Electromagnetic Induction and AC Circuits

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1 HAPTER22 C. Return to Table of Contents Electromagnetic Induction and AC Circuits This electric generator, shown being in - spected, will be used to provide electric power. Inside the generator a changing magnetic field induces an electric current in huge coils of copper wire. Fig. 22 Drawing from Faraday s diary of the first generator. Modern society relies on the electric generator for the abundant electrical power it produces. Fig. 22 shows a sketch of the first generator, invented by Michael Faraday. Today s large and elaborate generators operate on the same physical principle as Faraday s generator. Although realizing he had discovered a device with great practical potential, Faraday was not interested in developing it himself. When asked by a British government official of what use his invention was, he replied, I know not, but I wager that one day your government will tax it. In this chapter we shall study the electric generator and the alternating current it pro - duces. In preceding chapters we have described electric fields produced by stationary charge distributions (electrostatics) and magnetic fields produced by constant currents (magnetostatics). Electricity and magnetism were studied under conditions in which it was possible to have one field without the other either an electric field and no magnetic field or a magnetic field and no electric field. In this chapter and the next we shall con - sider the more general situation of charge distributions and currents that vary with time. As these sources vary, they produce time-varying electric and magnetic fields. We shall find that there are relationships between electric and magnetic fields that make it impossible to have a time-varying electric field without an associated magnetic field and impossible to have a time-varying magnetic field without an associated electric field. We shall often refer to these associated electric and magnetic fields as a single entity the electromagnetic field. 588

2 22 Faraday s Law 22 Faraday s Law Induction Experiments In 83 Michael Faraday observed an effect that even today, over 60 years later, is rather amazing. Faraday produced electric current in a coil of wire that was not connected to a battery and had no other obvious source of current. The current was induced in the coil when there was a changing electric current in a nearby circuit (Fig. 22 2). The induced current lasted only very briefly, as the switch was being closed. When there was a steady current in the circuit containing the battery, Faraday detected no current in the coil. Faraday also found that he could induce an electric current in a coil of wire simply by moving a permanent magnet through the coil (Fig. 22 3). This effect is the basis for electric generators. Faraday s discoveries followed years of experimentation, based on his intuitive belief that it should be possible to induce electric currents with other electric currents. (See the essay at the end of this chapter for a discussion of Faraday s life and work.) A particularly simple induction experiment is illustrated in Fig A copper ring is in a uniform time-varying magnetic field, directed perpendicular to the plane of the ring. When the magnetic field changes, current is induced in the ring. This current can be detected in various ways, for example, by the heat it produces in the ring. The presence of an electric current in a conductor means that there must be a source of energy; that is, there must be a source of emf. A battery is a common source of emf. However, in the case of the ring in Fig. 22 4, there is no battery. The source of emf must be related to the changing magnetic field. The time-varying magnetic field induces an emf in the ring as though there were a battery inside the ring. We can get a picture of what is happening within the ring on the microscopic level by noting that the presence of a current in the ring means that there must be an electric field within the ring (Eq. 9 26: E I/A). The changing magnetic field produces this electric field. From symmetry, the electric field is the same throughout the ring. The induced emf is the work per unit charge done by the induced electric field. The electric field is not limited to the metal ring. Electric field lines circulate throughout the region of space where the magnetic field is varying. 589 Fig As switch S is closed, there is a momentary deflection of the gal van - o meter needle, indicating a temporary current induced in the second circuit. An iron ring enhances the effect, which would otherwise be too small to be seen. The circuits are electrically insulated from the ring. Opening the switch also produces a brief current. Fig As a permanent magnet is moved toward a coil of wire, current is induced in the coil, as shown by the deflection of a galvanometer needle. When the magnet stops moving, the current also stops. Fig A time-varying magnetic field induces a current in a copper ring.

3 590 CHAPTER 22 Electromagnetic Induction and AC Circuits Magnetic Flux To discuss induced emf quantitatively, we must first introduce the concept of magnetic flux. Fig shows a magnetic field that is uniform over a surface of area A. For a uniform field such as this, the magnetic flux is defined as the product of the area A and the magnetic field component B, perpendicular to the surface.* B A BA cos (for a uniform field) (22 ) This definition implies that the magnetic flux through a surface is proportional to the number of magnetic field lines passing through that surface. For example, when a surface is oriented so that field lines are parallel to the surface (B 0), the flux equals zero and the number of lines passing through the surface also equals zero (Fig. 22 6a). Both magnetic flux and the number of field lines through a surface are maximized if the surface is oriented perpendicular to the field lines (Fig. 22 6b). Fig Magnetic flux B A. Faraday s Law; Constant E Faraday s law states that the emf induced around any loop equals the negative rate of change of outward magnetic flux through the surface bounded by the loop. If the flux changes at a constant rate / t, Faraday s law is expressed E (Faraday s law; constant E) (22 2) t (a) Positive emf means that the direction of the induced electric field in the loop will produce counterclockwise current, whereas negative emf means an induced field that produces clockwise current (Fig. 22 7). Rather than use Faraday s law to determine the direction of an induced emf, it is usually easier to use Faraday s law to determine only the magnitude of the induced emf. Taking the magnitude of both sides of Eq. 22 2, we obtain t E (constant E) (22 3) (b) Fig Magnetic flux and field lines. (a) Field lines parallel to a surface. Both the flux and the number of field lines passing through the surface equal zero. (b) Field lines perpendicular to a surface. Both the flux and the number of field lines through the surface are maximized. All six field lines pass through. Fig The same coil shown in Fig is viewed from above. As the magnet moves toward the coil, the magnet s B field produces an increasing outward magnetic flux through the coil ( / t 0). Faraday s law predicts a negative (clockwise) induced emf (E / t 0), resulting in clockwise current through the coil. *If the field is not uniform over a surface, we find the flux by dividing the surface into small elements of area A, computing the flux B A for each element, and summing to compute the total flux [ (B A)]. For a constant source of emf, like a battery, the emf was defined as a positive quantity the energy per unit charge supplied by a source, or, equivalently, the work per unit charge done by a source. Here we need to define a sign for emf in order to deal with situations in which the source is an induced electric field, which may be changing in direction. It is as though we were using a battery with the orientation of the positive and negative terminals being continually reversed.

4 22 Faraday s Law 59 Lenz s Law The direction of the induced current is easiest to find from Lenz s law, which may be derived from Faraday s law and which states that an induced current produces a magnetic field that opposes the change in magnetic flux that produced it. For example, the induced current in the coil shown in Fig produces its own magnetic field. If the right fingers curve around the coil in the direction of the current, the extended right thumb points in the direction of the magnetic field produced by the current, according to the rule formulated in Chapter 2 (see Fig. 2 28). Applying this rule to the coil in Fig we see that the magnetic field of the induced current shown in the figure is inward and therefore opposes the increase in the outward magnetic field of the approaching magnet, as predicted by Lenz s law. The magnetic field produced by the induced current is much weaker than the magnetic field of the permanent magnet. So there is still an increasing magnetic flux. It is simply reduced slightly by the field of the induced current. To find the direction and magnitude of an induced current, we use both Lenz s law and Faraday s law, as seen in the following example. EXAMPLE Current Induced in a Copper Ring The ring shown in Fig has a radius of 4.00 cm and a resistance of The magnetic field is increasing at a constant rate from T to T in a time interval of s. Find the current in the ring. SOLUTION We apply Lenz s law to find the direction of the current. The current must be counterclockwise so that the magnetic field of this induced current is directed outward, as shown in Fig. 22 8, opposing the increase in the inward flux. To find the magnitude of the induced current, we must first calculate the change in magnetic flux and apply Faraday s law to determine the induced emf. The external magnetic field is uniform over the area of the ring and perpendicular to this surface area. We apply Eq. 22 : B A BA Since the ring has a fixed area (A r 2 ), the change in mag - netic flux,, may be expressed in terms of the change in magnetic field, B, as Fig An increasing external magnetic field (directed inward) induces a counterclockwise current, which produces an outward magnetic field, opposing the change that produced it. B(A) B( r 2 ) (0.400 T T)( )( m) T-m 2 Applying Faraday s law (Eq. 22 3), we find.0 0 E 3 T-m V t s The magnitude of the current resulting from this emf may be calculated in the usual way for a single loop circuit consisting of a constant counterclockwise emf E 0.0 V and a resistance R (Fig. 22 9). Fig V 0 E IR 0 E 0.0 V I 0 A R

5 592 CHAPTER 22 Electromagnetic Induction and AC Circuits Fig shows a rather dramatic demonstration of induction. A long iron bar inserted in a coil with a time-varying electric current forms an electromagnet, with a time-varying magnetic field in the iron. An emf is induced in a conducting ring placed over the iron bar. The ring s induced current produces a magnetic field opposing the changing magnetic field in the iron (Fig. 22 0a). The induced current in the ring is opposite the current in the coil. Therefore the ring is repelled by the coil and flies upward* (Fig. 22 0b). (a) Motional emf Faraday s law and Lenz s law may also be applied to a moving circuit, for example, to a flexible loop of wire in a uniform, constant magnetic field (Fig. 22 ). As the sides of the loop are pulled outward, the area enclosed by the loop decreases. According to Lenz s law, there must be a clockwise current induced in the loop so that the induced current produces a magnetic field directed inward, in the same direction as the external field, and so the induced field opposes the decrease in flux through the loop. The emf induced by the motion of a conductor through a magnetic field is called motional emf. Another example of motional emf is illustrated in Fig A metal rod slides to the right at constant velocity v while in contact with a stationary, U-shaped conductor. A complete conducting loop is formed by the rod and the part of the U-shaped con - ductor to the left of the rod. A uniform, constant magnetic field B is directed downward, perpendicular to the plane of the loop (Fig. 22 2a). As the rod slides, the area enclosed by the loop increases, and so the flux through the loop increases. Thus there will be an induced emf and an induced current in the loop. From Lenz s law we see that the induced current is counterclockwise, producing inside the loop a magnetic field directed opposite the external field (Fig. 22 2b). Since B is uniform and perpendicular to the plane of the loop, the magnitude of the flux through the loop at any instant is the product of the field magnitude B and the area A enclosed by the loop at that instant. BA B x (b) Fig Magnetic propulsion. (a) (b) Fig (a) A metal rod slides over a stationary U-shaped conductor in a uniform, constant magnetic field. (b) Because the conducting path encloses an increasing area, a current is induced along the path. The magnetic field of the induced current is outward, opposing the increasing inward magnetic flux. Fig. 22 Reducing the area enclosed by a flexible loop of wire in a magnetic field induces a current in the loop. *This explanation accounts only for the magnetic force on the ring during that part of the AC cycle during which there is increasing current in the coil in the direction indicated. The repulsive force on the ring during other parts of the cycle is more difficult to explain.

