Unformatted text preview: Exam #3 practice problems solutions 1. The power source labeled I causes the current shown in the graph to flow in the coil. (Notice that the wire in the single loop leaves Physics 227 the negative terminal of the detector, goes behind the coil, curls around it, then returns.) Sketch a graph of your prediction of what the current in the single loop will be as a function of time as measured by the detector. In this case, define a positive current as a current that goes in the negative terminal and out the positive terminal of the detector. While the current in the coil is increasing, the magnetic field produced by that current will be increasing to the right. Therefore there will be a changing magnetic flux within the single loop. The current in the single loop will produce a magnetic field that points to the left to oppose the changing magnetic flux. Since the rate of change of the magnetic flux through the single loop is linear, the current in the single loop will be constant. Once the current in the coil becomes constant, the current in the single loop will go to zero since the magnetic flux through the single loop is no longer changing. When the current in the coil quickly drops there will be a spike in the current in the single loop as it tries to maintain the field lines that were pointing to the right. 2. In our study of AC circuits we learned that inductors have the effect of making the current in the circuit lag behind the voltage being provided by the power supply. Strangely, while you are working with a circuit that has only a single 3.00 light bulb connected to a single high frequency AC source you do see a small amount of this lag, even though there is no inductor in the circuit. The source produces a voltage of (t ) = max sin ( 2 ft ) with max = 3.00V and f = 1.00 MHz = 1.00 ! 106 Hz . Explain in detail the cause of this inductance. When the current initially increases from zero there will be an increasing magnetic flux through the surface defined by the circuit. This means a back EMF will be generated which attempts to generate magnetic field lines to cancel the new ones that are appearing. The direction of the EMF that does this is counterclockwise, which is opposite the direction of the current. This results in the current lagging behind the voltage across the power supply slightly This means even in a circuit without an inductor there is a slight bit of inductance in the circuit. I r B 3. Imagine that a space mission has been planned to send a probe to the planet Mercury. One of the goals of the mission is to measure the magnetic field of the planet as a function of altitude above the north magnetic pole of the planet. To do this, the space probe is being designed with a ring made of 1000 turns of conducting wire. This ring has a radius of 2.00m and a total resistance of 10.0 . As the probe descends this ring will be kept parallel to the surface of the planet. In order to design the ammeter that will measure the current in this ring you have been asked to estimate the current that could flow in the ring as the probe descends through the increasing magnetic field of the planet. You estimate that the probe will take 300 seconds to descend from orbit (where the magnetic field is zero) to the surface (where the magnetic field has an estimated magnitude of 10-5 T ). Make your estimate of the current that will flow in the ring as the probe descends to the surface. (You can make assumptions to simplify things as long as you justify them and they are reasonable.) Since the probe is descending on the north magnetic pole of the planet the magnetic field lines point directly away from the surface. I'll choose the direction of the area vector of the coil to also point in this direction. I'll assume the probe is descending with constant velocity and that the magnetic field changes at a constant rate with altitude. I= R =- 2 2 1 d B 1 d =- ( NBA cos ( 0 )) = - N r dB = - N r R dt R dt R dt R 2 B f - Bi t -t f i I =- (1000 ) ( 2.00m ) (10.0 ) 10 T - 0 -5 = -4.19 10 A 300s - 0 -5 The minus sign just indicated the direction of the current (clockwise as seen looking down on the probe from above.) 