`Unit 21 Induced Voltages and InductanceFaraday's Law of Electromagnetic InductionReview some basic concepts. 1. A charge moving in a magnetic field experiences a force given by J oe ;@F=38) and the direction of the force is given by the right-hand rule. It doesn't matter if the field is stationary and the charge is moving, or the charge is stationary and the field is moving, all that is required is some kind of relative motion.2.Next, run the PhET Faraday Law demo and make a few observations. Make sure that the check box for show magnetic field lines is selected.Observations 1. There is a coil with a few loops that's part of a circuit with no apparent emf., voltage source. As the magnet is plunged in and out of the coil, current flows through the circuit as evidenced by the flickering of the light bulb.2.More specifically, as the north pole of the magnet approaches from the left, the magnetic field lines coming out of the north pole enter the coil and the brightness of the bulb increases in intensity until the first coil reaches the middle of the magnet. When this happens, the bulb momentarily dims. Then, as the magnet continues to move to the left, magnetic field lines going into the south pole start to enter the coil, increase in number, and the bulb brightens. As the magnet moves from left to right the effect is the same as above, with the roles of the poles reversed. 3. Voltage increase and decrease, as well as changes directions. This helps explain why the brightness of the light varies.Next, let's view the coil from the side as the magnet moves from left to right,up to a certain point, the number of flux lines passing through the coil increase. That is we have a changing magnetic field, one that is growing in intensity.For each of the above cases, there is so much flux passing through the coil. The amount of flux passing through a coil at a given instant, is denoted by the Greek letter F. If we assume, for the sake of discussion, the respective magnetic field intensities of A, B, and C are 1 Wb , 2 Wb , and 3 Wb ; and, the area of the loop is 0.1 m# , then F, for each case, can m# m# m# be determinded. For case A: For case B: For case C: F oe S1 F oe S2 F oe S3Wb m# &lt;S0.1 Wb m# &lt;S0.1 Wb m# &lt;S0.1m# &lt; oe !Þ&quot; W m# &lt; oe !Þ2 W m# &lt; oe !Þ3 WIn general, F oe EF This result will be used later. Recall that 1 Wb oe &quot; , &quot;!) lines of flux Also, its ok for the loop to have a non-circular shape, such as with a rectangle.In the above example, it is assumed that the plane of the loop is perpendicular to the lines of flux. This is not always the case, as illustrated below.If this is the case then F oe EF becomes F oe EF cos) Refering back to the PhET demo, Michael Faraday discovered each of the following, While the magnetic field is changing, an emf is induced and if the loop is a conductor, then current passes through the loop. (In the above case, the magnetic field can change in one of two ways: The coil is stationary and the magnet moves; or, the magnet is stationary and the coil moves.) An emf is induced regardless if the magnetic field intensity is increasing or decreasing. Increasing the number of turns in the coil increases the induced emf. Changing the area of the loop, while keeping both it and the magnet still induces an emf. (It's possible, for example, to change the area of the loop by squashing it.)---The faster the field increases the or decreases, the greater the induced emf.Faraday's observations can be summed up in the following statement: The induced emf is directly proportional to the rate at which the magnetic flux is changing. That is, Ioe  R ?F ?&gt;The negative sign indicates the polarity of the induced EMF. That is, the induced EMF sets up a current in the coil, which in turn sets up its own magnetic field which opposes or has the opposite polarity of the original magnetic field. (More about this when Lenz's Law is covered.) In the problems we will be working ?F will occur in one of two ways. a. F will be held constant and the area of the conducting loop will change, such as when squashing it. That is ?F oe F ?E-9=)Þ It follows that the induced EMF iwould be given by I oe  R F ?E -9=) ?&gt; b. The area of the loop will stay constant and F will change. That is, ?F oe E?F-9=)Þ It follows that the induced EMF would be given by I oe  R A ?F -9=)Þ ?