6 22 Faraday s Law 593 The rod moves a distance x v t during the time interval t. The magnitude of the change in flux during this time interval is B x B v t Dividing by t and applying Faraday s law (Eq. 22 3: E / t ), we find E B v (22 4) EXAMPLE 2 Current Induced by Pulling a Metal Rod Through a Magnetic Field Suppose that the left side of the U-shaped conductor in Fig has resistance of 0.0 and that there is negligible resistance in each of the other three sides of the loop, including the rod, which has a length of 0 cm and moves at a constant speed of 4.0 m/s through a magnetic field of magnitude 0.50 T. Calculate the current induced in the loop. Fig SOLUTION Applying Eq. 22 4, we find for the magnitude of the emf E B v (0.50 T)(0.0 m)(4.0 m/s) 0.20 V The loop is equivalent to the circuit shown in Fig. 22 3, with counterclockwise induced current I, which we find by applying Kirchhoff s second rule. V 0 I(0.0 ) 0.20 V 0 I 0.20 V A Faraday s Law; Variable E In general, the rate of change of magnetic flux through a surface is not constant. Then Faraday s law must be expressed in its more general form, in which the emf equals the negative of the instantaneous flux rate: E limit t 0 t (Faraday s law; variable E) (22 5)

7 594 CHAPTER 22 Electromagnetic Induction and AC Circuits Generators The electric generator is an important practical application of motional emf. A simple alternating-current generator is shown in Fig A coil of wire is rotated in an external magnetic field. The magnetic flux through the coil varies as the coil rotates. When the plane of the coil is parallel to the magnetic field, the flux is zero. The flux through the coil is greatest when the plane of the coil is perpendicular to the field, so that the plane s normal n is aligned with the field. In the figure, the generator shaft is shown rotating counterclockwise at a point in the motion where n is aligning with the magnetic field. The magnetic flux through the surface is increasing; therefore, according to Faraday s law, there must be an emf induced in each loop of the coil. The induced current is directed as indicated in the figure (so that it produces a magnetic field that opposes the increase in flux), in accord with Lenz s law. As the shaft continues to turn, the normal n will point above the horizontal, and the flux through the surface will begin to decrease. The induced current will then be in the opposite direction. Thus the generator produces current that alternates in direction alternating current (AC). We can obtain an expression for the alternating induced emf by applying Faraday s law. If the shaft is rotated at constant angular velocity (that is, a fixed number of radians per second), the angle between B and n may be expressed t The flux through the surface of area A bounded by the coil is given by Eq. 22 : BA cos BA cos t In this case flux changes at a rate that is not constant. Thus, in applying Faraday s law to find the induced emf, we must use Eq. 22 5: E limit t 0 t Fig An alternating-current generator connected to a resistor R.

8 22 Faraday s Law 595 The methods of calculus must be used to calculate the limit. We present the result without proof:* E BA sin t This is the emf induced in each turn of the coil. For a coil with N turns, the total emf is N times the emf in a single turn. E NBA sin t (22 6) The maximum induced emf occurs when sin t, that is, when t 90, so that n is perpendicular to B. It is at this point that the flux is changing most rapidly. Denot - ing the maximum value of induced emf by E 0, we may express the induced emf as where E E 0 sin t (22 7) E 0 NBA (22 8) The generator s angular velocity is related to the induced emf s period T and frequency f by Eqs. 5 6 and 5 7: 9t 2 f 2 T Commercial generators are designed somewhat differently from the generator we have described here. These generators utilize an electromagnet that rotates inside a stationary coil, in which the current is induced. The principle is the same: changing magnetic flux through the coil induces the emf. The symbol ~ is used to represent an AC generator. Using this symbol, the generator and resistor shown in Fig are represented by the circuit diagram in Fig Fig shows a graph of the generator s emf as a function of time. Also shown in this figure, for selected values of t, are DC circuits with the same emf as the instantaneous emf in the AC circuit. To find the voltage drop V ab across the generator s terminals, we follow the rule developed for batteries in Chapter 9, which gives Fig The circuit representation of the generator and resistor in Fig V ab E E 0 sin t (22 9) Fig The emf produced by an AC generator. *We obtain this result by taking the time derivative of cos t, which equals sin t. One can also derive this result based on the formula for the power developed by an electric motor from Chapter 2. (See Problem 4.)

9 596 CHAPTER 22 Electromagnetic Induction and AC Circuits Fig Counterclockwise rotation of the generator coil induces a current and a magnetic moment m. The coil ex - per i ences a clockwise magnetic torque, which tends to align m with the external field. A counterclockwise torque must be applied to the shaft to cancel the mag - netic torque and maintain the counterclockwise rotation. Chemical energy is the source of the electrical potential energy provided by a battery. What then is the source of the electrical potential energy provided by a generator? The key to answering this question is to realize that the generator coil will not turn spontaneously. An applied external torque is required to balance the torque exerted on the coil by the magnetic field, as seen in Fig Thus the generator s energy source can be anything that provides a torque to turn the generator s shaft. For example, in a hydroelectric plant (Fig. 22 8a) the kinetic energy of water is used to turn turbine blades connected to generator coils. In other power plants coal is often used as the energy source (Fig. 22 8b). The coal is burned and used to produce steam, which turns turbine blades and generator coils. Nuclear fuel is used as the heat source in place of coal in nuclear power plants (Fig. 22 8c). In the desert outside Palm Springs, California, windmills utilize the kinetic energy of the wind to power electric generators (Fig. 22 8d). Human energy can even be used to generate electrical energy, for example, to power a small bicycle light (Problem 6). Fig Various ways of generating electricity. (a) A hydroelectric power plant, outside and inside. (b) A coal-powered plant. (c) A nuclear powered plant. (d) Wind-powered generators. EXAMPLE 3 Electric Power Supplied by a Generator An electric generator operating at 60.0 Hz has a coil consisting of 00 turns enclosing an area of 5.00 m 2 in a magnetic field of 0.00 T. (a) Calculate the maximum value of the emf generated. (b) Find the maximum instantaneous power supplied by the generator when the current is 50.0 A. SOLUTION (a) Applying Eq. 22 8, we find E 0 NBA (00)(0.00 T)(5.00 m 2 )( Hz) 8,800 V (b) Applying Eq. 9 8, we obtain The greater the current drawn from a generator, the more electric power is supplied by the generator, and of course the more mechanical power must be delivered to the generator to turn it. The current through the generator is determined by the load con nected to it. Every time someone switches on an electric device, a parallel branch is added to the vast electric network connected to a generator, and more current is drawn from it. For example, when you turn on your air conditioner, more current must pass through the generator providing your electricity, and some energy source such as coal must be used to provide additional energy to the generator. P IE 0 (50.0 A)(8,800 V) 940,000 W 940 kw MW

10 22 2 Inductance 22 2 Inductance Self-induction In previous examples of induction, the source of the changing magnetic field was external to the conductor in which an emf was induced. However, when a circuit carries a time-varying current, this creates a time-varying magnetic field in the space around the circuit and therefore induces an emf in the circuit. So, in this case, the circuit is the source of its own induced emf. This particular kind of induction is appropriately called self-induction. Lenz s law tells us that any induced emf opposes the change in magnetic flux that produces it. In the case of self-induced emf, the changing magnetic flux is created by a changing current in the circuit. This self-induced emf always opposes a change in current. Self-induced emf is sometimes referred to as back emf. Consider, for example, a simple DC circuit consisting of a 3 V battery, a resistor, and a switch S (Fig. 22 9). Initially the switch is open, and the current I 0. At t 0 the switch is closed. The current does not instantly change from zero to its steady-state value of 3 A, for this would imply an infinite rate of change in magnetic flux through the surface bounded by the circuit, which would produce an infinite back emf in the circuit. Instead the current grows gradually over a finite time interval from zero to its final value of 3 A. The time interval over which current changes is actually very short on the order of 0 7 s for this circuit. However, if a large solenoid were inserted into the circuit, it would take much longer to reach a steady-state current. We shall return to a detailed discussion of the time variation of current in such circuits, but first we must develop the precise relationship between back emf and the rate of change of current. The magnetic field produced by a circuit at any field point is proportional to the current I through the circuit, according to the Biot-Savart law (Eq. 2 2). The magnetic flux through the surface bounded by the circuit is proportional to the average magnetic field, and so the flux itself is also proportional to the current. I We can express the relationship between and I as an equality by introducing a constant of proportionality L. 597 Fig (a) When the switch S is first closed, there is no current in the circuit. (b) A short time after the switch is closed there is a steady current of 3 A. LI (22 0) The constant L, called the self-inductance, depends only on the size and shape of the circuit. We can see the significance of the self-inductance by applying Faraday s law to the equation above.* E t I E L (22 ) t The value of the self-inductance determines the size of the back emf E for a given rate of change of current I t. The back emf opposes the change in current. So the greater a circuit s self-inductance, the more it will resist a change in current. Thus selfinductance can be thought of as a measure of a circuit s electrical inertia, analogous to mass in mechanics. *For the remainder of this chapter we will omit the limit notation when referring to a varying emf; that is, we will take it to be understood that if / t is not constant what is really meant is the limiting value of this ratio.