4. You have been asked to build a circuit that will be part of a radio telescope that will be used in the search for extraterrestrial intelligence. The circuit needs to be able to tune to all frequencies between 200kHz and 500kHz . So, you connect an RLC series (resistance 0.100 , inductance 4.00mH , capacitance variable) circuit to the output of the telescope. The capacitor is a circular parallel plate capacitor (radius 0.300m ) whose plate separation can be adjusted, which changes its capacitance. a. Explain why the variable plate separation allows the circuit to accomplish what it is designed to do. b. What values of the separation are needed to be able to tune the circuit to all of the desired frequencies? Changing the plate separation will cause the capacitance of the circuit to change, and hence the resonant frequency of the circuit. The resonant frequency is the frequency that will maximize the peak voltage across the resistor. Connecting a voltmeter across the resistor will cause the circuit to behave as a band pass filter allowing a signal with the same frequency as the resonant frequency of the circuit to come through strongest. = 1 LC 2 f = 1 L ( 0 A d ) = 1 L ( 0 r 2 d ) 2 2 d = 4 3 f 2 L 0 r 2 d = 4 3 ( 200 103 to 500 103 Hz ) ( 4.00 10-3 H )(8.85 10-12 C 2 Nm2 ) ( 0.300m ) d = 1.58cm to 9.88cm 5. A circular parallel plate capacitor (radius 30.0cm , plate separation 2.00cm ) is connected to a power supply so that the magnitude of the electric field between the plates is increasing according to the following function: E (t ) = ( 2.00 106 N C ) t a. Predict the displacement current between the plates. b. A tiny compass is placed between the plates of the capacitor and a magnetic field (other than the earth's) is detected. Explain the presence of this magnetic field and why it is so significant. Id = 0 I d = (8.85 10 dE d d = 0 ( E ( t ) r 2 ) = 0 r 2 dt dt dt
-12 2 2 (( 2.00 10 6 N C )t )
-6 C Nm ) ( 0.300m ) ( 2.00 10 N C ) = 5.00 10 A
2 6 The presence of that magnetic field is significant since it means that a magnetic field can be created by a changing electric field. Since an electric field can be created by a changing magnetic field (electromagnetic induction) then it means that each field can be created by the other, without the need for electric charges or currents. This allows for the possibility of electric and magnetic fields to exist as a self-propagating disturbance. Maxwell found that Faraday's law and Ampere's Law do allow for this and that the speed of the electromagnetic disturbance is the speed of light. Maxwell then suggested that this is what light is, and electromagnetic wave. So, the existence of the magnetic field between the plates of the capacitor actually is an indication that light is an electromagnetic wave. 6. The RC circuit shown has an extremely strong power supply that provides a constant EMF of 1200V . The circuit has a resistance of 100 and a large capacitance of 2.00 10-3 F . a. Once it's fully charged, if it expended all its energy to fire a 50.0g bullet how fast would the bullet be traveling? b. How long does it take to charge the capacitor so that it contains half its maximum amount of energy? Once the current has stopped flowing the potential difference across the capacitor will be equal to the potential difference across the power supply (since the current is zero the potential difference across the resistor will be zero). The electric potential energy stored in the capacitor is U E = this is fully utilized to fire the bullet the bullet will be traveling with a speed of 1 2 C ( V ) . If 2 K = UE 1 2 1 2 mv = C ( V ) 2 2 C ( V ) m
2 v= = ( 2.00 10 F ) (1200V )
-3 2 0.050kg = 240 m s As for how long it takes for the capacitor to contain half it's maximum stored energy (not charge, energy)... U E (t ) = 2 1 (Q (t )) = 21 C (1 - e-t RC ) 2C C ( ) 2 2 2 1 = C 2 (1 - e -t RC ) = U E ,max (1 - e -t RC ) 2 But, U E (t ) U E ,max = 1 2 so, 2 1 = (1 - e -t RC ) 2 1 e - t RC = 1 - 2 1 1 1 -3 t = - RC ln 1 - = - RC ln 1 - = - (100 ) ( 2.00 10 F ) ln 1 - 2 2 2 t = 0.246 s 7. A coil of wire (diameter 20cm) with 100 windings is suspended between the poles of a strong horseshoe magnet by an axle that can rotate. The average strength of the magnetic field between the poles of the magnet is 3.0T. The axle is connected to a waterwheel which causes the coil to spin in between the poles of the magnet at a rate of 10 revolutions per second. This causes a current to flow in the coil. The coil is connected through a "floating contact" to the power grid which acts as a single resistor of 10 . a. Explain in detail why this causes a current to flow in the coil. Will this current always have the same value? Will it always flow in the same direction? b. Determine the average current that