because permittivity is constant, this means that the charge density in jets is much higher—an order of magnitude or more—than that of ordinary thunderstorms. Presumably, this occurs because the ash particles are charge carriers and they are found at high concentrations in the rising ash columns. In the main part of the plume, the charge density is much like a thunderstorm.


As rules of thumb, I've generally thought of myself as a 70 kΩ resistor to ground that feels pain at around 1 mA, which can be driven by 70 V or so. In my experience, the pain threshold is slightly above 48 V.

I can't say that I have any good medical science to back this up, but there are a few empirically obtained data points in that I'm not dead yet.

Here is an article titled "A review of hazards associated with exposure to low voltages" which I used as a reference when answering a medical safety question (I design embedded hardware and firmware for medical devices which go through FDA approval).

Because the body has a minimum resistance of around 550 ohms, to get enough current to do damage a minimum theoretical voltage level of around 16.5 Vpeak is required (corresponding to a current of 30 mApeak, which can induce respiratory paralysis if conducted across the chest for several minutes of contact at this low voltage). Based on the cases studied by the author, the single lowest voltage reported to cause transdermal electrocution in an adult is 25 volts.

For less than one minute of contact, currents >40 mA are required to cause ventricular fibrillation, corresponding to a theoretical voltage of 27.5 Vpeak. For less than one second exposure, >100 mApeak and 55 Vpeak are required. The author states that in all the cases he studied, there was no accidental electrocution from short-term exposures to voltages below 50 Vpeak.

Voltage doesn't really matter, it's a requirement to get a certain voltage to pass through the skin, but voltage doesn't have any impact on "damage".

Current is what does damage.

I've heard tons of claims as to what will kill you. In EE school is was 60mA AC and

100mA DC across your chest that would send your heart into fibrillation.

I've seen claims that < 10mA directly through your heart could do the same. Honestly both are probably correct. I don't know what a real electrical model of the body looks like, but I don't have a hard time believing that if there were 100mA running through my body from one hand to the other that only 10% would pass through my heart directly.

I've worked on live phone lines before (

58V DC with off hook) and that didn't pass through my skin initially. A half hour of being in the 105°F degree attic and sweaty hands later, it made my finger twitch and didn't feel good. On another occasion I was working on a phone line when someone dialed it. that sucked. the ring pulse is 120V AC (current limit though) and does not feel good at all.

It only takes a couple milliamps to seriously get your attention, 10+mA will lock up muscles, this is highly frequency-dependent though.

To get back to your point. greater than 100-200mA is when you'd expect to start to see flesh burning and things like that. But obviously from the heart discussion above, localized currents that are much smaller can be deadly.

I don't really know if there is a firm rule as to what's "safe". The current debate over the use of tasers, for example, would seem to indicate there isn't much conclusive evidence.


The Fatal Current

Strange as it may seem, most fatal electric shocks happen to people who should know better. Here are some electro-medical facts that should make you think twice before taking that last chance.

It's The Current That Kills

Offhand it would seem that a shock of 10,000 volts would be more deadly than 100 volts. But this is not so! Individuals have been electrocuted by appliances using ordinary house currents of 110 volts and by electrical apparatus in industry using as little as 42 volts direct current. The real measure of shock's intensity lies in the amount of current (amperes) forced though the body, and not the voltage. Any electrical device used on a house wiring circuit can, under certain conditions, transmit a fatal current.

While any amount of current over 10 milliamps (0.01 amp) is capable of producing painful to severe shock, currents between 100 and 200 mA (0.1 to 0.2 amp) are lethal. Currents above 200 milliamps (0.2 amp), while producing severe burns and unconsciousness, do not usually cause death if the victim is given immediate attention. Resuscitation, consisting of artificial respiration, will usually revive the victim.

From a practical viewpoint, after a person is knocked out by an electrical shock it is impossible to tell how much current has passed through the vital organs of his body. Artificial respiration must be applied immediately if breathing has stopped.

The Physiological Effects of Electric Shock

The chart shows the physiological effects of various currents. Note that voltage is not a consideration. Although it takes voltage to make current flow, the amount of shock-current will vary, depending on the body resistance between the points of contact.

As shown in the chart, shock is relatively more severe as the current rises. For currents above 10 milliamps, muscular contractions are so strong that the victim cannot let go of the wire that is shocking him. At values as low as 20 milliamps, breathing becomes labored, finally ceasing completely even at values below 75 milliamps.

As the current approaches 100 milliamps, ventricular fibrillation of the heart occurs - an uncoordinated twitching of the walls of the heart's ventricles which results in death.