11 598 CHAPTER 22 Electromagnetic Induction and AC Circuits The SI unit of self-inductance is the henry, abbreviated H, after the nineteenth cen - tury American physicist Joseph Henry.* We can relate this unit to other electrical units, using Eq. 22. Since the units for E and I/ t are respectively V and A/s, we have Fig The magnetic field pro duced by a coil is proportional to N, the number of turns in the coil. The coil s self-induct - ance is proportional to N 2. or V H-A/s H V-s/A (22 2) Unless a circuit contains some kind of coil, such as a solenoid, its self-inductance will be extremely small and, for most purposes, can be ignored. For example, the selfinductance of a circular loop of wire of radius 0 cm is on the order of 0 7 H. However, when a circuit does contain a coil, the self-inductance can be quite significant. The magnetic field is much stronger inside a coil, such as that shown in Fig , than it is near a single conducting wire; the field is enhanced by a factor of N, where N is the number of turns in the coil. This means that when current changes at a certain rate I t the changing magnetic flux inside the coil is enhanced by a factor of N and so also is the induced electric field circulating around the changing magnetic field lines. Each turn of wire in the coil experiences the induced electric field, and so the total emf induced in the N turns is N times the emf around a single turn, or N 2 times the emf induced by a single turn on itself. Thus the self-inductance of a coil with N turns is proportional to N 2. Coils with thousands of turns obviously have much greater self-inductance than that of a single loop. Inductors An inductor is a coil whose primary function is to provide self-inductance to a circuit in which it is inserted (Fig. 22 2). Whenever there is a changing current through the inductor, it provides the circuit with back emf. Inductors are often used in electric circuits; for example, an inductor is used as part of a tuning circuit in a radio. Since an inductor consists of many turns of wire, it usually has a significant resistance. However, we can think of a real inductor as a resistor in series with an ideal inductor, which has no resistance. An ideal inductor is represented in circuits by the symbol. We shall obtain an expression for the self-inductance of a long, thin solenoid of length and cross-sectional area A with N turns. The magnetic field inside the solenoid is, to a good approximation, given by the equation found in Chapter 2 for an ideal (infinite) solenoid with n N/ turns per unit length. The magnetic field is directed along the solenoid s axis and has magnitude given by Eq. 2 2: NI B 0 ni 0 The flux through any cross section of the solenoid is the product of the field magnitude B and the cross-sectional area A (Eq. 22 ): Fig An inductor. or BA NA 0 I *Henry built the first large electromagnets, including one that could lift two tons of iron. He also apparently discovered induction independently of Faraday, though he published his results somewhat later.

12 22 2 Inductance 599 According to Faraday s law, there is an emf E / t in each turn of the solenoid. For all N turns, the total emf is N times the value for one turn: E N t Using the preceding expression for to compute, we find NA 0 I and N E 2 A 0 I t Comparing this expression with Eq. 22, we see that the solenoid s self-induct ance is given by 0 N 2 A L (22 3) EXAMPLE 4 A Solenoid s Self Inductance Find the self-inductance of a solenoid of length 0 cm, with 000 turns and a cross-sectional area of 2.0 cm 2. SOLUTION Applying Eq. 22 3, we find 0 N L 2 A (4 0 7 T-m /A)(000) 2 ( m 2 ) 0.0 m H 2.5 mh The fact that an inductor produces an emf in a circuit means that the voltage drop across the terminals of an (ideal) inductor must have a magnitude equal to the magnitude of the emf, just as for a battery or a generator. I V ab L t To determine the sign of the voltage drop across an inductor, we note that its back emf means that the inductor acts like a battery, with terminals that are always oriented so that they oppose a change in the current, as indicated in Fig As shown in the figure, when one crosses the inductor terminals in the direction of positive current, the voltage drop has the same sign as I/ t: positive if I is increasing and negative if I is decreasing. Using this result and the preceding expression for the magnitude of the voltage drop, we see that the voltage drop across an inductor can be expressed I V ab L (for positive current directed from a to b) (22 4) t Fig Back emf of an inductor.

13 600 CHAPTER 22 Electromagnetic Induction and AC Circuits Fig An RL circuit. RL Circuits Next we shall describe the time variation of current in a simple circuit containing a resistor R and an ideal inductor L an RL circuit. Fig shows a resistor and an inductor connected to a battery through a switch, which is closed at t 0. We apply Kirchhoff s second rule around the circuit in the clockwise direction (starting at point a), and use the rules for the voltage drops across an inductor, a resistor, and a source of emf. V 0 I L IR E 0 (22 5) t This equation has the same form as the equation describing the charge on a capacitor being charged by a battery in an RC circuit (discussed in Section 20 6). Like the charge on the capacitor, the current in the RL circuit is an exponential function: E I ( e t/ ) (22 6) R where is the time constant of the RL circuit and is given by L (22 7) R The methods of calculus are used to verify that Eq is the solution to Eq When t 0, Eq predicts that E I ( ) 0 R This is as we expected. The current cannot change instantaneously from zero before the switch is closed to a nonzero value at the instant the switch is closed. At t, I E E ( e ) R R This is the final constant value of the current. The time constant is a measure of the time required to approach the final current. It is easy to find I as a fraction of the final current E/R at times that are multiples of. E E t : I ( e ) 0.63 R R E E t 2 : I ( e 2 ) 0.86 R R E E t 5 : I ( e 5 ) 0.99 R R E E t 7 : I ( e 7 ) R R Thus when t, the current is very nearly equal to E/R.

14 22 2 Inductance 60 EXAMPLE 5 Current in an RL Circuit A very large coil has self-inductance of 2.0 H and resistance of only 0.* The coil is connected across a 2 V battery at t 0. (a) Find the steady-state current. (b) Find the current through the coil at t 0.50 s, and indicate this point on a graph of I versus t. SOLUTION The coil can be treated as an ideal 2.0 H in - ductor in series with a 0 resistor. The circuit is that shown in Fig (a) The steady-state current is simply E/R, as for a circuit containing only a battery and a resistor. E 2 V I.2 A R 0 (b) Applying Eqs and 22 7, we find Fig L 2.0 H 0.20 s R 0 E I ( e t/ ) (.2 A)( e 0.50 s/0.20 s ) R. A Fig shows a graph of I versus t. *The ratio L/R for a coil usually has a value of 0 3 s or less. Suppose that in an inductive circuit, in which a steady current has been established, a switch is suddenly opened. The current will not instantly change from its initial value to zero. The circuit s back emf tends to maintain the current. The continuing current causes charge to accumulate on either side of the switch until the potential difference across the switch equals the applied emf. If the voltage across the switch is great enough, there may be a spark resulting from dielectric breakdown of the air. For example, when you turn off a toaster by unplugging it, you can see a spark between the plug and the electrical outlet.

15 602 CHAPTER 22 Electromagnetic Induction and AC Circuits Magnetic Energy When electric current passes through a resistor, electrical potential energy is converted to heat within the resistor. The loss of electrical potential energy equals the product of the charge passing through the resistor and the voltage drop across the resistor. Similarly, there is a loss of electrical potential energy when there is an increasing electric current I through an inductor ( I/ t 0). This follows from the fact that positive current flows through a positive voltage drop V ab L ( I/ t). Thus the charge q passing through the inductor loses electrical potential energy, where I Electrical energy lost qv ab q L (22 8) t This loss of electrical potential energy differs, however, from the loss in a resistor in two respects: The energy is not converted to heat, since no heat is generated in an ideal inductor. 2 The energy can be recovered when the current decreases; when I/ t 0, the equation above predicts a negative loss of electrical potential energy, meaning that there is a gain in electrical potential energy. Conservation of energy requires that the loss (or gain) of electrical potential energy of charge passing through an inductor must be accompanied by an equal gain (or loss) in some other form of energy. We shall find that this other energy is associated with the magnetic field inside an inductor. It is therefore called magnetic energy, denoted by U m. Equating the gain in magnetic energy, U m, to the loss in electrical potential energy expressed in Eq. 22 8, we obtain an expression for U m. I q U m q L L I t t As the current increases from zero to a final value I, more and more magnetic energy is stored in the inductor. The total magnetic energy U m is found from the equation above by inserting for the change in current, I I, and for q t the average current, I. 2 U m LI 2 (22 9) 2 This equation shows that magnetic energy stored in the inductor at any instant depends on the current in the inductor at that instant. Notice that this expression has the same form as the equation for the electrostatic energy stored by a capacitor C with a voltage V across its plates (Eq. 8 9: U e CV 2 2 ). Next we shall show how the magnetic energy may be written in terms of the magnetic field for the special case of a long, thin solenoid. The expression we shall obtain, though derived only for this simple special case, turns out to be valid in general. Inserting the expression for the self-inductance of the solenoid (Eq. 22 3) into the equation above, we obtain N U m 2 AI We can think of this energy as being stored inside the solenoid of length and crosssectional area A. The magnetic energy per unit volume, which we shall denote by u m, is found when we divide the expression above by the volume of the solenoid A, which gives

16 22 3 Alternating Current Circuits 603 N 2 I 2 u m Since the magnitude of the uniform magnetic field inside the solenoid is given by B 0 NI/ (Eq. 2 2), we may substitute B/ 0 for NI/ in the equation above to obtain the general expression for the magnetic energy per unit volume at any point in space: B u m (22 20) Notice that this expression for magnetic energy density in a magnetic field is similar in form to the expression for electric energy density in an electric field, obtained in Chapter 8 (Eq. 8 2: u e 0 E 2 2 ). The magnetic energy in an inductor may be compared with the electrostatic energy in a capacitor. When a capacitor is charged, electric potential energy is stored in the electric field between the capacitor plates; the energy is released as the capacitor discharges and the electric field is reduced. When current is established in an inductor, electrical potential energy is converted to magnetic energy, which is stored in the magnetic field inside the inductor. This energy is released when the inductor s current and its associated magnetic field decrease. EXAMPLE 6 Large-Scale Energy Storage in a Magnetic Field Energy storage is important for electrical power companies because of fluctuations in the rate of electrical energy consumption. Superconducting coils have been used for large-scale energy storage. (a) Find the magnetic energy density inside a superconducting coil that produces a magnetic field of 0.0 T. (b) Find the volume of space necessary to store kwh of energy at the density found in part (a). (This is enough energy to supply the electrical energy for roughly 5000 households for one day.) SOLUTION (a) Applying Eq , we find B u m 2 (0.0 T) J/m T-m/A This is a very high energy density compared to the energy densities attainable with other methods of storing energy. See Problem 29. (b) Applying the definition of magnetic energy density (magnetic energy per unit volume), we solve for the volume V required to store kwh of energy. u m V U m V U m um ( kwh)( J/kWh) J/m m 3 This volume corresponds to a cube with edge length V / m. *22 3 Alternating Current Circuits Although Edison s first power company supplied electrical energy in the form of direct current, today electricity is supplied by power companies as alternating current. We will discuss the practical advantage of alternating current, or AC, when we explain transformers at the end of this section. But first we shall describe a few simple AC circuits, containing an alternating source of emf, such as a generator, and either a resistor, a capacitor, an inductor, or all three together.