flows in the coil as it rotates by 180 degrees. This is a generator! It will generate a sinusoidally varying EMF, or alternating current. This occurs because the magnetic flux through the coil is changing with time according to B = BA cos = BA cos (t ) where B is the average strength of the magnetic field and is the angular frequency of the coil's rotation. The average EMF can be determined using Faraday's law, and the average current with some help from Kirchhoff's loop rule and Ohm's law. When the coil rotates by 180 degrees the magnetic flux through it changes sign. av 1 B 1 B , f - B , f 1 NBA cos0 - ( -NBA cos180 ) =- =- =- R R t R t R t 2 2 2NB ( d 2 ) 2 (100 )(3.0T ) ( 0.2m 2 ) 2NBA Iav = - =- =- = 37.7 A Rt Rt (10 )( 0.05s )
Iav = 8. The dielectric strength of an insulator is the maximum external electric field the material can withstand before ionizing and becoming a conductor. You have been contracted to determine the dielectric strength of an unknown crystal. To do this you assemble an RC series circuit (resistance 10.0 , capacitance 8.0F ) connected to a strong DC power source (voltage 710V ). The distance between the capacitor plates is 0.1mm , so you fashion a sample of the crystal 0.1mm thick and slip it between the plates. You then turn on the power supply and find that after every 3.41 10-4 s a spark jumps across the capacitor. a. Sketch a graph of the current in the circuit as a function of time and explain why the capacitor is repeatedly sparking. b. Determine the dielectric strength of the crystal. The capacitor is sparking because as the capacitor charges up the potential difference across it increases. This causes the electric field between the plates to increase as well. Eventually this electric field exceeds the dielectric strength of the crystal and a spark occurs. This electric field is related to the potential difference across the capacitor by I 70 60 50 40 30 20 10 E =- Vc d Using the loop rule 0.0000 0.0002 0.0004 0.0006 0.0008 0.0010 t VC = -VR - V = - ( -I (t ) R ) - = e -t RC R - = (e -t RC - 1) R 710V E = (1 - e -t RC ) = -4 (1 - e d 10 m - 3.4110-4 s (10 ) 810-6 F ( ) ) = 7.0 106 V m 9. It is the year 2042. You are an engineer designing a powerful electromagnet. Due to advances in superconductivity it is now possible to construct superconducting wires that operate at room temperature. You begin thinking that if you wrapped this superconducting wire 2000 times around a cylindrical iron core (radius 3.0cm, length 30cm) and connected it to a power supply (voltage 100V, resistance 0.1 ) you could generate an extremely strong magnetic field. a. How strong would this magnetic field be? Remember, the presence of the iron makes the magnetic field roughly 1000 stronger within the solenoid. Incidentally, this also makes the inductance of the solenoid roughly 1000 times larger as well. b. How long would it take after the power supply turns on for the magnetic field to reach 90% of its maximum strength? c. How much power would be dissipated by the internal resistance of the power supply once the electromagnet has been on for a long time? This is what will make this design impractical... The magnetic field of this solenoid is B = 1000 0 (2000 )(100V ) = 8378T N N I = 1000 0 = 1000 ( 4 10 -7 T m A ) L LR ( 0.3m )( 0.1 ) The current in an inductor as a function of time is -t I (t ) = Im 1 - e L R To use this equation to find how long it will take the current to reach 90% of its maximum value I need to determine the inductance of the solenoid. -7 1000 0N 2 r 2 1000 ( 4 10 T m A ) (2000 ) ( 0.03m ) L= = = 47.4H l 0.3m 2 2 I (t ) Im e = 0.9 = 1 - e
-t LR -t LR = 0.1 L 48.4H t = - ln ( 0.1) = - ln ( 0.1 ) = 1091s R 0.1
This electromagnet will take more than 18 minutes to reach 90% of its maximum strength. The power dissipated in the resistance of the power supply once the current nears its maximum value will be 2 (100 ) P = Im R = = = 100,000W R 0.1