Above 200 milliamps, the muscular contractions are so severe that the heart is forcibly clamped during the shock. This clamping protects the heart from going into ventricular fibrillation, and the victim's chances for survival are good.

Danger - Low Voltage

It is common knowledge that victims of high-voltage shock usually respond to artificial respiration more readily that the victims of low-voltage shock. The reason may be the merciful clamping of the heart, owing to the high current densities associated with high voltages. However, lest these details be misinterpreted, the only reasonable conclusion that can be drawn is that 75 volts are just as lethal as 750 volts.

The actual resistance of the body varies depending upon the points of contact and the skin condition (moist or dry). Between the ears, for example, the internal resistance (less the skin resistance) is only 100 ohms, while from hand to foot is closer to 500 ohms. The skin resistance may vary from 1000 ohms for wet skin to over 500,000 ohms for dry skin.

New Jersey State Council of Electrical Contractors Associations, Inc.
Bulletin VOL. 2, NO. 13
February, 1987
Submitted by Paul Giovinazzo


Introduction

Noninvasive electrical stimulation of the eyes has been studied as a promising therapeutic tool to recover visual functions in patients suffering from various eye diseases 1 . There are two well-known methods that deliver the electric current to the eye noninvasively. One is transcorneal electrical stimulation that delivers the currents via a contact-lens-type electrode attached right above the cornea 2 , 3 . Previous studies reported that transcorneal electrical stimulation has beneficial effects on the improvement of visual functions in patients with optic neuropathy 2 and retinitis pigmentosa (RP) 4 – 6 . According to the studies that used animal models with eye diseases, improvement of visual functions resulting from transcorneal electrical stimulation was closely associated with the survival of the retinal ganglion cells (RGCs) and photoreceptors preserved from the degeneration, suggesting that the neuroprotective effect on retinal cells determines the outcome of the transcorneal electrical stimulation 7 . Additionally, it was found that the increase in the survival of RGCs after transcorneal electrical stimulation is related to an increase in the insulin-like growth factor 1 (IGF-1), brain-derived neurotrophic factor (BDNF), and ciliary neurotrophic factor (CNF), which are released from Müller cells in the retina 8 , 9 .

The other method is transorbital electrical stimulation (tES) that delivers weak electrical current to the eye via electrodes attached to the skin around the eye. The stimulation parameters such as electrode configurations, current waveforms, injection current intensities differed among studies. Generally, square pulses in bursts with the frequency range of 5� Hz were applied for tES 10 . Compared to transcorneal electrical stimulation, tES is less invasive with no side effects such as dry-eye and punctate keratitis and easier to apply 11 . Repetitive tES, applied to patients with optic nerve damage, has been reported to improve visual field size, visual acuity, and detection ability 12 , 13 . Repetitive tES has also reportedly strengthened the alpha-band functional connectivity in patients with chronic prechiasmatic visual system damage 14 . Another study has demonstrated that the tES-treated group showed a significant improvement in visual fields and reaction times during the visual-field-related task compared to the sham stimulation group 15 . Furthermore, tES has also been effective in improving visual function in patients with RP 16 . A previous study reported that the effectiveness of tES was related to the synchronization of cortical activities after retinal cells were stimulated 10 . Another study insisted that improvement of visual functions together with changes in the spectral EEG alpha band power and connectivity in the occipital lobe after tES might be caused by a retinofugal entrainment through firing of RGCs 15 . Indeed, a previous in vivo experimental study with rats also demonstrated that electrically evoked, tES-induced responses stemmed from the retina 17 . These series of findings suggest that a stronger electric field should be delivered to the cells in the retina to increase the therapeutic effect of tES.

Generally, the conventional electrode configuration used for tES comprises two active electrodes attached to the skin near the orbital cavity and a single reference electrode placed on the occipital pole or extra-cephalic regions like the wrist and neck 10 , 12 , 18 . According to a numerical simulation study with the conventional electrode montage, most electric fields were delivered to the anterior part of the eye 15 . Therefore, the conventional tES dominantly stimulated the anterior part of the retina despite a large number of retinal cells, including RGCs and Müller cells, being densely distributed in the posterior part of the retina, particularly around the fovea 19 . Therefore, considering the abovementioned action mechanisms of both tES, a stronger electric field should be delivered to the posterior part of the retina to increase the effectiveness of tES. In the conventional tES, however, electric field delivered to the peripheral side of the retina (anterior retina) reaches an individual phosphene threshold 20 , which represents the maximally allowable injection current in tES that does not evoke phosphenes in an individual, before a sufficient amount of stimulation current is delivered to the posterior retina. Therefore, it is necessary to reduce the electric field delivered to the anterior retina in relative to that delivered to the posterior retina to maximize the overall therapeutic effects of tES.