17 604 CHAPTER 22 Electromagnetic Induction and AC Circuits In this section we shall adopt the notation that lower-cased letters i and v will be used to denote instantaneous, time-varying currents and voltages respectively. Capital letters I and V will be used to denote constant values of current and voltage. Fig A resistor connected across an AC source. Fig (a) Time variation of i and v for the circuit in Fig (b) The phasor representation for the current and voltage in (a). Rotating phasors V and I represent current and voltage. Resistor and an AC Source Fig shows an AC circuit consisting of a resistor R connected across the terminals of a source of alternating emf E. For example, this circuit could represent a lamp plugged into an electrical outlet. The voltage drop v across the source s terminals is given by Eq. 22 9: v V sin t (22 2) where we have used V instead of E 0 to denote the amplitude of the voltage drop. (House- hold line voltage has an amplitude V 70 V, a frequency f 60 Hz, and angular fre - quency 2 f 20 rad/s.) The alternating voltage will produce in the circuit an instantaneous current, which alternates in direction, but which at any instant is the same throughout the loop. We wish to find the current i in the counterclockwise direction, as indicated in the figure. The voltage drop across the resistor equals the voltage drop v across the source. So we simply apply Ohm s law to find the current through the resistor. v i (22 22) R Inserting the preceding expression for v, we obtain or where i V R sin t i I sin t (22 23) V I (22 24) R Fig a shows graphs of i and v versus time. Notice that both i and v are proportional to sin t. The current i through the resistor and the voltage v across the resistor both oscillate at the same frequency. Furthermore, both i and v go to zero at the same time, and both reach their peak values at the same time. The current and voltage are in phase. This is a feature of the current through a resistor and the voltage across a resistor that is satisfied in any circuit. Another way of representing current and voltage is by means of a phasor diagram (Fig b). Current and voltage are represented by phasors I and V respectively. Phasors are like vectors, in that they have both a direction and a magnitude. The mag nitude of a phasor is the amplitude of the time-varying quantity it is used to represent. The respective magnitudes of I and V are amplitudes I and V. The direction of a phasor varies with time. Phasors rotate at angular velocity, starting at t 0 in the x direction. At any time t the phasors make an angle t with the x-axis. The instantaneous values of either current or voltage are found when the respective phasor is projected onto the y-axis, as indicated in the figure. Phasor diagrams are useful in the analysis of more complex circuits. Notice that Fig a shows the counterclockwise current i oscillating between positive and negative values, with an average value equal to zero. The current changes direction every half-cycle. Electrons oscillate back and forth, and on the average go nowhere! This does not mean that there is no electrical energy supplied to the resistor. As the electrons move back and forth, they collide with the lattice and produce thermal energy. The rate at which this thermal energy is produced is given by Eq. 9 6 for the power loss in a resistor: P i 2 R (22 25)

18 22 3 Alternating Current Circuits 605 Since P is proportional to i 2, the instantaneous power is always positive. Substituting Eq for the current into Eq , we obtain P I 2 R sin 2 t Fig shows a graph of sin 2 t versus t. The function sin 2 t is always positive and varies from 0 to. The average value of sin 2 t is 2, as indicated in the figure. This means that we can find an expression for the average power dissipated by the resistor, P av, by replacing sin 2 t by in the equation above. 2 P av I 2 R (22 26) Another way to express this result is in terms of a quantity called the root-meansquare current, abbreviated rms current. The rms current is defined to be the square root of the mean (or average) current squared. Applying this definition to Eq and using the fact that the average value of sin 2 t equals, we find 2 2 Fig The value of sin 2 t varies between 0 and, oscillating above and below its average value of. 2 I rms (i 2 ) av (22 27) (I 2 sin 2 t) av I 2 2 or Substituting I 2 rms resistor as 2 I I rms (22 28) 2 I 2 into Eq , we can express the average power loss in the P av I 2 rms R (22 29) The rms voltage is defined in the same way, and since v has the same time dependence as i, the rms voltage V rms is related to the voltage amplitude V in the same way that I rms is related to I; that is, V V rms (22 30) 2 AC ammeters and voltmeters read rms values. Thus, if a voltmeter measuring household line voltage reads 20 V, this means that the root mean square of the instantaneous voltage across the lines is 20 V. Then, according to Eq , the amplitude of the line voltage is 2 V rms 2 (20 V) 70 V. We have seen that Ohm s law can be expressed either in terms of instantaneous values of current and voltage (Eq : i v/r) or in terms of amplitudes (Eq : I V/R). It is easy to show that Ohm s law can also be expressed in terms of rms values. Substituting I 2 I rms and V 2 V rms into Eq , we find or I rms V rms R V rms I rms R (22 3)

19 606 CHAPTER 22 Electromagnetic Induction and AC Circuits EXAMPLE 7 An AC Voltmeter s Reading Find the reading of an AC voltmeter placed across a 50.0 resistor, which carries a current i 5.00 sin (20 t), where i is in amps. SOLUTION The current amplitude I 5.00 A. Applying Eq , we find the rms current. I 5.00 A I rms 3.54 A 2 2 We apply Ohm s law (Eq. 22 3) to find the rms voltage, which gives the voltmeter s reading. V rms I rms R (3.54 A)(50.0 ) 77 Fig A capacitor connected to an AC source. Capacitor and an AC Source Fig shows an AC circuit consisting of a capacitor connected across the terminals of a source of alternating emf, producing across the capacitor a voltage drop v V sin t (22 32) The figure shows positive counterclockwise current, which results in positive charge on the capacitor s left plate. The charge q stored on this plate at any instant is the product of the capacitance C and the instantaneous voltage v. q Cv CV sin t The current i flowing through the connecting wires equals the rate of change of q. i q t The methods of calculus can be used to find an expression for i from the preceding equation for q. The result is i I cos t (22 33) where I CV (22 34) Fig a shows graphs of v and i versus t. Notice that although the current and voltage both oscillate at the same frequency they are not in phase. The current leads the voltage by 4 cycle, or 90. This is also shown in the corresponding phasor diagram (Fig b). (a) (b) Fig (a) Time variation of i and v for the circuit in Fig (b) The phasor representation for the current and voltage in (a).

20 22 3 Alternating Current Circuits 607 Although no steady-state direct current can exist in a circuit containing a capacitor, our solution indicates that an alternating current can continue indefinitely. Of course no current actually crosses the capacitor s plates. The current exists in the wires connected to the capacitor, as the charge on its plates alternates in sign, in response to the alternating applied voltage. Initially there is a positive current in the counterclockwise direction, meaning that positive charge is being deposited on the left plate of the capacitor. But when the voltage reaches its peak value and starts to reduce (when t 90 ), the capacitor begins to discharge. This means that the current is positive in the clockwise direction (or negative in the counterclockwise direction, as the negative value of i in Fig a indicates). Next we shall express the relationship between V and I given in Eq in a different form. First we define the capacitor s capacitive reactance, denoted by X C, as Substituting C /X C in Eq , we obtain X C (22 35) C V I (22 36) XC This equation has the same form as Ohm s law for a resistor. The reactance of a capacitor is related to the capacitor s current and voltage in the same way that the resistance of a resistor is related to its current and voltage. Thus reactance must have the same units as resistance, units of ohms. The same relationship exists between rms values of current and voltage, since each equals 2 times the respective amplitude. V rms I rms XC (22 37) We can understand the physical reason for the current s dependence on reactance by considering why increasing the value of either capacitance or frequency should tend to increase the current, as predicted by Eqs and During a quarter cycle, from t 0 to t 90, charge CV is deposited on the left plate of the capacitor, as a result of the positive current. If C increases, both the charge and the current must increase. And if increases, the charge is deposited in a shorter time interval, meaning that again the current increases. The electrical power supplied to the capacitor at any instant is given by Eq. 9 5: P iv Substituting for i and v from Eqs and 22 33, we obtain P IV sin t cos t (22 38) or, using the trigonometric identity sin 2 t 2 sin t cos t, we get P 2 IV sin 2 t The average value of sin 2 t is zero, and so the average electrical power supplied to the capacitor is zero. P av 0 (capacitor) (22 39)

21 608 CHAPTER 22 Electromagnetic Induction and AC Circuits A charged capacitor stores energy in the field between its plates. As the capacitor is being charged, the source of emf is supplying energy to the capacitor. But when the capacitor discharges, the energy that was stored in the electric field is delivered back to the source. So the result is like a battery that is alternately discharging (giving up energy) and being charged (being reenergized). EXAMPLE 8 The Frequency at which a Capacitor s Reactance is 50 V At what frequency would a 5.00 F capacitor produce the same current as the 50.0 resistor in Example 6, when connected to the same voltage source as the resistor? Applying Eq , we solve for f. X C C 2 fc SOLUTION If the current and voltage amplitudes are the same as for the resistor, it follows that the ratio V/I must be the same: V X C 50.0 I f 2 ( CXC F)(50.0 ) 637 Hz Fig An inductor connected to an AC source. Inductor and an AC Source Fig shows an AC circuit consisting of an inductor connected across the terminals of a source of alternating emf, producing across the inductor a voltage drop v V sin t (22 40) This voltage drop is related to the rate of change of current through the inductor by Eq. 22 4: v L Equating the two expressions for v and solving for i t, we obtain i t V L i t sin t The methods of calculus can be used to solve this equation for i. The result is where i I cos t (22 4) I V L We define inductive reactance, denoted by X L, as X L L (22 42) The preceding equation can then be expressed in the form V I (22 43) XL As usual, the same relationship exists between rms values of current and voltage.