2 2 This is a huge amount of energy per second. Unless the power supply has a very good cooling system it will heat up and melt itself. 10. You are a graduate student working on SETI (the Search for Extraterrestrial Intelligence). While you are analyzing the signals coming in from the radio telescope you find two signals that look like they are not being produced by any natural phenomenon. One of the signals is at a frequency of 1kHz while the other is at 100kHz. Quickly you design an LRC series circuit and plan on using it to filter out one signal at a time so you can analyze them separately. a. Explain how you will use this circuit to analyze one of these signals at a time. Explain why this circuit can be used for this purpose. b. Determine a possible choice of values for L, R, and C that will allow the circuit to function as desired. This is challenging. Most of the points for this question will come from part a. Connect the input (the combined signals) to the LRC series circuit. Then, place your detector in parallel across the capacitor. The low frequency signal will produce a large potential difference across the capacitor (that gives the charge on the capacitor time to build up), so your detector will primarily see only the 1kHz signal there. The high frequency signal will produce a large potential difference across the inductor (the rapidly varying signal causes a large back-EMF), so your detector will primarily see only the 100kHz signal there. To determine a set of values for L, R, and C you need to choose value where the following will be satisfied: VC = Vsignal VL = Vsignal 1 C 1 R 2 + L - C L 1 R 2 + L - C 2 is high for the 1kHz signal and low for the 100kHz signal 2 is low for the 1kHz signal and high for the 100kHz signal This takes some playing around with, but if you choose typical values for these circuit elements ( L = 3 10-3 H , R = 10 , C = 6 10-7 F ) you get the following results: VC 1.076 Vsignal VL 0.076 Vsignal = 2 103 Hz = 2 105 Hz 0.001 1.001 11. James C. Maxwell found a problem with Ampere's Law. Explain this problem, what Maxwell did to correct it, and how this led to a new understanding about the nature of light. Ampere's Law prior to Maxwell's correction is ! ! " !#$ = "% !"
Maxwell realized that there was a way to violate Ampere's law by looking at a charging capacitor circuit. If he threaded the surface used to evaluate the right hand side of Ampere's law through the capacitor plates the left hand side would not be zero, but the right hand side would be. Maxwell suggested that the changing electric flux through the surface would exactly resolve this contradiction and added a new term to the equation to include this. The improved Ampere- Maxwell law is ! " " !#$ = & % + # %
" ! ! $ " # ! !' ' !#() #& " ( This new term together allowed Maxwell to realize that not only could changing magnetic fields create electric fields (Faraday's law) but also that changing electric fields could create magnetic fields. Using this idea Maxwell was able to find solutions to the equations of electricity and magnetism that represented traveling waves that traveled at what was experimentally known at the time to be the speed of light. Maxwell suggested that this was not a coincidence and that light was in fact an electromagnetic wave. 12. You are a newly hired engineer for an alternative power company that specializes in wind power. You are asked to evaluate a design for an electric generator that will use wind to produce an electric current. Here's how it works: A coil of wire (2000 windings, radius 0.5m , length 2.0m ) is suspended between the poles of two bar magnets. The magnets are connected through a gearing mechanism to the spinning wind blades so that each magnet rotates 180 degrees every tenth of a second. Effectively, this means the poles of each magnet swap places every 0.1s . When the magnets are aligned so that the north pole of one magnet is facing the south pole of the other (as in the diagram), the average magnetic field within the coil is 3.5T . a. Sketch a graph of what the current in the coil will be as a function of time. If the current is flowing upwards on the side of the coil facing you indicate that with a positive current on the graph. Indicate a current flowing the other way as negative. The coil is connected to the power grid at the top, so it is part of a complete circuit. b. Predict the magnitude of the EMF (voltage-like effect) this design will produce during the time it takes the magnets to rotate 180 degrees. S N S N The magnetic field lines within the coil point to the right, I 1.0 but as the poles of the two magnets rotate away the number of field lines pointing to the right will decrease. The induced magnetic field will try to restore these field 0.5 lines as they disappear. Right hand rule #2 (grab a wire on the front face of coil with fingers inside the coil pointing to the right) predicts that the induced current will be flowing t 0.1 0.2 0.3 0.4 0.5 0.6 downward on the front face of the coil. This is a negative current