AAST/AEDT. Electromagnetic Induction. If the permanent magnet is at rest, then - there is no current in a coil.

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Esmond Small
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1 1 AP PHYSICS C AAST/AEDT Electromagnetic Induction Let us run several experiments. 1. A coil with wire is connected with the Galvanometer. If the permanent magnet is at rest, then - there is no current in a coil. If the magnet is in motion, then the current flows. Its magnitude depends on the magnet s velocity. Its direction depends on the direction of the magnets motion and polarity. 2. A coil with a current is moving relatively to another coil, connected with an ammeter.

2 3. Analyzing all data we come to a conclusion, that when a permanent magnetic field is penetrating through the closed circuit, there is no current in the circuit.

2 2 3. Analyzing all data we come to a conclusion, that when a permanent magnetic field is penetrating through the closed circuit, there is no current in the circuit. If the flux is variable, then the current appears. The process of the current generation by a variable magnetic field is defined as the ELECTROMAGNETIC INDUCTION. MAGNETIC FLUX As we can observe the effect of electromagnetic induction takes place when the number of the magnetic field lines through the closed circuit varies. To describe that number physicists use a new quantity - magnetic flux. Let us assume that the loop with the area of A is located in a magnetic field with induction of B and that the angle between the direction of the field and the normal to the In case of a uniform filed and flat surface, the magnetic flux is defined as the dot product of the magnetic field induction times the area of the loop B. A As you may recall dot product of the vectors is a scalar quantity, that is equal to the product of the vectors times the cosine of the angle between them. In our case the second vector is a normal to the area. Thus for uniform field.

3 3 B A cos Magnetic flux through the loop is at maximum when the normal is parallel to the field (pic. 1). Magnetic flux through the loop is 0 if the loop s normal is perpendicular to the direction of the magnetic field.(pic.2) In a case of non-uniform field and/or curved surface similar to one shown on a diagram, magnetic flux has to be calculated as. The unit of the magnetic flux is Weber = T*m 2. = B da FARADAY S LAW Electromagnetic induction can be described with the electromotive force created inside the closed circuit in the time of the effect.. As we mentioned before, the magnitude of the current is proportional to the rate at what the magnetic field does change. That rate can be described as a derivative of the magnetic flux over time. d dt Where - is the magnetic flux. Let us run a dimensional analyses of the quantity. That quantity is measured in Volts and that is why it can be used for the EMF measurements. Thus, we have d dt Sign minus is introduced because of the Len z law that we will discuss below. LENZ S LAW The goal of the Lenz s research was to investigate the direction of EMF and thus the direction of the induced current. To obtain that goal let us run an imaginary experiment.

4 4 We move a bar magnet toward the coil. The magnetic flux through the coil increases. The variable flux created the EMF and the induced current starts to flow through the coil. That current creates its own magnetic field. We do know that the magnetic field of the wire coil with the current has the same shape as the field of the bar magnet. That means, that on the edge B of the coil we have one pole and on the edge C another one. Let us initially assume that S pole is located at the edge B of the wire coil. That means that if we initially push the magnet somewhere far away from the coil, it will infinitely continue to move, because of the attraction between the unlike poles and that the current will be created without any energy input. That is impossible, because it contradicts to the energy conservation law. So, we have to have pole N on the edge B. Than, if you move the magnet toward the coil, you have to do some work to overcome the repulsion between the like poles. That work is transforms into the current s energy. If we draw the magnetic field lines for the magnetic field created by the induced current, we can observe that that field oppose the increase of the flux through the coil.

5 5 If we move the magnet out of the coil, then the flux through the wire coil decreases. We can repeat our reasoning and prove that the pole S has to appear on the edge B. That time the magnetic field created by the induced current opposes the decrease of the flux through the coil. In general Lenz s law states: The current is induced in a direction such that the magnetic field produced by the current oppose any change in flux that induced the current. We can observe the effects of the Lenz s law with the real experiment. If we move the magnet toward the aluminum ring, it moves away. If we move the magnet out of the ring, it follows the magnet. Explanation: When the magnet moves toward the ring the flux through the ring increases and the induced current create a magnetic field that tries to prevent that increase. So it moves the ring out of the magnet. When the magnet moves away from the ring the flux through the ring decreases and the induced current create a magnetic field that attempts to prevent that decrease. Thus, the ring follows the magnet.

6 THE NATURE OF ELECTROMAGNETIC INDUCTION (INDUCED ELECTRIC FIELD) The goal of this section is to explain the mechanism of the phenomenon of electromagnetic induction.