22 22 3 Alternating Current Circuits 609 I rms XL (22 44) Like capacitive reactance, inductive reactance has units of ohms. Fig. 22 3a shows graphs of v and i versus t for the inductor circuit. As in the case of the capacitor circuit, though the current and voltage both oscillate at the same frequency, they are not in phase. But here the current lags the voltage by 4 cycle, or 90. This is also shown in the corresponding phasor diagram (Fig. 22 3b). The electrical inertia of the inductor causes this current lag. V rms (a) (b) Fig (a) Time variation of i and v for the circuit in Fig (b) The phasor representation for the current and voltage in (a). The electrical power supplied to the inductor at any instant is found by applying Eq. 9 5: P iv Substituting for i and v from Eqs and 22 4, we obtain P IV sin t cos t Except for the minus sign, this is the same expression as Eq , which gives the instantaneous power supplied to a capacitor. As we found for the capacitor, the average power supplied is zero. P av 0 (inductor) (22 45) As the current increases, the magnetic field and magnetic energy inside the inductor increase, with the energy being supplied by the source. As the current decreases, the magnetic energy decreases, with the energy being returned to the source. EXAMPLE 9 Finding the Current Through an Inductor A 20.0 mh inductor is connected across a 60.0 Hz source with an rms voltage of 20 V. Find the rms current. SOLUTION reactance First we apply Eq to find the inductive X L L 2 fl 2 (60.0 Hz)( H) 7.54 The rms current is found by applying Eq V rms 20 V I rms XL 5.9 A 7.54

23 60 CHAPTER 22 Electromagnetic Induction and AC Circuits Fig A resistor, a capacitor, and an inductor connected in series with an AC source. RCL Series Circuit Fig shows an alternating current source connected across a series combination of a resistor, a capacitor, and an inductor. The source produces a voltage drop v across the combination, where v V sin t (22 46) This voltage drop equals the sum of the voltage drops across the circuit elements R, C, and L. v v R v C v L (22 47) Inserting into this expression the preceding equation for v and the expressions for the voltage drops across R, C, and L, we obtain q i V sin t ir L (22 48) C t This equation may be solved for i, using methods of calculus. The result is a cur rent that oscillates at the same frequency as the source but not necessarily in phase with it. i I sin ( t ) (22 49) The amplitude I and phase angle are determined by V, R, C, L, and. Fig a shows graphs of v and i versus t, and Fig b shows the corres - ponding phasor diagram. We can find the phase angle between the current and voltage, as well as the relationship between the amplitudes V and I, by inserting into our phasor diagram phasors representing voltages across each of the elements R, C, and L. This expanded phasor diagram, shown in Fig , makes use of the fol lowing phase relationships (the same as in the simpler circuits containing only one element: R, C, or L):. The voltage v R across the resistor is in phase with the current. 2. The current leads the capacitor voltage v C by The current lags the inductor voltage v L by 90. (a) (b) Fig (a) Time variation of the voltage v across the source and the current i for the circuit in Fig (b) The phasor representation for the voltage and current in (a). In order for the projections of the phasors on the y-axis to satisfy Eq , the phasors themselves must satisfy the corresponding vector relationship: V V R V C V L (22 50)

24 22 3 Alternating Current Circuits 6 Fig A phasor diagram used to relate the amplitudes of v and i and the phase angle in Fig Since (V L V C ) has magnitude (V L V C ) and is perpendicular to V R, as indicated in Fig , the magnitude of phasor V is related to the magnitudes of the other phasors by the equation V 2 V R2 (V L V C ) 2 The voltage amplitudes are related to the current amplitude by the equations V R IR V L IX L V C IX C Substituting these three equations into the preceding equation, we find or We may express this result as V 2 I 2 R 2 I 2 (X L X C ) 2 V I R 2 ( X L X C ) 2 where Z, called the impedance, is defined as V IZ (22 5) Z R 2 ( X L X C ) 2 (22 52) Once again the rms values of current and voltage satisfy the same equation as the amplitudes; that is: V rms I rms Z (22 53) We can think of impedance as a generalization of resistance. The impedance appears in Eq the same way that resistance appears in Ohm s law. And, according to Eq , Z reduces to R when X L and X C both equal zero.

25 62 CHAPTER 22 Electromagnetic Induction and AC Circuits EXAMPLE 0 Voltmeter Readings Across RCL Circuit Elements Suppose the RCL circuit shown in Fig has the following values for the circuit elements: R 300, C 5.60 F, and L 200 mh. Let the rms voltage across the 60.0 Hz source be 20 V. (a) Find the rms current. (b) Find the rms voltage drop across each of the circuit elements. SOLUTION (a) First we calculate the capacitive and inductive reactances, using Eqs and 22 42: X C C 2 fc 2 (60.0 Hz)( F) 474 X L L 2 fl 2 (60.0 Hz)( H) 75.4 Next we insert these values, along with the resistance, into Eq to find the impedance. Z R 2 ( X L X C ) 2 (3 0 0 ) 2 ( ) Finally, we apply Eq to find the rms current. I rms V rms Z 20 V A (b) We find the rms voltage drop across each circuit element by multiplying the rms current times that element s impedance, either R, XC, or XL. V R, rms I rms R (0.240 A)(300 ) 72.0 V V C, rms I rms X C (0.240 A)(474 ) 4 V V L, rms I rms X L (0.240)(75.4 ) 8. V These values correspond to the readings of an AC voltmeter placed across each of the circuit elements. If we place the voltmeter across all three elements, it will not read the sum of these voltage drops (204 V). Instead it will read the rms voltage of the source, 20 V. The rms voltage drops are not additive because the voltages are not in phase. Resonance Consider an RCL circuit with fixed values of resistance, capacitance, and inductance, but with a source frequency f, which can be varied. Since both inductive and capacitive reactance depend on frequency, varying the frequency causes the circuit s impedance and current to vary. According to Eq. 22 5, current is maximized when impedance is minimized. From Eq , we see that, at a frequency such that X L X C 0, the impedance will have its smallest value, Z R. Setting the two expressions for reactance equal to each other, and solving for the corresponding frequency f 0, we find X L X C 2 f 0 L 2 f0 C f 0 (22 54) 2 L C This frequency at which the current is maximized is the circuit s resonant frequency. See Chapter 5, Section 5 5, for a discussion of resonance. Resonance is used in tuning a radio or television. Tuning adjusts the receiving circuit so that it is in resonance at a resonant frequency equal to the broadcast frequency, so that only the signal from that one source is amplified.

26 Transformers Electrical power companies provide electrical energy by means of the alternating potential difference maintained between power lines. The potential difference between the lines entering a residence must be great enough to operate the typical electrical devices to which they are connected but not so great as to be extremely dangerous. (For example, you would not want to plug an appliance into a 00,000 V outlet.) So residences are supplied with lines at 20 V and 240 V, rms. See Chapter 20, Section On the other hand, long transmission lines leading from a distant power generator need to be at very high voltage to minimize energy loss in the lines. Some of the electrical energy supplied by a generator is wasted as heat in the power lines. These lines are thick wires, but since the lines are very long, their total resistance becomes significant. (The resistance might be on the order of per mile.) The average rate of production of heat in lines having resistance R and carrying rms current I rms is given by P av, waste I 2 rmsr (22 55) The average rate at which energy is supplied to lines, carrying rms current I rms and with an rms potential difference V rms between the lines, is given by P av, supply I rms V rms (22 56) To minimize the production of waste heat while supplying the required power, it is best to maintain the lines at as high a voltage as possible, so that the current is as small as possible. For example, from Eq , we see that lines transmitting 00,000 W of power could carry 00 A with a potential difference of 000 V, or the lines could transmit the same power by carrying A with a potential difference of 00,000 V. According to Eq , in the first case the rate of production of waste heat equals (00 A) 2 R, or 0 4 W per of line resistance, whereas in the second case the rate of heat production is ( A) 2 R, or W per of line resistance. Clearly the higher the voltage, the more efficient will be the transmission of power. But this voltage requirement differs from the requirements for energy utilization. So what is needed is a device to change voltages, so that power can be transmitted over long distances at high voltage and then, close to the point of energy use, transmitted at much lower voltages. The transformer is such a device Alternating Current Circuits 63

27 64 CHAPTER 22 Electromagnetic Induction and AC Circuits Fig Electrical distribution system. Fig shows a schematic diagram indicating how voltages are first increased and then decreased in lines leading from a power plant to points of energy consumption. Fig shows a schematic diagram of a transformer consisting of two electric - ally insulated coils wound around a common iron core. The primary coil, consisting of N turns, is shown connected to a generator that maintains an rms voltage V,rms across it. The alternating current in the primary coil produces in the iron core a time-varying magnetic field, which induces an emf in the secondary coil, and a voltage V 2,rms across the secondary, which is different from V,rms. Each coil of the transformer acts like an inductor. The magnetic flux is the same through each turn of either coil. Therefore, according to Faraday s law, any one turn of either coil must have the same emf, which we shall designate E turn. The primary coil, with N turns, has a total emf E, where E N E turn (22 57) and the secondary coil, with N 2 turns, has a total emf E 2, where Fig Schematic diagram of a transformer. E 2 N 2 E turn (22 58) Equating two expressions for E turn obtained from Eqs and 22 58, we obtain E 2 E N2 N