according to the convention described in the 0.5 problem. As the magnets complete their 180 degree rotation the magnetic field lines will be pointing to the left since the north and south poles have changed places. Still 1.0 the induced current will be down in order to produce an induced field that cancels the newly appearing leftward field lines. Once the poles begin rotating away again these leftward field lines start to disappear. To try to restore these lines as they are disappearing a positive current is needed (use RHR #2 with the field within the coil pointing to the left.). The induced current continues to alternate directions every 0.1s . To predict the magnitude of the EMF I'll use Faraday's law. At t = 0 s the magnetic field lines within the coil point to the right. I'll define that to mean a positive magnetic flux through the coil. =N B A cos f - Bi Ai cos i B (1) - B ( -1) 2BN r 2 B =N f f = N r 2 = t t f - ti t f - ti t f - ti 2 ( 3.5T )(2000 ) ( 0.5) 0.1s - 0 s
2 = = 1.1 105V 13. Research is underway to build the world's first 100 tesla electromagnet. When switched on it will behave essentially like a horseshoe magnet and produce an approximately uniform magnetic field of magnitude 100 T between the poles of the magnet. Once the magnet is turned off, the field collapses in 3.0 s. Your goal is to determine if the 100 T magnetic field was actually achieved. To do this you suspend a small circular wire coil (5 turns, 3.0 cm radius) between the poles of the magnet. The coil is connected to an ammeter. The entire circuit (coil plus wires plus ammeter) has a total resistance of 0.1 . During the time the magnet is shutting down, the ammeter measures a current of 4.6 A flowing in the circuit. a. Draw a large picture showing the magnet, its poles, its magnetic field, the coil, and direction the current is flowing in the coil while the magnet is shutting down. b. Detemine if the magnet achieved its goal of producing a 100 T magnetic field. Make sure you show all your work. Since the magnetic field of the electromagnet is weakening, the induced magnetic field will attempt to compensate for that weakening. The current in the coil is being caused by electromagnetic induction, so I'll use Faraday's law to analyze what is happening. I'll choose the orientation vector of the surface defined by the loop to point to the left. S r Bind I ind N r B 1 dB 1 d =- ( NBA) R R dt R dt N r 2 dB N r 2 0 - B I =- =- R dt R t 2 N r B I= Rt IRt ( 4.6 A )( 0.1 )( 3.0 s ) B= = = 97.6 T 2 N r 2 (5) ( 0.03 m ) I= =-
Unfortunately, the new electromagnet did not reach its target field strength of 100 T. 14. Capacitors are sometimes used to temporarily supply power to a circuit when the primary power supply briefly fails. Imagine a circuit that has a single light bulb (resistance 1440 , making it a 10 W light bulb) and a single capacitor in parallel connected to a 120 V DC power supply. This "backup capacitor" has a large capacitance of 10-3 F . a. The power supply has been turned on for a long time. What is the value of the potential difference across the capacitor? Explain how you know. b. The power supply abruptly fails. Treat this as a break in the circuit where the power supply is. What is the power output of the light bulb one-tenth of a second after the power supply fails? Since this is a parallel circuit, the potential difference across the capacitor will equal minus the potential difference across the power supply, or -120 V. This comes from Kirchhoff's loop rule. The sum of the potential differences around a circuit loop must be zero. Once the power supply fails, the circuit now becomes a discharging RC circuit. The current in such a circuit decays exponentially with time. t VC - RC I (t ) = e R One-tenth of a second after the power supply fails the power output of the light bulb will be - t 2t VC - RC VC 2 - RC ( -120 V ) (1440 )(10-3 F) 2 P = I (t ) R = e R= e = e = 8.7 W R 1440 R 2 2 2( 0.1 s ) 15. A hydroelectric power plant produces a potential difference of (t ) = max sin (t ) = (5000 V ) sin (( 2 60 Hz ) t ) a. Describe the transformer that is needed to convert this to a 10 million volt alternating current so that it can be efficiently transmitted to nearby cities. b. Explain why our power grid is based on AC current rather than DC current. Describe the features of AC that make it desirable. The transformer that is needed to do this is a "step-up" transformer. Using the transformer equation Vp Np = Vs Ns N Vs 107 V s = = = 2000 N p Vp 5000 V In other words, this transformer will work as required as long as the number of windings on the secondary, or output side of the transformer is 2000 times greater than the number of windings on the primary, or input side of the transformer. This ability of a transformer to increase the potential difference produced by an AC power supply dramatically allows for electric potential energy to be transmitted long distances with very low currents. This minimizes power losses in transmission lines which would be given by P = I 2 R . Transformers rely on electromagnetic induction to function, which requires a changing magnetic field. Only currents that are changing can produce a changing magnetic field, which means that DC cannot accomplish this. 16. An RLC series circuit connected to an input that contains a wide range of signal frequencies can be used to filter out some of those frequencies. To do this, you connect your detector (a voltmeter, oscilloscope, or something more sophisticated) in parallel with one of the three circuit elements. a. How you would use this circuit as a low-pass filter? Explain why this works. A low-pass filter allows for the detection of low frequencies, while reducing the strength of high frequencies. b. How you would use this circuit as a high-pass filter? Explain why this works. A high-pass filter allows for the detection of high frequencies, while reducing the strength of low frequencies. Connecting the detector across the capacitor creates a low-pass filter. Low frequencies give the charge on the capacitor sufficient time to become large, resulting in a large potential difference. High frequency signals do not result in a large response by the capacitor. On the other hand, connecting the detector across the inductor creates a high-pass filter. High frequencies (rapidly changing currents) cause a large back-EMF across the inductor, resulting in a large potential difference. Low frequencies (slowly varying currents) do not result in a large response across the inductor. 17. Answer these questions about the interplay between electric and magnetic fields. a. We learned that the magnetic field can be used to create an electric field. Give an example of a situation where this happens. Draw a large, clear, and labeled diagram of the situation. Make sure you include the magnetic field and the resulting electric field in your diagram. b. We also learned that the electric field can be used to create a magnetic field. Give an example of a situation where this happens. Draw a large, clear, and labeled diagram of the situation. Make sure you include the electric field and the resulting magnetic field in your diagram. c. Explain the significance of this ability of the electric and magnetic fields to create each other. Look back to problem 1 for an example of a changing magnetic field. When the current in the coil is changing, so is the magnetic field that it produces. This changing magnetic field produces an electric field which drives the current in the wire loop connected to the ammeter. The electric field lines form are circular and point in the direction the current flows in that loop. To produce a changing electric field, create a circuit with a capacitor connected to a DC power supply. As the capacitor charges the electric field between the plates increases with time. According to the Ampere-Maxwell law, this increasing r electric flux (pointing to the right) produces a magnetic field similar to an actual B current flowing to the right, even though there are no charges passing between the plates. Using the right hand rule determines the direction of the magnetic - + field produced by the changing electric field, shown in the diagram. I I + Since each field, when changing with time, has the ability to produce the other, r - E it allows for the two fields to mutually produce one another in a repeating pattern as they travel. Maxwell realized these were traveling waves made of the electric and magnetic field. He also determined that these waves will travel at a speed equal to the speed of light. This led him to the suggestion that light in fact was a traveling electromagnetic wave. 18. Answer two of the following three conceptual questions (you choose). Each of your answers should be no longer than a few sentences. Use diagrams if helpful. a. What is a phasor and why are phasors useful? b. Give an example of a situation where the original form of Ampere's law : " B " ds = I !