6 6 THE NATURE OF ELECTROMAGNETIC INDUCTION (INDUCED ELECTRIC FIELD) The goal of this section is to explain the mechanism of the phenomenon of electromagnetic induction. Let us assume we have conductor without a current. Free electrons inside the conductor are moving randomly as it shown on a diagram. Let us assume that a variable magnetic field directed into the paper is turned on. It does not matter what is the direction of B. We choose one to avoid ambiguity. Out next step, is to apply the left hand rule and determine the direction of the force exerted on each electron. As it is obvious from a diagram, the distribution of forces exerted on electrons is random and thus, those forces can not force the electrons to participate in a directed motion, i.e. create current. Thus, we have a contradiction. We do know that variable magnetic field produces current, but we also know that this can not be done by magnetic forces, because their direction depends on the direction of the velocities and thus is random. The only other known force that can be exerted on a charged particle is an electric force F=qE, that is independent on the velocity of the particle. Thus the existence of the induced current proves that such force does exist and that proves that electric field did appear. The only source of creation of electric field is a variable magnetic field. Thus, we come to conclusion: Variable magnetic field produces in a surrounding space electric field and that electric field serves as a source of the induced current. The only important difference between the induced field and the field created by charges is that electric filed lines of the induced field are continuous, because of the absence of the source charges. EMF in a straight wire Let us assume that a straight wire with a length L is moving in a magnetic field with the velocity V. We also assume,. that the direction of the magnetic field is normal and up to that sheet of paper. Magnetic field does exert a force on every electron in a moving wire. According to the left-hand rule that force is directed toward point B. Thus, all electrons would move toward the edge A and it becomes negative. At the same time the lack of electrons on edge B will create a positive charge there.

7 7 Let us assume that at time t, the wire will travel distance d=vd t. It will cross all magnetic lines in a rectangle ABA*B*. That means - the magnetic flux change is and the magnitude of EMF is equal to d BLd =BLvdt = d dt BLvdt BLv Another question to be discussed is the energy transformations throughout the induction. Let us assume that instead of single conductor we move a rectangular loop with a resistance of R out of the field and that one side of the loop is already out as it shown on a diagram. We also assume that magnetic field induction is directed out of the paper. Each electron in a rod MN will be moving to the right with the loop and will experience the Lorenz force, directed from M to N. Thus, the electrons in the loop will be traveling clockwise and that means that the current in the loop would be counterclockwise. As we aware the current in magnetic filed experiences Ampere force. We apply left hand rule and determine that the force exerted on rod MN is directed to the left. Forces on the top and the bottom part of the loop are exerted up and down respectively and cancel each other. Thus to move the loop uniformly away with a speed of V we have to apply a force that is equal to Ampere force exerted on MN part of the loop. That force is equal to F=IBl, and the power to be developed by the hand to move the loop should be P=FV= IBLV ( * ). The current I can be estimated as I= E/R (**), where E is the induced EMF, which is equal to E=BVL(***) If we substitute (***) into (**) and the result into (*), we obtain the final equation for the power necessary to move the loop out of the field. P B2 l 2 V 2 R dt

8 8 Self-Inductance Let us run an experiment. We design a circuit and turn the power on. We do observe that bulb 2 starts to glow immediately and bulb 1 experiences a certain delay. The explanation is simple. When we turn the current on, it starts to rise and it creates the rising magnetic flux through the coil. That variable flux will create an induced current. The induced current according to the Lenz s law resists to that rise and so we observe the delay in lighting of the bulb 1. We can observe the similar effect when we turn the current off. The schematic of the experiment is in the diagram below. *When we turn the current off, decreasing current creates decreasing magnetic flux. That flux pierces the coil and creates the induced current. That current according to the Lenz s law resists the initial decrease of the main current. So, it has the same direction in the coil but opposite direction through the ammeter. That effect we can observe. That phenomenon, when the current itself creates an EMF that resists to the current change is called self-inductance. There is no difference in principal between the effects of electromagnetic induction and self-inductance. It is more question of terminology. In case of electromagnetic induction, the EMF is created as the result of the external magnetic field. change In the effect of self-induction the EMF is created by the change of the internal magnetic field, i.e. created by its own current. During the effect of self-induction the magnetic flux through the conductor is proportional to the current through it. = L I L is the coefficient that depends on the geometry of the conductor (for example in the case of the coil it depends on the number of, cross-section area and on the surrounding medium). That coefficient is called an inductance. The unit of inductance is Henry (H) Henry = Weber/Amp As we know, according to Faraday s law, the EMF = / t. If instead of we substitute LI, we obtain the final formula for the EMF of the self-inductance t L I t

9 9 INDUCTANCE of SOLENOID As it was mentioned above inductance depends on a geometry and properties of a conductor. It can be estimated for different conductors. To illustrate the concept let us estimate the inductance of a solenoid. As we derived above, magnetic field of a solenoid can be estimated as B o ni where n- is the turn s density, i.e. number of turns per unit of length. Thus, a magnetic flux through each turn should be equal to turn o nia where A is the cross-section area. The total flux through the entire solenoid should be N times more where N is the total number of turns, which can be expressed as N=nl, where l is the length of a solenoid. Thus, o n 2 lia If we compare the formula, with LI we can conclude that Inductance for solenoid can be presented as L o n 2 la The product A is a volume of a solenoid -V. Thus the final expression for the inductance of a solenoid is L o n 2 V It is apparent from the formula, that inductance depends only on a geometry, size and medium. It is independent on the current. RL CIRCUIT The goal of the section is to describe the process of self-inductance quantitatively, i.e define the equations that describe the function I(t) when the current is turned on and off.