This study proposes a novel tES montage with eight active electrodes, with the diameter of 1਌m, attached around the eye (approximately 2਌m away from the center of the cornea) and a reference electrode on the occipital pole to reduce the difference in the electric field intensities delivered to the anterior and posterior retina. In other words, the study aims to maximize the electric field delivered to the posterior retina when that delivered to the anterior retina reaches the individual phosphene threshold. As aforementioned, short duration square pulses at a specific frequency are generally employed for tES. Although the electrical conductivity values of tissues are dependent on the frequency of injected current 21 – 23 , we employed tissue electrical conductivity values at DC frequency and solved a quasi-static Laplace equation because the frequency range used for tES (5 – 30 Hz) was low enough for the quasi–static approximation. Indeed, it was reported that there was no difference between the electric fields calculated assuming DC and AC with a relatively high frequency (

𠂑 kHz) 24 . The optimal injection currents of the active electrodes were determined to maximize the electric field delivered to the posterior retina, near the fovea, by employing a constrained convex optimization approach. The efficacy of the new stimulation conditions was evaluated by comparing it with the conventional electrode montage.


Section Summary

  • where is the distance from A to B, or the distance between the plates.
  • In equation form, the general relationship between voltage and electric field is

  • where is the distance over which the change in potential, , takes place. The minus sign tells us that points in the direction of decreasing potential.) The electric field is said to be the gradient (as in grade or slope) of the electric potential.

Conceptual Questions

1: Discuss how potential difference and electric field strength are related. Give an example.

2: What is the strength of the electric field in a region where the electric potential is constant?

3: Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.

Problems & Exercises

1: Show that units of V/m and N/C for electric field strength are indeed equivalent.

2: What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of ?

3: The electric field strength between two parallel conducting plates separated by 4.00 cm is . (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

4: How far apart are two conducting plates that have an electric field strength of between them, if their potential difference is 15.0 kV?

5: (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air if the plates are separated by 2.00 mm and a potential difference of is applied? (b) How close together can the plates be with this applied voltage?

6: The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are discussed in Chapter 19.5 Capacitors and Dielectrics and Chapter 20.7 Nerve Conduction—Electrocardiograms.) You may assume a uniform electric field.

7: Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Chapter 20.7 Nerve Conduction—Electrocardiograms.) What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.

8: Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?

9: Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be .

10: A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

11: An electron is to be accelerated in a uniform electric field having a strength of . (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?


19.2 Electric Potential in a Uniform Electric Field

where d d size 12 <> is the distance from A to B, or the distance between the plates in Figure 19.5. Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb thus the following relation among units is valid:

Voltage between Points A and B

Example 19.4

What Is the Highest Voltage Possible between Two Plates?

Strategy

Solution

The potential difference or voltage between the plates is

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Discussion

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

Example 19.5

Field and Force inside an Electron Gun

(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a 0.500 μC 0.500 μC charge that gets between the plates?

Strategy

Solution for (a)

The expression for the magnitude of the electric field between two uniform metal plates is

Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for V AB V AB size 12 > > <> and the plate separation of 0.0400 m, we obtain

Solution for (b)

The magnitude of the force on a charge in an electric field is obtained from the equation


Problems & Exercises

What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?

Find the intensity of an electromagnetic wave having a peak magnetic field strength of 4 . 00 × 10 − 9 T 4 . 00 × 10 − 9 T size 12 <4 "." "00"´"10" rSup < size 8<-9>> " T"> <> .

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.

An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this distance?

A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of 7 . 50 μ V/m 7 . 50 μ V/m size 12 <7 "." "50" mV/m><> . (See [link].) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of 1 . 50 × 10 13 m 2 1 . 50 × 10 13 m 2 size 12 <1 "." "50"´"10" rSup < size 8<"13">> " m" rSup < size 8<2>> > <> (a large fraction of North America), how much power does it radiate?

Satellite dishes receive TV signals sent from orbit. Although the signals are quite weak, the receiver can detect them by being tuned to resonate at their frequency.

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1 . 00 × 10 11 V / m 1 . 00 × 10 11 V / m size 12 <1 "." "00"´"10" rSup < size 8<"11">> " V"/m> <> for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on a 1 . 00 -mm 2 1 . 00 -mm 2 size 12 <1 "." "00""-mm" rSup < size 8<2>> > <> area?

