# Chapter 30 Mastering Physics 10

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The AmpÃ¨re-Maxwell Law
To show that displacement current is necessary to make AmpÃ¨re's law consistent for a charging capacitor AmpÃ¨re's law relates the line integral of the magnetic field around a closed loop to the total current passing through that loop. This law was extended by Maxwell to include a new type of "current" that is due to changing electric fields: âˆ®Bâƒ— â‹…dlâƒ— =Î¼0(Icharge+Idisplacement). The first term on the right-hand side, Icharge, describes the effects of the usual electric current due to moving charge. In this problem, that current is designated I(t) as usual. The second term, Idisplacement=Ïµ0dÎ¦Edt, is called the displacement current; it was recognized as necessary by Maxwell. His motivation was largely to make AmpÃ¨re's law symmetric with Faraday's law of induction when the electric fields and magnetic fields are reversed. By calling for the production of a magnetic field due to a change in electric field, this law lays the groundwork for electromagnetic waves in which a changing magnetic field generates an electric field whose change, in turn, sustains the magnetic field. We will discuss these issues later. (Incidentally, a third type of "current," called magnetizing current, should also be added to account for the presence of changing magnetic materials, but it will be neglected, as it has been in the equation above.) The purpose of this problem is to consider a classic illustration of the need for the additional displacement current term in AmpÃ¨re's law. Consider the problem of finding the magnetic field that loops around just outside the circular plate of a charging capacitor. The cone-shaped surface shown in the figure has a current I(t) passing through it, so AmpÃ¨re's law indicates a finite value for the field integral around this loop. However, a slightly different surface bordered by the same loop passes through the center of the capacitor, where there is no current due to moving charge. To get the same loop integral independent of the surface it must be true that either a current or an increasing electric field that passes through the AmpÃ¨rean surface will generate a looping magnetic field around its edge. The objective of this example is to introduce the displacement current, show how to calculate it, and then to show that the displacement current Idisplacement(t) is identical to the conduction current Icharge(t). Assume that the capacitor has plate area A and an electric field E(t) between the plates. Take Î¼0 to be the permeability of free space and Ïµ0 to be the permittivity of free space.
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First find âˆ®RBâƒ— â‹…dlâƒ— , the line integral of Bâƒ— around a loop of radius R located just outside the left capacitor plate. This can be found from the usual current due to moving charge in AmpÃ¨re's law, that is, without the displacement current.
âˆ®RBâƒ— â‹…dlâƒ— = Î¼0I(t)
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Now find an expression for âˆ®RBâƒ— â‹…dlâƒ— , the same line integral of Bâƒ— around the same loop of radius R located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the AmpÃ¨re-Maxwell law above. (Figure 2) Express your answer in terms of the electric field between the plates E(t), dE(t)/dt, the plate area A, and any needed constants given in the introduction.
âˆ®RBâƒ— â‹…dlâƒ— = Î¼0AÏµ0dE(t)/dt
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We now have two quite different expressions for the line integral of the magnetic field around the same loop. The point here is to see that they both are intimately related to the charge q(t) on the left capacitor plate. First find the displacement current Idisplacement(t) in terms of q(t). Express your answer in terms of q(t), dq(t)/dt, and any needed constants given in the introduction.
Idisplacement(t) = dq(t)/dt
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Now express the normal current Icharge(t) in terms of the charge on the capacitor plate q(t). Express your answer in terms of q(t), dq(t)/dt, and any needed constants given in the introduction.
Icharge=dq(t)/dt Using Gauss's law, you have shown that the displacement current from the changing electric field between the plates equals the current from the flow of charge through the wire onto that plate. This means that the AmpÃ¨re-Maxwell law can consistently treat cases in which the normal current due to the flow of charge is not continuous. This realization was a great boost to Maxwell's confidence in the physical validity of his new displacement-current term.
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The Magnetic Field in a Charging Capacitor
When a capacitor is charged, the electric field E, and hence the electric flux Î¦, between the plates changes. This change in flux induces a magnetic field, according to AmpÃ¨re's law as extended by Maxwell: âˆ®Bâƒ— â‹…dlâƒ— =Î¼0(I+Ïµ0dÎ¦dt). You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero.
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A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. Express your answer in terms of Î¼0 and given quantities.
B(r) = Î¼0rI/2Ï€a2
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Electromagnetic Waves from a Light Bulb
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 m from the bulb, the light intensity is S0, the average energy density of the waves is u0, and the rms electric and magnetic field values are E0 and B0, respectively.
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At 2.0 m from the bulb, what is the light intensity?
1/4S0
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At 2.0 m from the bulb, what is the average energy density of the waves?
1/4u0
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At 2.0 m from the bulb, what is the rms magnetic field value?