28 22 3 Alternating Current Circuits 65 As for an ideal battery, inductor, or generator, the voltage drop across either coil of an ideal transformer (one with negligible energy loss) equals the emf induced in the coil. Thus we can substitute voltages for emf s in the equation above, which yields V 2,rms N2 V,rms N or V 2,rms N V,rms (22 59) If the number of turns in the secondary is greater than that in the primary, N 2 /N, and so the output voltage is greater than the input voltage. Such a transformer steps up voltage. If, however, there are fewer turns in the secondary than in the primary, N 2 /N, the output voltage is less than the input voltage; the transformer steps down voltage. A transformer depends on time-varying voltage for its operation, since it is based on the principle of induction. There is no comparable device for changing the voltage of a direct-current source. Hence alternating current, rather than direct current, is used today to supply society s electrical needs. The fact that a step-up transformer can actually put out a higher voltage than is input to the transformer might lead you to believe that you can get more energy out of a transformer than you put in. This would be true if the currents in the primary and secondary coils were equal. However, they are not. In an ideal transformer (one with negligible energy loss from heat in the windings and so forth), conservation of energy requires that the power input and power output are the same. Since the average power is the product of rms current and voltage, we have the following relationship between primary and secondary currents and voltages: N 2 I,rms V,rms I 2,rms V 2,rms (22 60) EXAMPLE Power Loss in Transmission Lines at Different Voltages A power station transmits electrical power a long distance This is one fifth of the 400 kw of power generated a significant loss. through lines having a total resistance of 200. Calculate the power lost through heating of the lines, if kw of (b) Repeating our calculations at the higher line voltage, we power is supplied by the generator to lines at a potential difference of (a) V; (b) V. find P av,supply W I rms Vrms A SOLUTION (a) First we calculate the current carried by V the lines, applying Eq P av,waste I 2 rms R (0.800 A) 2 (200 ) P av,supply I rms V rms 28 W I rms P av,supply Vrms W V 20.0 A Applying Eq , we find the power wasted as heat in the lines. P av,waste I 2 rmsr (20.0 A) 2 (200 ) W 80.0 kw This is a very small fraction of the power generated. Stepping up the voltage from 20,000 V to 500,000 V can be accomplished by a transformer in which the ratio of secondary turns to primary turns, N 2 /N, equals the ratio of the voltages, V 2 /V, according to Eq Thus N 2 N 500, ,000 25

29 In Perspective Michael Faraday Michael Faraday (79 867). Fig. 22 A Faraday lecturing at the Royal Institution. In 805 Michael Faraday, the 4-year-old son of a London blacksmith, began work as an apprentice bookbinder. During the next several years, young Faraday read many of the books that passed through the shop where he worked. Encyclopædia Britannica articles on electricity and chemistry stimulated his interest, and he began to perform simple experiments and to attend scientific lectures. At 2 years of age Faraday was hired as a laboratory assistant at the Royal Institution, an important center of scientific research. Having only the most elementary formal education, Faraday had gained access to the best laboratory and library facilities, as well as to the institute s director, the chemist Sir Humphrey Davy, who tutored him. Faraday was given in - creas ing responsibility, and over a period of years gradually progressed from assisting others in their work to performing his own original research. By age 30 his discoveries had made him world famous. Eventually Faraday became the director of the institute. In that capacity, he often lectured to the public on a broad range of scientific subjects. Faraday was a skillful and popular lecturer. At his annual Christmas lecture for children, he loved to inspire and entertain his young audience with dramatic demon strations. Faraday s research was directed by his belief in unifying principles in nature and by his search for symmetries in natural phe nomena. He believed, for example, that gravitational, electric, and magnetic forces were all related.* Faraday s vision led him to perform important experiments in chemistry, electricity, magnetism, optics, and sound. Faraday studied and repeated other scientists experiments in electricity and magnetism. Oersted s demonstration of the force exerted on a magnet by a conducting wire had been the first step toward the unification of electricity and magnetism. Ampère, Biot, and Savart then studied the interaction of electric currents. *The unification of the gravitational force with the other fundamental forces was a goal of Einstein; physicists today still search for a grand unified field theory. Faraday s attention was particularly drawn to the experiments and theories of Ampère, who proposed that the mag ne - tism displayed by permanent magnets is due to hidden electrical currents within the magnet, as dis cussed in Chapter 2. This suggested to Faraday that the mag - netism induced in iron by a permanent magnet might be an indication of a general property of induction, which all electric currents should possess. Thus Faraday thought that an electric cur rent in one conductor might induce a current in another nearby conductor, and in 825 he tried to observe such an effect. One current-carrying wire was placed close to a second wire, which was a part of a com - plete circuit that included a sensitive gal - vanometer G, used to detect any cur rent that might be induced in the second circuit (Fig. 22 B). No current was detected. This induced magnetism is actually caused by alignment of magnetized, randomly oriented domains with an external magnetic field, as discussed in Section 2 6.

30 In Perspective Fig. 22 B The constant current in the circuit on the left induced no current in the circuit on the right, contrary to Faraday s expectations. However, Faraday was never satisfied with one experiment. He was an extremely careful, thorough, and open-minded experimentalist, who constantly tried to challenge his own theories and to discover any faulty preconceptions. Over the next several years, he performed many experiments in the search for induced current. Finally in 83 Faraday observed induced electric current using the apparatus illustrated in Fig Two coils of electrically insulated wire were wound around an iron ring. A battery provided the current in the primary circuit on the left, and a galvan - ometer was used to detect any current induced in the secondary circuit on the right. Faraday correctly supposed that the iron would enhance any induced current that was produced, making it easier to detect. While the primary circuit containing the battery was being connected, Faraday noticed a momentary deflection of the galvanometer. With a steady current through the primary circuit, no deflection of the galvanometer was observed. But when the primary circuit was disconnected, there was again a momentary deflection of the galvanometer needle, this time in the opposite direction. Thus Fara - day discovered that although a steady current produces no induction effect a changing current in one circuit does induce a current in a nearby circuit. He then began a series of other experiments in which he showed that electric currents were induced in conductors moving near permanent magnets, as illustrated in Fig Using this principle, Faraday built the first primitive electric generator; by rotating a conducting disk in a magnetic field, he produced a continuous current in wires connected to the disk (Fig. 22 ). Eventually the electric generator would make possible the production of cheap, abundant electric current, which batteries could never do. One of Faraday s most important con - tributions was the conceptual framework he used in observing electric and magnetic phenomena. In the early nineteenth cen - tury, electric and magnetic forces were described mathematically, but, lacking an understanding of atomic structure, the conceptual basis of the description was confused, with various theories competing to explain the phenomena. Faraday did not understand all of the mathematics that was being used, but he had a great intuitive sense. He was the first to use the concept of a field and to picture field lines, which he referred to as lines of force. This approach was to flourish in the hands of James Maxwell, who possessed the mathematical education that Faraday lacked. Maxwell read Faraday s work, as well as the more mathematical work of others. His synthesis resulted in the unification of electricity and magnetism, using the first field theory. Field theory has been an important part of theoretical physics ever since. In the introduction to his great work A Treatise on Electricity and Magnetism, in 873 Maxwell wrote: As I proceeded with the study of Faraday, I perceived that his method of conceiving the phenomena was also a mathematical one, though not exhibited in the conventional form of mathematical symbols. For instance, Faraday, in his mind s eye, saw lines of force traversing all space where the mathematicians saw centres of force attracting at a distance: Faraday saw a medium where they saw nothing but distance: Faraday sought the seat of the phenomena in real actions going on in the medium, they were satisfied that they had found it in a power of action at a distance.

31 C HAPTER 22 SUMMARY The magnetic flux through a surface of area A is defined as the product of the area and the field s component B, perpendicular to the surface. B A The flux is proportional to the number of field lines passing through the surface. When the magnetic flux changes, an electric field is created or induced in the surrounding space. The result of this induced field can be an emf and a current induced in a conducting loop. According to Faraday s law, the magnitude of the emf induced around any loop equals the magnitude of the rate of change of magnetic flux through the surface bounded by the loop. E (constant E) t or E limit (variable E) t 0 t The direction of the induced current resulting from the emf is found from Lenz s law, which states that the induced current produces a magnetic field that opposes the change in magnetic flux that produced it. Rotating a coil of wire in an external magnetic field at an angular velocity results in an induced emf E and a terminal potential difference V ab, where V ab E E 0 sin t The amplitude E 0 is the product of the number of turns N, the field strength B, and the coil s area A. E 0 NBA A time-varying current in a coil produces a self-induced emf, opposing the emf E in the coil. The magnitude of the induced emf depends on a property of the coil called its self-inductance, denoted by L and defined by the relationship between the current I through the coil and the magnetic flux through the coil resulting from that current: LI A coil designed to have large self-inductance is called an inductor. The self-inductance of a long, tightly wound solenoid of length and cross-sectional area A, with N turns, is given by 0 N 2 A L The voltage drop V ab across an inductor is given by V ab L I t if positive current flows through the inductor in the direction from point a to point b. When a resistor R and an inductor L are initially con - nected in series across a battery of emf E, current is not immediately established in the circuit because of the electrical inertia of the inductor. Instead the current grows with time according to the equation E I ( e t/ ) R where the time constant is given by L R An inductor L carrying current I stores magnetic energy U m, where U m LI 2 The magnetic energy per unit volume, u m, inside the in - ductor (or anywhere there is a magnetic field) may be expressed in terms of the field strength as B u m An RCL series circuit is formed when a resistor R, a capacitor C, and an inductor L are connected in series across an AC source supplying a voltage V sin t across the combination. The current through the circuit is given by i I sin ( t ) The amplitude I is determined by the voltage amplitude V and the impedance Z from the equation V IZ where Z is related to the resistance R and to the capacitive reactance X C and to the inductive reactance by the equation X L L Z R 2 ( X L X C ) 2 It is often convenient to express the current by its rms (root-mean-square) value, I rms, which is related to the cur - rent amplitude I by the equation I rms 2 C I 2 68