&gt; Problems Faraday's Law 1. A circular loop of radius 20 cm is placed in an external magnetic field of strength 0.20 T so that the plane of the loop is perpendicular to the field. The loop is then pulled out of the field in 0.30 s. Find the induced emf.2.A square, single-turn coil 0.20 m on a side is placed with its plane perpendicular to a constant magnetic field. An emf of 18 mV is induced in the winding when the area of the coil decreases at a rate of 0.10 m/s# . What is the magnitude of the magnetic field?3.The flexible loop in the figure below has a radius of 12 cm and is in a magnetic field of strength 0.15 T. The loop is grasped at points A and B and is stretched until it closes. If it takes 0.20 s to close the loop, find the magnitude of the induced emf.4.The plane of a rectangular coil, 5.0 cm by 8.0 cm, is perpendicular to the direction of a magnetic field, B. If the coil has 75 turns and a total resistance of 8.0 H, at what rate must the magnitude of B change to induce a current of 0.10 A in the coil?5.A 500-turn circular loop coil 15 cm in diameter is initially aligned so that its axis is parallel to the Earth's magnetic field. In 2.77 ms the coil is flipped so that its axis is perpendicular to to the Earth's magnetic field. If an average voltage of 0.166 V is thereby induced in the coil, what is the value of the Earth's magnetic field?6.A wire loop of radius 0.30 m lies so that an external magnetic field of strength +0.3 T is perpendicular to the loop. The field changes to -0.20 T in 1.5 s. (The plus and minus signs refer to opposite directions in the loop.) Find the magnitude of the induced emf.7.A square, single-turn wire of coil 1 cm on a side is placed inside a solenoid that has a circular cross section radius of 3 cm, as shown below. The solenoid is 20 cm long and wound with 100 turns of wire. (a) If the current in the solenoid is 3 A, find the flux through the coil. (b) If the current in the solenoid is reduced to zero in 3 s, find the magnitude of the emf.Motional EMF 8. Consider the arrangement below. Assume that R = 6.0 H and L = 1.2 m, and that a uniform 2.5 T magnetic field is directed into the page. At what speed should the bar be moved to produce a current of 0.50 A in the resistor?9.Over a region where the vertical component of the earth's magnetic field is 40.0 µT directed downward, a 5 m length of wire is held in an east-west direction and moved horizontally to the north with a speed of 10.0 m/s. Calculate the potential difference between the ends of the wire, and determine which end is positive.10.A 12.0 m long steel beam is accidently dropped by a construction crane from a height of 9.0 m. The horizontal component of the Earths magnetic field over the region is 18.0 µT. What is the induced emf in the beam just before impact with the earth, assuming that its long dimension remains in a horizontal plane, oriented perpendicularly to the horizontal component of the earth's magnitude field?The homework is continued after the following material on the generator.The GeneratorThe above picture is a screen shot taken from the following web site: http://www.walter-fendt.de/ph11e/generator_e.htm This site has a java applet that simulates a generator. The basic idea behind a generator is that an emf is induced in a conductor whenever the conductor cuts across lines of flux in a magnetic field. When the conductor is a loop rotating in a magnetic field, a voltage is induced and current flows. The loop is called the armature. The ends of the loops are connected to slip rings which rotate as the armature is turned. Graphite or carbon brushes ride on each slip ring, connecting the armature to the external circuit. An electric generator converts mechanical energy into electric energy. We would like a formula that gives the induced emf at a specific time &gt; as the loop rotates. We would also like such a formula for current. To get these formulas, consider the picture on the next page. Charges in segmnet T U experience a magnetic force given by the formual J oe ;@F=38)Þ (The blue arrow is the direction of the magnetic field and the black one is the direction of the armature.)The magnetic force causes the negative charges in T U to accumlate at U. This leaves the portion of the segment at T positively charged, this sets up an electric field in the segment.The magnetic force continues to deposit electric charges at end U until the magnetic force is balanced by the electric field force. That is until ;I oe ;@F=38 ) Dividing both sides by ; , it follows that the electric field in T U is given by I oe @F=38 ) The potential difference between points T and U is given by Z oe IH H is the same as 6, the length of T U . So we can writeZ oe I6 Substituting from the electric field formula from above, we can write Z oe @F6=38 ). Using % for instead of Z for the induced voltage, we can write%oe @F6=38 ).There is also an induced voltage in the segment of conductor parallel to T U that is also given by%oe @F6=38 ).Therefore, the total amount of induced voltage at a given angle is given by%oe #@F6=38 ).Before proceeding, study the diagrams below and read their captions.In the previous diagram, on the rightß U traces out a circle, as does every other point in T UÞ As it rotates, U undergoes a tangential velocity given by @ oe &lt;=, where = is the angular speed measured in radians per second. And since, from the diagram on the left, &lt; oe +Î#ß the tangential velocity can be written by @ oe +  =Þ This allows use to write the # above induced voltage formula as%oe # +  =F6=38 ) oe +6F ==38 ) oe EF ==38), #where E is the area of the loop. Now if the loop has R turns, then % is given by%oe R FE==38)We need to express % a= a function of time, &gt;Þ We know that ) oe =&gt;Þ So we can write%oe R FE==38=&gt;The period and frequency of the rotating loop is given by X oe &quot;Î0 Þ So = in the above &lt;+. forrmula can be replaced by = oe ) oe #1X oe #&quot;1 oe #10 and % becomes &gt;0%oe R FE==38#10 &gt;Go back to the generator applet and note that the induced voltage is zero when ) oe !°àand, that it is at a maximum when ) oe *!°ÞSince sine takes on it= maximum value at 90°, we know that the maxium induced voltage % oe R FE==38(90) oe R FE=  &quot; oe R FE=. That is, the maximum induced voltage is given by%7+B oe R FE=ÞNext, we would like to come up with a formula for the induced current. We will do this using Ohm's Law.% 3oe V oeRFE==38#1 0&gt; VoeRFE= V =38#10 &gt; oe %7+B  =38#10 &gt; oe M7+B =38#10 &gt; VThat is, 3 oe M7+B =38#10 &gt;ÞThe homework problems are now continued. 11. The alternating voltage of a generator is represented by the equation Z oe #%!=38&amp;!!&gt;Þ Find the frequency of the voltage and the maximum voltage output.12.A semicircular conductor of radius 0.25 m is rotated about axis EG at a constant rate of 120 rpm. A uniform magnetic field in all of the lower half of the figure is directed out of the page and has a magnitude of 1.3 T. Determine the maximum of the emf induced in the conductor.13.A 100 turn square wire coil of area 0.040 m# rotates about a vertical axis at 1500 rpm. The horizontal component of the Earth's magnetic field at the location is 2 , 10% T. Calculate the maximum emf induced in the coil.14.A loop of area 0.10 m# is rotating at 60 rev/s with its axis of rotation perpendicular to a 0.20 T magnetic field. a. If there are 1000 turns on the loop, what is the maximum voltage induced in the loop?b.When the maximum induced voltage occurs, what is the orientation of the loop with respect to magnetic field.15.A generator consists of 40 turns of wire with an area of 0.12 m# and a total resistance of 30 H. The loop rotates in a magnetic field of 0.10 T at a constant frequency of 60 Hz. Find a. the maximum induced emf.b.the maximum induced current.c.express voltage as a function of &gt; and current as a function of &gt;Þ Then graph both of these functions on the same set of axes for just one period. Copy the graphs in the space below. (calculators in radian mode)d.the angular speed of the rotating loop.e.find the period of the rotating loop.Motors and Back EMF Motors are devices that convert electrical energy to mechanical energy. Essentially, a motor is a generator that runs in reverse. Instead of a current being generated by a rotating loop, a current is supplied to a loop by a source of emf, and the magnetic torque on the current carrying loop causes it to rotate. A motor can perform useful mechanical work when a shaft connected to its rotating coil is attached to some external device. As the coil in the motor rotates however, the changing magnetic flux through it induces its own emf, which acts to reduce the current in the coil. The phrase back emf is used for an emf that tends to reduce the applied current. The back emf increases in magnitude as the rotational speed of the coil increases. We can picture this state of affairs as the equivalent circuit pictured below.When the motor is first turned on, the back emf is zero and the current in the motor is at a Z maximum and is given