Suppose a source of electromagnetic waves radiates uniformly in all directions in empty space where there are no absorption or interference effects. (a) Show that the intensity is inversely proportional to r 2 r 2 size 12 > > <> , the distance from the source squared. (b) Show that the magnitudes of the electric and magnetic fields are inversely proportional to r r size 12 <> .

(b) I∝E 0 2 , B 0 2 ⇒ E 0 2 , B 0 2 ∝ 1 r 2 ⇒ E 0 , B 0 ∝ 1 r I∝E 0 2 , B 0 2 ⇒ E 0 2 , B 0 2 ∝ 1 r 2 ⇒ E 0 , B 0 ∝ 1 r


196 Energy in Electromagnetic Waves

Anyone who has used a microwave oven knows there is energy in electromagnetic waves . Sometimes this energy is obvious, such as in the warmth of the summer sun. Other times it is subtle, such as the unfelt energy of gamma rays, which can destroy living cells.

Electromagnetic waves can bring energy into a system by virtue of their electric and magnetic fields . These fields can exert forces and move charges in the system and, thus, do work on them. If the frequency of the electromagnetic wave is the same as the natural frequencies of the system (such as microwaves at the resonant frequency of water molecules), the transfer of energy is much more efficient.

The behavior of electromagnetic radiation clearly exhibits wave characteristics. But we shall find in later modules that at high frequencies, electromagnetic radiation also exhibits particle characteristics. These particle characteristics will be used to explain more of the properties of the electromagnetic spectrum and to introduce the formal study of modern physics.

Another startling discovery of modern physics is that particles, such as electrons and protons, exhibit wave characteristics. This simultaneous sharing of wave and particle properties for all submicroscopic entities is one of the great symmetries in nature.

Energy carried by a wave is proportional to its amplitude squared. With electromagnetic waves, larger -fields and -fields exert larger forces and can do more work.

But there is energy in an electromagnetic wave, whether it is absorbed or not. Once created, the fields carry energy away from a source. If absorbed, the field strengths are diminished and anything left travels on. Clearly, the larger the strength of the electric and magnetic fields, the more work they can do and the greater the energy the electromagnetic wave carries.

A wave’s energy is proportional to its amplitude squared ( or ). This is true for waves on guitar strings, for water waves, and for sound waves, where amplitude is proportional to pressure. In electromagnetic waves, the amplitude is the maximum field strength of the electric and magnetic fields. (See (Figure).)

Thus the energy carried and the intensity of an electromagnetic wave is proportional to and . In fact, for a continuous sinusoidal electromagnetic wave, the average intensity is given by

where is the speed of light, is the permittivity of free space, and is the maximum electric field strength intensity, as always, is power per unit area (here in ).

The average intensity of an electromagnetic wave can also be expressed in terms of the magnetic field strength by using the relationship , and the fact that , where is the permeability of free space. Algebraic manipulation produces the relationship

where is the maximum magnetic field strength.

One more expression for in terms of both electric and magnetic field strengths is useful. Substituting the fact that , the previous expression becomes

Whichever of the three preceding equations is most convenient can be used, since they are really just different versions of the same principle: Energy in a wave is related to amplitude squared. Furthermore, since these equations are based on the assumption that the electromagnetic waves are sinusoidal, peak intensity is twice the average that is, .

On its highest power setting, a certain microwave oven projects 1.00 kW of microwaves onto a 30.0 by 40.0 cm area. (a) What is the intensity in ? (b) Calculate the peak electric field strength in these waves. (c) What is the peak magnetic field strength ?

In part (a), we can find intensity from its definition as power per unit area. Once the intensity is known, we can use the equations below to find the field strengths asked for in parts (b) and (c).

Solution for (a)

Entering the given power into the definition of intensity, and noting the area is 0.300 by 0.400 m, yields

Here , so that

Note that the peak intensity is twice the average:

Solution for (b)

To find , we can rearrange the first equation given above for to give

Entering known values gives

Solution for (c)

Perhaps the easiest way to find magnetic field strength, now that the electric field strength is known, is to use the relationship given by

Entering known values gives

As before, a relatively strong electric field is accompanied by a relatively weak magnetic field in an electromagnetic wave, since , and is a large number.

Section Summary

  • The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as

where is the average intensity in , and is the maximum electric field strength of a continuous sinusoidal wave.

and in terms of both electric and magnetic fields as

Problems & Exercises

What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?

Find the intensity of an electromagnetic wave having a peak magnetic field strength of .

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.

(a)

(b)

(c)

An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this distance?

Suppose the maximum safe intensity of microwaves for human exposure is taken to be . (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)

A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of . (See (Figure).) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of (a large fraction of North America), how much power does it radiate?

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on a area?