32 Questions 69 Similarly, rms voltage V rms is related to the voltage amplitude V by the equation V V rms 2 The rms current and voltage are related in the same way as the current and voltage amplitudes: V rms I rms Z A resistor converts electrical energy to heat at an average rate P av I rms2 R (for a resistor) whereas neither an inductor nor a capacitor uses electrical power: The current in the circuit is maximized at a frequency f 0, called the resonant frequency, which is given by f 0 2 L C Transformers transform voltages from an input voltage V,rms to an output voltage V 2,rms, where the voltages are related to the number of turns of wire in the primary and secondary, N and N 2, by the equation V 2,rms N 2 N V,rms P av 0 (for a capacitor or inductor) Questions Do more of the earth s magnetic field lines pass outward through the earth s surface or inward, or is there an equal number in each direction? 2 Can there ever be more magnetic field lines passing out - ward through a spherical surface than passing inward? 3 What is the direction of the current through resistor R in Fig as switch S is (a) closed; (b) opened? 5 Fig shows a metal ring moving upward toward the south pole of a permanent magnet. Is the current induced in the ring in the clockwise or counterclockwise direction, when viewed from above? Fig Is the direction of the induced current in the circular loop in Fig clockwise or counterclockwise as switch S is (a) closed; (b) opened? Fig Fig

33 620 CHAPTER 22 Electromagnetic Induction and AC Circuits 6 Fig shows a square coil of wire falling through a magnetic field. Indicate for each of the three positions shown in the figure the direction of the current (if any) induced in the coil and the direction of the net magnetic force (if any) acting on the coil. This illustrates how induced currents can tend to decelerate a conductor moving through a magnetic field. 8 A coil of wire is connected to the pedals of a bicycle, so that the coil turns as the bicycle is pedaled (Fig ). Pedaling will generate an emf in the coil if a magnetic field is present, unless the magnetic field is in which of the following directions: (a) vertical; (b) horizontal, from the front to the back of the bicycle; (c) horizontal, from one side of the bicycle to the other? For which of these directions of the magnetic field will the force required to pedal the bicycle be greater than when there is no magnetic field? For which directions will the force be less than when there is no magnetic field? Fig Fig Fig shows a coil of wire in the xz plane, with a magnetic field, directed along the x-axis. Around which of the three coordinate axes should the coil be rotated in order to generate an emf and a current in the coil? 9 When an iron bar is inserted into a solenoid, the selfinductance of the solenoid: (a) increases; (b) decreases; (c) remains the same. 0 Switch S in Fig is initially closed. (a) Find the sign of the voltage drop V ab across the in - ductor just after the switch is opened. (b) Estimate the current through the resistor 0 2 s after the switch is opened. Fig Fig Suppose you insert an ideal inductor in series with a resistor in a DC circuit. What effect will the inductor have on the steady-state current through the resistor? 2 Does a very high-frequency, single-loop AC circuit con - taining an inductor behave like (a) an open circuit or (b) a short circuit?

34 Questions 62 3 Does a very high-frequency, single-loop AC circuit con - taining a capacitor behave like (a) an open circuit or (b) a short circuit? 4 The primary of a transformer is connected to a 2 V battery. What is the voltage across the transformer s sec ondary if it has 0 times as many turns as the primary? 5 The current through the coil of a stereo speaker determines the intensity of the sound it produces, as dis - cussed in Chapter 2 (Example 4 and Problem 55). A typical large speaker box contains two speakers, as indicated in Fig The larger one, called a woofer, is better at producing high-quality low-frequency, longwavelength sound. The smaller one, called a tweeter, is better at producing high-frequency, short-wavelength sound. The electric current from the stereo amplifier is a time-varying signal, with both high-frequency and lowfrequency components. The two speakers are connected in parallel as shown in the figure. (a) To reduce the amplitude of the low-frequency com - ponent of the current through the tweeter, relative to the high-frequency component, should a capacitor or an inductor be inserted at point P, in series with the tweeter s coil? (b) Should a capacitor or an inductor be inserted at point Q in series with the woofer s coil? 6 Consider plugging each of the following directly into an electrical outlet in a home: (a) a 00 resistor; (b) a 00 F capacitor; (c) a 00 H ideal inductor. Which would be more likely to blow a fuse or trip a circuit breaker? 7 Which of the following 20 V AC devices costs more to operate: (a) one that draws an rms current of 0 A, or (b) one that draws a peak current of 2 A? 8 Which of the following 20 V AC devices, each of which has an impedance of 00, requires more electrical power: (a) one with 50 of inductance, or (b) one with 0 of inductance? 9 In a certain RCL series circuit, L 30 H and C 0 F. The rms current is 0 A and the rms voltage across the capacitor is 20 V. Is the average energy stored by the inductor or the capacitor greater? 20 An airport metal detector consists of a large coil con - nected as part of an RCL series circuit. When one walks through the coil with a significant amount of metal, an alarm is set off (Fig ). The detector is sensitive to small changes because the circuit is initially at resonance, and a small change in the circuit produces a large drop in current, which is used to activate the alarm. Which circuit parameter does one change by bringing metal through the detector: (a) resistance; (b) capa ci - tance; (c) inductance; or (d) frequency? Fig Stereo speakers. Fig Answers to Odd-Numbered Questions equal number in each direction; 3 (a) to the right; (b) to the left; 5 clockwise; 7 z-axis only; 9 a; none; 3 b; 5 (a) capacitor; (b) inductor; 7 a; 9 capacitor

35 622 CHAPTER 22 Problems (listed by section) Electromagnetic Induction and AC Circuits 22 Faraday s Law A magnetic field of magnitude 0.40 T is directed vertically upward. Find the magnetic flux through a flat sur - face of area 20 cm 2, if the surface is (a) vertical; (b) hor - izontal; (c) at an angle of 60 with the horizontal. 2 Find the flux of the earth s magnetic field through a hor izontal section of the earth s surface of area 00 m 2 in a region where the magnetic field has a magnitude of gauss and a horizontal component of gauss. 3 You stand in the earth s magnetic field, directly facing its horizontal component (in other words, facing mag - netic south). The field has a magnitude of gauss and is directed 60.0 below the horizontal. (a) Calculate the inward magnetic flux through your body s front surface, of area m 2. (b) Compute the outward flux through your body s back surface. (c) What is the net inward magnetic flux through your body s entire surface? 4 The Houston Astrodome encloses a ground area of 37,000 m 2. Find the flux through the dome of the earth s magnetic field, which has a magnitude there of 0.50 gauss and is directed 60 below the horizontal. *5 A magnetic field of magnitude 0.0 T is directed along the positive x-axis and passes through a wedge-shaped object, as indicated in Fig Find the outward directed magnetic flux through (a) the bottom; (b) the front (triangular) side; (c) the left side; (d) the right (inclined) side. 6 A circular loop of wire of radius 0 cm moves at a speed of 5.0 m/s in the direction of a 0.0 T uniform magnetic field, perpendicular to the plane of the loop. What is the emf induced in the loop? 7 In a time interval of s, the magnetic flux through the circular loop shown in Fig increases from zero to T-m 2, as switch S is closed. What is the average emf induced in the loop? *8 A small rectangular loop of wire is located close to a long straight wire initially carrying a current of 00 A (Fig ). Find the average emf induced in the loop if the current in the wire reverses direction in a time interval of s. 20 Fig *9 A small circular loop of wire of radius 5.0 cm and resistance is centered inside a large circular loop of wire of radius 50 cm (Fig ). The larger loop, which initially carries a current of 8.0 A, is cut and its current is reduced to zero over a time interval of s. Find the average current in the smaller loop dur - ing this time interval. (The magnetic field of the larger loop is approximately constant over the smaller loop.) Fig Fig

36 Problems The area enclosed by the wire loop in Fig. 22 is reduced from 30 cm 2 to 0 cm 2 in a time interval of 0.0 s. The magnitude of the magnetic field is 0.50 T. Find the emf induced in the wire. Suppose that the rod in Fig moves to the left, so that the area enclosed by the conducting path decreases at the rate of cm 2 /s. Find the magnitude and direction of the current if the resistance of the con duct - ing path is 0.0 and the magnetic field strength is 0.50 T. **2 A horizontal copper rod of mass 5.0 g and length 0 cm slides downward with negligible friction as it maintains contact with vertical copper wires, which connect it to a resistor (Fig ). The rod and wires have negligible resistance. A uniform magnetic field of 0.60 T is directed horizontally, perpendicular to the plane of the circuit. Initially the rod accelerates downward but soon approaches a constant terminal speed v. Find v. *5 Suppose that the electric motor discussed in Example 5, Chapter 2, is operated in reverse, so that it acts as an elec tric generator (Fig ). At any time during the rotation of the coils, the single coil that is connected through the commutator to the external circuit is approximately perpendicular to the magnetic field. Thus, in - stead of an emf that varies as E 0 sin t, we have an emf E E 0 sin 90 E 0, with E 0 NBA. This means that the generator acts as a nearly constant source of emf, a DC generator. Suppose that the generator supplies 20 W of electric power while connected to a 50.0 resistor. The coil has negligible resistance. Find the gen - erator s emf and the current in the circuit. Fig **3 Fig shows a square metallic loop of edge length 0.0 cm, mass 4.00 g, and resistance moving downward at a speed of 4.00 cm/s, and entering a uniform magnetic field B. (a) What is the strength of the field if the loop experiences no acceleration initially? (b) When will its speed begin to change? *4 When a torque turns the shaft of an electric generator at an angular velocity, the mechanical power input is given by P (the rotational analog of Eq. 7 40: P Fv). Equating the mechanical torque to the mag netic torque on the current loop and equating the mech anical power input to the electrical power supplied by the generator, derive Eq Fig DC generator. *6 Suppose that the electric generator shown in Fig consists of 000 turns of wire enclosing an area of 00 cm 2 in a magnetic field of 0.30 T. The generator is connected to an 8 resistor. By pedaling the bicycle you are able to supply power of 75 W ( 0. hp) to the generator. (a) Find the current and emf generated, and the rate of pedaling in rev/s. (b) Suppose you decrease your rate of pedaling to half the original rate. How much power is supplied to the generator?