by Ohm's Law M oe Z oe &quot;#!H oe &quot;# EÞ At the maximum speed, V &quot;! the back emf has its maximum value and the effective supplied voltage is now that of the external source minus the back emf, and the current is reduced to MoeZ Z,+-5 Voe&quot;#! Z (! Z &quot;! Hoe&amp;! Z &quot;! Hoe &amp;Þ! EÞProblems continued &quot;'Þ A motor has coils with a resistance of 30 H and operates from a voltage of 240 V. When the motor is operating at its maximum speed, the back emf is 145 V. Find the current in the coils a. when the motor is first turned on.b.when the motor has reached maximum speed.c.If the current in the motor is 6 A at some instant, what is the back emf at this time?Self-Inductance An inductor is a circuit element in which a self-induced emf accompanies a changing current. The circuit symbol for an inductor is shown below.Assume that the any current passing through the coil changes with time. Current sets up a magnetic field. A changing current sets up a changing magnetic field; and, according to Faraday's Law, this changing field induces an emf (back emf) which opposes the intial changing current. This phenomenon is called self-induction and the induced emf is called self-induced emf. Inductance is a property that causes opposition to a changing current. Once the current reaches a certain constant value, it quit changing and the magnetic field that it creates quits changing, so the back emf goes away. Inductance can happen if the current is increasing or decreasing, all that is required for inductance to occur is for the amplitude or the direction of a current to change.The back emf caused by inductance is given by the formula % oe  P ?M ?&gt; That is, the back emf is directly proportional to the rate at which the current is changing. The constant of proportionality, P, is called inductance and it is given by Poe.! ER # ß 6where .! oe %1 , &quot;!( Tm , R oe number of turns, 6 oe length of coil, and E is the crossA sectional area of the coil. Recall that 8 oe R ß the number of turns per meter. 6 Proglems Continued 17. A coil has an inductance of 3 mH, and the current through it changes from 0.20 A to 1.5 A in 0.20 s. Find the magnitude of the average induced emf in the coil during this period.18.A Slinky toy spring as a radius of 4 cm and an inductance of 275 µH when extended to a length of 1.50 m. What is the total number of turns in the spring?19.A solenoid of radius 2.5 cm has 400 turns and a length of 20 cm. Find a. its inductance.b.the rate at which current must change through it to produce and emf of 75 mV.RC &amp; RL Circuits Remember that a capacitor is a device that stores electric charge. When the switch, in the diagram below, is closed to the left, the capacitor starts to charge. The electrons from one of its plates, say the top plate, are attracted to the positive terminal of the battery. The electrons are then &quot;pumped&quot; through the battery and deposited on the bottomt plate. This process continues until the potential difference across the plates is the same as that of the battery. At first, there isn't much opposition to this process; but, as the process continues, the electrons that build up on the plate resist those that are being deposited . In effect, this is a resistance. So as a capacitor charges, the current through the circuit decreases, and the potential difference across the plates increases.The increasing voltage across the plates, is a function of time. The function is exponential. Z Ð&gt;Ñ oe %S&quot;  /&gt;ÎVG &lt; Here is the graph of this function.% is the emf of the battery, &gt; time, and VG is called the capacitive time constant.It is the amount of time that it takes the capacitor to become 63% charged. This constant is often denoted by 7 oe VGÞ 7 is the Greek letter tau. 20. A 6 V battery is connected in series with a resistor and an an inductor. The series circuit has a time constant of 600 µs, and the maximum current is 300 mA. What is the value of the inductance?21.A 25 mH inductor, an 8 H resistor, and a 6 V battery are connected in series. The switch is closed at &gt; oe ! =. Find the voltage drop across the resistor a. at &gt; oe !.b.after one time constant.Find the voltage drop across the inductor c. at &gt; oe !.d.after one time constant.22.The switch in a series VP circuit in which V oe ' H, P oe \$ H, and Z oe #% @96&gt;= is closed at &gt; oe !=Þ a. What is the maximum current in the circuit?b.What is the current when &gt; oe !Þ&amp; =?`