0 ! ! leads to a contradiction. Explain your reasoning. c. Explain why it is not possible to have a wave made only of vibrating electric field. A phasor is a vector-like entity used to represent an oscillating quantity. Its tail is located at the origin and it rotates around in the plane. Its projection onto the horizontal axis is the physical quantity that the phasor represents. Phasors allow differential equations involving oscillating quantities to be solved geometrically. In a circuit with a charging capacitor, this version of Ampere's law will fail when the integration path encircles the wire but the surface bounded by that path passes between the capacitor plates. The right hand side of Ampere's law will be zero but the left hand side will not. An electromagnetic wave must consist of both an oscillating electric field and an oscillating magnetic field. The fields create each other as the wave propagates, but since they cannot create themselves, a wave of just electric or just magnetic field cannot exist. 19. You are a safety inspector hired by the Princeton Plasma Physics Laboratory to determine if their new experimental nuclear fusion reactor is generating too strong a magnetic field in nearby offices. You go to the office nearest the experiment which turns out to be directly above the main fusion reactor. The scientists who designed the reactor tell you that the device will create a magnetic field in the office that points directly upward and will have approximately the same strength everywhere in the room. You take out a circular coil of wire (radius 0.500m , number of turns 50, resistance 0.200 ) and orient it so the plane of the coil is parallel to the floor. You connect the coil to an ammeter as shown in the diagram. When the nuclear fusion reactor is turned on the ammeter deflects to the right1 and reads a value of 30.0mA for 2.00s a. Were the scientists correct about the direction of the magnetic field? Explain your reasoning. 1 The ammeter will deflect to the right when the current is flowing into the right terminal and will deflect to the left when the current is b. Estimate the magnitude of the magnetic field in the office. No, the scientists were not correct. If the magnetic field were pointing upwards and increasing in strength then the magnetic field produced by the induced current would point down to resist the increasing field. The current that would produce this induced field is one that flows through the ammeter from left to right, which would have caused the needle to deflect left. But, since it actually deflected right the scientists must have been wrong about the direction of the magnetic field produced by the fusion reactor. It must be pointing downward. To estimate the magnitude of the magnetic field in the office: = Bf - 0 dB d B IR = ( NB r 2 ) = N r 2 = N r 2 dt dt t t -3 IRt ( 30.0 10 A ) ( 0.200 )( 2.00 s ) Bf = = = 3.06 10-4 T 2 N r 2 (50 ) ( 0.500m ) This is only about 10 times stronger than the magnetic field of the earth, so the office will not need to be moved. 20. After taking your last physics exam you decide to go home I (t ) = ( 2.00 A) sin (120 rad s ) t and relax out on your deck. Looking up you see that single high voltage power line that has always stretched out across the sky above your neighborhood. Then, you have an idea! Remembering that there is a rapidly oscillating current in the power line you decide to try to get some free power from the power company2. You head out to Radio Shack and buy a spool of wire. You then wind it into a rectangular coil of 1000 turns and orient it beneath the high voltage line as shown in the diagram. (The rectangular coil and the power line are both in the plane of the page. 0.200m Also, the diagram is not drawn to scale.) Estimate the EMF you can expect to be generated in your coil. (You can make assumptions to simplify things as long as you justify them and they are reasonable) ( ) 30.0m 1.00m The current in the power line will create a changing magnetic field in the region of the rectangular coil, which means electromagnetic induction will potentially induce an EMF in it. Since the height of the coil is small compared to its distance from the power line it would be reasonable to take the magnetic field within the loop to be uniform. It is varying, just not very much. I'll just use the value of the magnetic field in the middle as the average value. The area vector of the coil will point out of the page (in the same direction as the magnetic field at its location.) =- N 0 lw d d B d d I (t ) = - ( NBA cos ( 0 ) ) = - N 0 lw = - I (t ) dt dt dt 2 r 2 r dt =- =- N 0 lw d N lwI ( I max sin (t )) = - 02 rmax cos (t ) 2 r dt (1000 ) ( 4 10-7 T m A) (1.00m )( 0.200m )( 2.00 