37 624 CHAPTER 22 Electromagnetic Induction and AC Circuits 7 In a region where there is a vertical magnetic field of T, a ring of radius.00 cm is flipped in the air as one would flip a coin, so that it begins rotating about a horizontal axis at the rate of 50.0 rev/s. (a) Find the maximum instantaneous emf induced in the ring and the maximum current if the ring has a resistance of (b) What effect does the magnetic field have on the rate of rotation? *8 Suppose you are stranded on a desert island and have no magnets, but you do have plenty of 0.50 mm radius copper wire and flowing water as a source of power. You also have various electrical devices that require for their operation an electric generator producing 70 V peak emf (or an rms emf of 20 V) at 60 Hz. You hope to use the earth s magnetic field to generate the electricity you need, by turning a meter square coil of wire at 60 Hz about an axis perpendicular to the earth s field of magnitude 0.50 gauss. (a) How many turns of wire are required? (b) What is the total resistance of the coil? (c) What is the maximum (short-circuit) peak current provided by the generator, and what peak instan - tan eous power is required to turn the generator in this case? (d) Could your generator be used to operate a device requiring 50 W of electrical power? 22 2 Inductance 9 The self-inductance of the circuit in Fig is H. Find the back emf when the current is increasing at the rate of A/s. 20 Find the cross-sectional area of a 5.0 cm long solenoid with 500 turns and a self-inductance of 40 H. 2 A solenoid with 200 turns and a cross-sectional area of.00 cm 2 is 3.00 cm long. How much current must the solenoid carry in order that the flux produced by its own magnetic field is equal to the flux produced by earth s magnetic field of magnitude G, directed along the solenoid s axis? 22 (a) Find a possible set of values for the number of turns, length, and cross-sectional area of a solenoid with a self-inductance of.00 H. (b) What is the back emf produced by such an inductor when the current through it is increasing at the rate of 00 A/s? 23 Find the voltage drop V ab across each of the inductors in Fig Fig *24 An extension cord consists of two long, thin wires sep - arated by a small distance d, carrying current of equal magnitude but in opposite directions. When the current I changes, there is a changing magnetic flux through the surface area between the wires. Show that if one approximates the magnetic field by its value at the midpoint between the wires the cord s self-inductance is approximately 2 0 /, where is the length of the cord. Evaluate for a cord of length 0 m. *25 Find the rate of change of current in the circuit shown in Fig at t 0 just as the switch is closed, if L 2.00 mh and E 6.00 V. *26 At t 0 an inductor is connected to a 9.0 V battery. At t 2.0 s, the current through the inductor is 0.37 A, and after about 0 s, the current reaches a stable value of.0 A. Find the resistance and the inductance of the inductor. 27 Find the magnetic energy stored by a 5.00 mh inductor when the current through it is 0.0 A. 28 Estimate roughly the magnetic energy stored in the earth s magnetic field by calculating the magnetic energy in a sphere with twice the earth s radius, within which is a uniform magnetic field of 0.5 G. For comparison, the earth s rotational kinetic energy is J (Example 7, Chapter 9.) 29 Calculate the gravitational potential energy stored in.00 m 3 of water raised a vertical distance of 00 m. Find the ratio of this gravitational energy density to the magnetic energy density found in Example 6.

38 Problems 625 *22 3 AC Circuits 30 The rms current through a light bulb is A when the bulb is connected to a standard 20 V rms source. Find (a) the bulb s resistance; (b) the peak instantan - eous current; (c) the peak instantaneous voltage; (d) the average electrical power used by the bulb. 3 An electric heater is rated at 000 W, 20 V rms. Find the heater s: (a) resistance; (b) rms current; (c) peak current; (d) peak instantaneous power; (e) minimum instantaneous power. 32 Find the reactance of a 2.00 F capacitor at a frequency of (a) 60.0 Hz; (b) 600 Hz; (c) 6000 Hz. 33 Find the rms current that results from connecting a 2.00 F capacitor directly across a 20 V rms source at a frequency of (a) 60.0 Hz; (b) 600 Hz; (c) 6000 Hz. *34 A capacitor draws a 0.0 A rms current from a 20.0 V rms source. Find (a) the maximum instantaneous electrical power supplied to the capacitor; (b) the minimum instantaneous electrical power supplied to the capacitor. 35 Find the reactance of a 2.00 mh inductor at a frequency of (a) 60.0 Hz; (b) 600 Hz; (c) 6000 Hz. 36 Find the rms current that results from connecting a 2.00 mh inductor directly across a 20 V rms source, at a frequency of (a) 60.0 Hz; (b) 600 Hz; (c) 6000 Hz. 37 For the circuit in Fig , the AC voltmeter V reads 50.0 V rms. Find the readings of the AC ammeter A and of voltmeters V 2 and V 3. Assume that the ammeter and voltmeters operate ideally; that is, the ammeter has zero impedance, and the voltmeters have infinite impedance. Fig For the circuit shown in Fig , the AC ammeter A reads 4.00 A rms. Find the readings of the four AC voltmeters. Assume that the ammeter has zero impedance and the voltmeters have infinite impedance. Fig *39 An RCL series circuit has the values R 8.00, C 4.00 F, and L 80.0 H. Find the average power dissipated when the rms voltage across the three circuit elements is 5.0 V, and the frequency is Hz. 40 A 0.0 resistor, a 0.0 F capacitor, and a 30.0 mh inductor are connected in series with a 20 V (rms) source with variable frequency. (a) At what frequency will the current be maximum? (b) Find the maximum rms current. (c) Suppose you now replace the 0.0 F capacitor by a 20.0 F capacitor. What is the new value of the rms current? 4 Find the resonant frequency for the circuit in Fig (a) Sketch the phasor diagram corresponding to the RCL series circuit in Fig , with a voltage amp litude of 00 V, a frequency of 60 Hz, a resistance of 00, a capacitance of 25 F, and an induct ance of 200 mh. Draw the phasors at t 0, and again at t s. 240 (b) What are the values of the instantaneous current at t 0 and at t s? 240 *43 (a) Find the current in the circuit shown in Fig if R 6.00, C F, and L 0, and the source supplies an emf of 40.0 V (rms) at a fre - quency of 3.00 khz. (b) What circuit element (other than a source) can we insert into this series circuit to increase the current as much as possible? (c) Find this maximum rms current.

39 626 CHAPTER 22 Electromagnetic Induction and AC Circuits 44 A transformer is used to operate a model train. The trans former, which is plugged into a standard electrical outlet, produces an output voltage of 20.0 V (rms), applied to the train s tracks. (a) What is the ratio of the number of turns in the primary to the number of turns in the secondary? (b) If the train uses an average power of 0.0 W, what are the rms currents in the secondary and in the primary? Assume an ideal transformer, with no energy loss. 45 Power lines at 20,000 V (rms) are connected to a utilitypole transformer. The output of the transformer s sec - ondary, which leads to a residence, is at 240 V (rms). Find the ratio of the number of turns in the primary to the number of turns in the secondary. *46 An ideal step-up transformer has 20 times as many turns in its secondary as in its primary. An rms current of 0 A passes through the primary when the secondary is connected to a circuit with an impedance of What is the rms voltage across the primary? 47 Neon lights are an example of light produced as a result of an electrical discharge in a tube filled with a gas, which emits a characteristic color (Fig ). The light emitted by discharge tubes is important in the study of atomic physics. (See Sections 23 2 and 29.) To pro - duce the electrical discharge, a high voltage must be applied to electrodes sealed in the ends of the tube. The resistance of a tube is and the rms current through it is 3.00 ma. The transformer used to produce the high voltage is plugged into a standard electrical outlet. Find the ratio of the number of turns in the secondary to the number of turns in the primary. 48 A 20 V (rms) power line is connected to the primary of a step-down transformer, with a turn ratio of 2:. The transformer s secondary is connected to a doorbell, which requires 0.0 W of electric power for its operation. What is the peak instantaneous current in the primary? Additional Problems *49 A square loop 0 cm on a side is placed in a uniform, time-varying magnetic field perpendicular to the plane of the loop, as indicated in Fig The field has an instantaneous magnitude of 0.50 T and is increasing at the rate of 0.20 T/ms. The loop has a resistance of (a) Find the loop s magnetic moment. (b) Find the torque on the loop. Fig Fig Neon lights and other gas discharge tubes require a high-voltage source.

40 Problems 627 **50 A plane loop of wire enclosing an area of 5.00 cm 2 and having a resistance of is placed inside a long solenoid. The plane of the loop makes an angle of 60 with the solenoid s axis. The solenoid has 00 turns per cm of length, and at some instant carries a current of 5.0 A. The current is increasing at the rate of A/s. Find the torque on the loop. *5 A solenoid of length.00 m and cross-sectional area m 2 has a self-inductance of.00 H. Find (a) the number of turns; (b) the radius of the copper wire used to form the solenoid (assuming one layer of wire); (c) the solenoid s resistance; (d) the time constant. *52 The edges of an inclined plane are made of rigid con - ducting rods separated by 20.0 cm and connected at the base by a resistor (Fig ). The plane makes an angle of 20.0 with the horizontal. There is a vertical downward magnetic field of magnitude T. A rigid rod slides down the plane at a constant speed of 60.0 cm/s. Neglect friction and the resistance of the rods. (a) What current flows in the resistor? (b) What is the mass of the rod? 53 A small canal diverts water from the Perlbach, a river in Bavaria. Water flowing through the canal drops through a small distance and turns a waterwheel, which powers an electric generator, providing electricity for the house shown in Fig (a) Calculate the maximum electrical power that can be generated if water moves through the canal at the rate of kg/s and drops through a vertical distance of 2.00 m. (Only about kw is used by the household, the remainder being sold to the local power company.) (b) The generator s coil consists of 00 turns enclosing an area of 400 cm 2. The magnetic field strength is T, and the generator operates at 60.0 Hz. Find the rms value of the emf generated. (c) Find the maximum rms current through the gen erator. Fig Fig Water flowing through the canal provides the energy necessary to operate a small generator and generate electricity. *54 Suppose that the resistance of the woofer and tweeter coils shown in Fig are each Suppose that a 0.0 F capacitor is inserted at point P, and a.00 mh inductor is inserted at point Q in the figure. (a) Calculate the rms current through the woofer and tweeter coils respectively when a 20.0 V (rms), 500 Hz voltage is input to the speaker. (b) Repeat the calculation for a frequency of 5000 Hz.

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