1/2B0
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Satellite Television Transmission
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
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Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use Ïµ0=8.85Ã—10âˆ’12C/(Vâ‹…m) for the permittivity of space and c=3.00Ã—108m/s for the speed of light. Express the amplitude of the electric field vector in microvolts per meter to three significant figures.
E0 = 7.00 Î¼V/m
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Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver? For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal. Give your answer in centimeters, to two significant figures.
R = 18 cm
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Graphical Analysis of AC Source EMF Conceptual Question What is the maximum emf Vmax of the source?
Vmax = 3 V https://session.masteringphysics.com/problemAsset/1943183/2/1012515.jpg
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What is the average emf Vavg of the source?
Vavg = 0 V
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What is the root-mean-square emf Vrms of the source? Express your answer to three significant figures.
Vmax/sqqrt(2(=Vrms = 2.12 V
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What is the period T of the source? Express your answer in seconds to two significant figures.
T = 0.08 s
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What is the frequency f of the source? Express your answer in hertz to three significant figures.
f = 12.5 Hz
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What is the angular frequency Ï‰ of the source? Express your answer in radians per second to three significant figures.
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Oscillations in an LC circuit.
To understand the processes in a series circuit containing only an inductor and a capacitor. Consider the circuit shown in the figure. (Figure 1) This circuit contains a capacitor of capacitance C and an inductor of inductance L. The resistance of all wires is considered negligible.
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Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=âˆ’kq, where k is a constant. Find the value of the constant k. Express your answer in terms of L and C.
k=1/LC
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From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=âˆ’kmx=âˆ’Ï‰2x, such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(Ï‰t+Ï•), where Ï•, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of Ï• based on the initial conditions described in the problem. Express your answer in terms of q0, L, and C. Use the cosine function in your answer.
q(t) = q0cos(t/root(LC))
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What is the current I(t) in the circuit at time t after the switch is closed? Express your answer in terms of q0, L, C, and other variables given in the introduction.
I(t) = q0/(rootLC) x sin(t/root(LC))
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Recall that the top plate of the capacitor is positively charged at t=0. Once the switch is closed, the current will start to increase. In what direction will this current go?
clockwise
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Immediately after the switch is closed, what is the direction of the EMF in the inductor? (Recall that the direction of the EMF in the inductor refers to the direction of the back-current or the induced electric field in the inductor.)
counterclockwise
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At what time does the current reach its maximum value for the first time? t=0
2.0 ms
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At what moment does the EMF in the inductor become zero for the first time?
2.0 ms
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At what moment does the current reverse direction for the first time?
4.0 ms
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At what moment does the EMF in the inductor reverse direction for the first time?
2.0 ms
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At what moment does the energy stored in the inductor reach its maximum for the first time?
2.0 ms
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At what time t do the capacitor and the inductor possess the same amount of energy for the first time? Express your answer in milliseconds.
1.0m/s
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What is the direction of the current in the circuit 22.0 milliseconds after the switch is closed?
counterclockwise
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What is the direction of the EMF in the inductor 42.0 milliseconds after the switch is closed?
The EMF is zero
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Inductive Reactance
To understand the concept of reactance (of an inductor) and its frequency dependence. When an inductor is connected to a AC source that varies sinusoidally, a sinusoidal current will flow through the inductor, its magnitude depending on the frequency. This is the essence of AC (alternating current) circuits used in radio, TV, and stereos. Circuit elements like inductors, capacitors, and resistors are linear devices, so the amplitude I0 of the current will be proportional to the amplitude V0 of the source emf. However, the current and source emf may not be in phase with each other. This new relationship between source emf and current is summarized by the reactance, the ratio of source emf amplitude and current amplitudes, V0, and I0: XL=V0/I0, where the subscript L indicates that this formula applies to an inductor.
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To find the reactance XL of an inductor, imagine that a current I(t)=I0sin(Ï‰t), is flowing through the inductor. What is the potential difference V(t) across this inductor? Express your answer in terms of I0, Ï‰, and the inductance L.
V(t) = I0cos(Ï‰t)Ï‰L
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What is the reactance XL of an inductor? Express your answer in terms of Ï‰ and the inductance L.
XL = Ï‰L
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In thinking of an inductor as a circuit element, it is helpful to consider its limiting behavior at high and low frequencies. At one extreme, the inductor might behave like a short circuit, that is, like a resistor with almost no resistance (an ideal wire) having essentially no potential difference across it no matter what the current. Alternatively, the inductor might behave like an open circuit, that is, like a resistor with large resistance so that essentially no current will flow no matter what the source emf amplitude. Based on the formula you obtained for the reactance, how does an inductor behave at high and low frequencies?