(b)

Show that for a continuous sinusoidal electromagnetic wave, the peak intensity is twice the average intensity (), using either the fact that , or , where rms means average (actually root mean square, a type of average).

Suppose a source of electromagnetic waves radiates uniformly in all directions in empty space where there are no absorption or interference effects. (a) Show that the intensity is inversely proportional to , the distance from the source squared. (b) Show that the magnitudes of the electric and magnetic fields are inversely proportional to .

(a)

(b)

An circuit with a 5.00-pF capacitor oscillates in such a manner as to radiate at a wavelength of 3.30 m. (a) What is the resonant frequency? (b) What inductance is in series with the capacitor?

What capacitance is needed in series with an inductor to form a circuit that radiates a wavelength of 196 m?

Police radar determines the speed of motor vehicles using the same Doppler-shift technique employed for ultrasound in medical diagnostics. Beats are produced by mixing the double Doppler-shifted echo with the original frequency. If microwaves are used and a beat frequency of 150 Hz is produced, what is the speed of the vehicle? (Assume the same Doppler-shift formulas are valid with the speed of sound replaced by the speed of light.)

Assume the mostly infrared radiation from a heat lamp acts like a continuous wave with wavelength . (a) If the lamp’s 200-W output is focused on a person’s shoulder, over a circular area 25.0 cm in diameter, what is the intensity in ? (b) What is the peak electric field strength? (c) Find the peak magnetic field strength. (d) How long will it take to increase the temperature of the 4.00-kg shoulder by , assuming no other heat transfer and given that its specific heat is ?

(a)

(c)

On its highest power setting, a microwave oven increases the temperature of 0.400 kg of spaghetti by in 120 s. (a) What was the rate of power absorption by the spaghetti, given that its specific heat is ? (b) Find the average intensity of the microwaves, given that they are absorbed over a circular area 20.0 cm in diameter. (c) What is the peak electric field strength of the microwave? (d) What is its peak magnetic field strength?

Electromagnetic radiation from a 5.00-mW laser is concentrated on a area. (a) What is the intensity in ? (b) Suppose a 2.00-nC static charge is in the beam. What is the maximum electric force it experiences? (c) If the static charge moves at 400 m/s, what maximum magnetic force can it feel?

(a)

(b)

(c)

A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field of the incoming electromagnetic wave is perpendicular to the coil and has a maximum strength of . (a) What power is incident on the coil? (b) What average emf is induced in the coil over one-fourth of a cycle? (c) If the radio receiver has an inductance of , what capacitance must it have to resonate at 100 MHz?

If electric and magnetic field strengths vary sinusoidally in time, being zero at , then and . Let here. (a) When are the field strengths first zero? (b) When do they reach their most negative value? (c) How much time is needed for them to complete one cycle?

(a)

(b)

(c)

A researcher measures the wavelength of a 1.20-GHz electromagnetic wave to be 0.500 m. (a) Calculate the speed at which this wave propagates. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

The peak magnetic field strength in a residential microwave oven is . (a) What is the intensity of the microwave? (b) What is unreasonable about this result? (c) What is wrong about the premise?

(a)

(b) Much too great for an oven.

(c) The assumed magnetic field is unreasonably large.

An circuit containing a 2.00-H inductor oscillates at such a frequency that it radiates at a 1.00-m wavelength. (a) What is the capacitance of the circuit? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

An circuit containing a 1.00-pF capacitor oscillates at such a frequency that it radiates at a 300-nm wavelength. (a) What is the inductance of the circuit? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

(a)

(c) The wavelength is unreasonably small.

Create Your Own Problem

Consider electromagnetic fields produced by high voltage power lines. Construct a problem in which you calculate the intensity of this electromagnetic radiation in based on the measured magnetic field strength of the radiation in a home near the power lines. Assume these magnetic field strengths are known to average less than a . The intensity is small enough that it is difficult to imagine mechanisms for biological damage due to it. Discuss how much energy may be radiating from a section of power line several hundred meters long and compare this to the power likely to be carried by the lines. An idea of how much power this is can be obtained by calculating the approximate current responsible for fields at distances of tens of meters.

Create Your Own Problem

Consider the most recent generation of residential satellite dishes that are a little less than half a meter in diameter. Construct a problem in which you calculate the power received by the dish and the maximum electric field strength of the microwave signals for a single channel received by the dish. Among the things to be considered are the power broadcast by the satellite and the area over which the power is spread, as well as the area of the receiving dish.


Glossary

Solutions

Problems & Exercises

3: (a) 3.00 kV

5: (a) No. The electric field strength between the plates is , which is lower than the breakdown strength for air .


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