A) (120 rad s -1 ) = - (1.03 10-3 V ) cos (120 rad s -1 ) t ( 2 ( 30.0m ) cos (t ) ) 2 By the way, this procedure can work but is actually illegal and you can be taken to court for stealing power in this way. 21. You just bought an expensive home entertainment system for your living room but didn't have enough money for a surge suppressor. So, you decide to make one out of a long spool of wire you have laying around. You do a little research and find that the most powerful lightning strikes can cause a spike in the potential difference provided by a wall outlet3 on the order of one hundred thousand volts for approximately one thousandth of a second. You decide that if you add enough inductance to the circuit containing the wall outlet and your equipment you should be able to protect it from lightning strikes. You do some more research and learn that your home entertainment system has a total resistance of 50.0 and cannot safely deal with potential differences more than 10% above standard house AC. You reason that if you can build an inductor that will prevent the potential difference across your equipment from increasing by more than 170V 10% = 17V then your equipment will be safe. You decide to model this situation as follows: The circuit is a power source (the lightning strike), a resistor (your entertainment system), and an inductor (the one you are going to build), all in series. The lightning strike is an EMF of = 100, 000V lasting from t = 0 s to t = 1 10-3 s If the potential difference across the resistor never exceeds 17V then your equipment will be adequately protected. What must the minimum inductance of your homemade inductor be if it is going to successfully protect your home entertainment system? Is the result reasonable? Okay, this really is just an RL circuit connected to a DC source, except that the source turns off after 1 10-3 s . So, if by 1 10-3 s the voltage across the resistor hasn't exceeded 17V then it never will. The voltage across the resistor is -t -t L R L R VR = IR = 1 - e R = 1 - e R e -t L R = 1- VR -1 1 VR L = - ln 1 - Rt 1 17V L = - ln 1 - 5 -3 ( 50 ) (1 10 s ) 10 V L = 294 H -1 This is a ridiculously large inductance. The inductor would have to be very large, too large to fit in a surge suppressor. Real surge suppressors use an entirely different approach to preventing large spikes in potential difference. 22. You are a photographer that has been hired to take pictures of some hieroglyphics deep within an ancient Egyptian tomb. You've decided to buy a portable generator that can be brought with you into the tomb to power your lamp and camera. The generator burns white gas (so the tomb doesn't become polluted) and produces an alternating EMF of ( t ) = (170V ) sin 120 s -1 t . Your lamp has a resistance of 144 . After a quick calculation you find that 100W of power will be delivered to the lamp. This is good news since you know that if the power drops below 95.0W it will be too dark to take quality pictures. You turn on the generator and the lamp and are frustrated to find out that the power output of the lamp is only 90.4W . Flipping through the instruction manual for the generator you notice a warning: This generator has a very large inductance. (( )) 3 Recall that normally the potential difference produced by standard wall outlets is Vrms = 120V , Vmax = 170V . a. Draw a circuit diagram for this situation. b. On a single graph, sketch the EMF and the current in the circuit as a function of time. c. What is the inductance of the generator? d. Since you can't modify the generator or the lamp, what can you do to make the lamp bright enough? In your stash of spare parts you have various other resistors, capacitors, and inductors. Be specific! If you decide to add anything to the circuit, give the value of its resistance/capacitance/inductance. E t
It H L H L Since the circuit has a large inductance, the current in the circuit lags behind the EMF of the power supply. Pav = I 2 rms 2 2 2 rms max R max R R = 2 R = = 2 2Z 2 2 ( L ) + R 2 Z ( ) 2 Pav 1 = 2 max R ( L )2 + R 2 1 L= 2 max R 2 Pav 1 -R = 120
2 (170V ) (144 ) - 144 2 ( ) 2 ( 90.4W )
2 L = 0.127 H
If you add some capacitance to the circuit you can cancel out the inductive effect that is causing the lamp to be dimmer. This will happen at resonance, which is when L - 1 =0 C 1 1 C = 2 = 2 L (120 ) ( 0.127 H ) C = 5.55 10-5 F So add a capacitor with this capacitance in series with the generator and lamp and the lamp will provide 100W of illumination. ...
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This note was uploaded on 03/18/2012 for the course 12312 123123 taught by Professor 123 during the Spring '12 term at Antioch University New England.
- Spring '12