like an open circuit at high frequencies and a short circuit at low frequencies
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Â± The R-L Circuit: Responding to Changes
To understand the behavior of an inductor in a series R-L circuit. In a circuit containing only resistors, the basic (though not necessarily explicit) assumption is that the current reaches its steady-state value instantly. This is not the case for a circuit containing inductors. Due to a fundamental property of an inductor to mitigate any "externally imposed" change in current, the current in such a circuit changes gradually when a switch is closed or opened. Consider a series circuit containing a resistor of resistance R and an inductor of inductance L connected to a source of emf E and negligible internal resistance. The wires (including the ones that make up the inductor) are also assumed to have negligible resistance. (Figure 1) Let us start by analyzing the process that takes place after switch S1 is closed (switch S2 remains open). In our further analysis, lowercase letters will denote the instantaneous values of various quantities, whereas capital letters will denote the maximum values of the respective quantities. Note that at any time during the process, the loop rule holds and is, indeed, helpful: Eâˆ’vRâˆ’vL=0.
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Immediately after the switch is closed, what is the current in the circuit?
zero
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Immediately after the switch is closed, what is the potential difference across the resistor?
zero
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Immediately after the switch is closed, what is the potential difference across the inductor?
E MF
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Shortly after the switch is closed, what is the direction of the current in the circuit?
counterclockwise
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Shortly after the switch is closed, what is the direction of the induced EMF in the inductor?
clockwise
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Eventually, the process approaches a steady state. What is the current in the circuit in the steady state?
EMF/R
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What is the potential difference vL across the inductor in the steady state?
zero
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What is the potential difference vR across the resistor in the steady state?
EMF
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Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. Note that we can use the loop rule (going around counterclockwise): Eâˆ’vRâˆ’vL=0. Note as well that vR=iR and vL=Ldidt. Using these equations, we can get, after some rearranging of the variables and making the subsitution x=ERâˆ’i, dxx=âˆ’RLdt. Integrating both sides of this equation yields x=x0eâˆ’Rt/L. Use this last expression to obtain an expression for i(t). Remember that x=ERâˆ’i and that i0=i(0)=0. Express your answer in terms of E, R, and L. You may or may not need all these variables. Use the notation exp(x) for ex.
i(t) = E/R(1âˆ’e^âˆ’R/Lt)
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Just as in the case of R-C circuits, the steady state here is never actually reached: The exponential functions approach their limits asymptotically as tâ†’âˆž. However, it usually does not take very long for the value of i to get very close to its presumed limiting value. The next several questions illustrate this point. Note that the quantity L/R has dimensions of time and is called the time constant (you may recall similar terminology applied to R-C circuits). The time constant is often denoted by Ï„. Using Ï„, one can write the expression i(t)=ER(1âˆ’eâˆ’Rt/L) as i(t)=ER(1âˆ’eâˆ’t/Ï„). Find the ratio of the current i(t) at time t=6Ï„ to the maximum current Imax=ER. Express your answer numerically, using three significant figures.
i(6Ï„)Imax = 0.998
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Find the time t it takes the current to reach 99.999% of its maximum value. Express your answer numerically, in units of Ï„. Use three significant figures.
t = 11.5 Ï„
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Find the time t it takes the current to reach 99.999% of its maximum value. Assume that R=10 ohms and L=50 millihenrys. Express your answer in seconds, using three significant figures.
t = 5.76Ã—10âˆ’2 seconds
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Find the time t it takes the current to reach 99.999% of its maximum value. Assume that R=0.0100 ohms and L=5.00 henrys. Express your answer in seconds, using three significant figures.
t = 5760 seconds
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What fraction of the maximum value will be reached by the current one minute after the switch is closed? Again, assume that R=0.0100 ohms and L=5.00 henrys. Use three significant figures in your answer.
i(1minute)/Imax = 0.113
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What is the direction of the current in the circuit?
counterclockwise
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What is happening to the magnitude of the current?
The current is decreasing.
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What is the direction of the EMF in the inductor?
counterclockwise
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Which end of the inductor has higher potential difference (i.e., to which end of the inductor should the positive terminal of a voltmeter be connected in order to yield a positive reading)?
left
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For this circuit, the loop rule gives iR+Ldidt=0. Note that didt<0, since the current is decreasing. Use this equation to obtain an expression for i(t). Express your answer in terms of E, L, and R. Use exp(x) for ex.
i(t) = E/Re^âˆ’R/Lt
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A radio can be tuned into a particular station frequency by adjusting the capacitance in an L-C circuit. Suppose that the minimum capacitance of a variable capacitor in a radio is 4.16 pF .
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What is the inductance L of a coil connected to this capacitor if the oscillation frequency of the L-C circuit is 1.64 MHz , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? Express your answer in henrys to three significant figures.
L = 2.26Ã—10âˆ’3 H
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The frequency at the other end of the broadcast band is 0.534 MHz . What is the maximum capacitance Cmax of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band? Express your answer in farads to three significant figures.
Cmax=Cmax = 3.92Ã—10âˆ’11 F
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Alternating EMFs and Currents