10 10 a) Let us assemble a circuit that consists of a battery, a resistor, an inductance coil and a switch, as it shown on a diagram. We turn the switch on and apply Kirchgoff s loop rule. We start our clockwise round trip at point A. We travel through the battery from - to + and thus the emf sign would be +. EMFself induction for the rising current according to Lenz s law should oppose the current change and its sign should be -. We travel through the resistor R along the current and thus the voltage drop on a resistor IR should be negative. The final result is di L IR 0, dt To solve it, we perform several simple algebraic manipulations. First we divide each term by R and move first and third term to the right.. L di R dt I R The next step is to multiply both parts of equation by Rdt. We obtain L Rdt di ( I )( ) R L Then we separate variables I and dt by dividing both parts by I R di Rdt I L R Now we can integrate both parts. The result is di Rdt I L R The result of integration is Rt Ln( I ) const R L The value of const can be determined from the following boundary condition. At t=0, the current I =0, and the value of the const is Ln( ). The equation becomes R Ln(I R ) Rt L Ln( R ) or Ln(I R ) Ln( R ) Rt L I or Ln R R Rt L In exponential shape the equation transforms to

11 11 I R e Rt L or I R Rt L e or I R R Rt L e R R The final result is: I R (1 e R L t ) (**) To test the solution we can take a derivative di/dt from (**) and plug it into (*). Graph of I(t) according to (**) is presented on a diagram. To determine the physical meaning of the coefficient in front of the time in an exponent function, let us estimate the units of the ratio R/L. Thus the reciprocal L/R has a dimensions of time. That quantity is defined as inductive time constant L L R And thus the equation (**) can be expressed as t I R (1 e L ) The physical meaning of inductive time constant is that if t= L, the value of power in exponent is one and the current reaches 0.63 of its maximum value, i.e.

12 12 b) This time we investigate the current vs.time function for a circuit presented at a diagram at a moment when the circuit is turned off. The induced current as we discussed above resists the decrease of the main current and travels around the loop that consists of inductance and resistor. Thus if we apply the Kirchgoff s loop rule traveling counterclockwise(i.e against the induced current) from point A, the result would be. L di dt IR 0, or di I R L t The solution of the differential equation is I I o e R L t t I o e L where L is the same inductive time constant we discussed above. It shows what time it takes for the induced current to drop e (2.7) times. The graph for the current at the shut down is presented on a diagram. ENERGY OF A MAGNETIC FIELD We will derive the formula for the energy of the magnetic field two times. First using the simple method of analogy and than exact mathematical method a) Let us compare two effects. Inertia in mechanics and self-induction in magnetism. Inertia describes the bodies unwillingness to change its velocity. Self-induction describes the conductors unwillingness to change current through itself. That lead us to conclusion that those effects are analogous. That means that velocity(v) is analogous to the current (I), and mass (m) is analogous to the inductance (L).

13 13 As we know in mechanics the bodies energy can be expressed as mv 2 /2. In analogy the energy of the magnetic field, created by current can be expressed as E LI2 2 To prove that our formula is correct, we can complete a dimensional analyses b) Let us derive the same formula using the exact laws. Just as we did above we apply Kirchgoff s loop rule for a circuit on a diagram L di dt IR 0, or L di dt IR Let us imagine that a charge dq=idt, traveled through the circuit. Out next step is to multiply left part of equation by dq and the right part by Idt. The result is. dq LIdI I 2 Rdt Now we integrate the equation. The result is q LI2 2 I2 Rt As we know (Joule s Law) expression q = W is work done by EMF. i 2 Rt - is the amount of heat released on the resistor. Thus LI 2 /2 should also be an energy of the inductance coil, i.e. the energy of the magnetic field. ENERGY DENSITY OF THE MAGNETIC FIELD Let us assume that a current i is flowing through the solenoid inductance of L. We also assume that a solenoid has a length of l and cross-section area of A.

14 14 We define magnetic field density as the ratio of the energy created by the field over the volume of solenoid. B Earlier we derived an expression for the solenoid s inductance - L= µ o n 2 la (**) We also know the expression for the field induction in solenoid B = µ o In (***) If we isolate I from (*** ), plug it in (**) and substitute the result into (*), we obtain final expression for the energy density of a magnetic field. B B2 2 o As we can see the expression for magnetic field density resembles the one for electric field density. The results show us the close relationship between the fields. LI 2 2 V LI2 2 Al Home assignment: Ditto, Chapter 30, Web Assign: Electromagnetic Induction-01 Problems #1,2,3,6, 9, 11(6 edition),13,15,23,27, 29,32,33, Questions 2,6 Web Assign: Electromagnetic Induction-02, RL circuits Problems: #38,40,42,44,49,53, 52,89,62,63,Questions 8,10

Chapter 27, 28 & 29: Magnetism & Electromagnetic Induction. Magnetic flux Faraday s and Lenz s law Electromagnetic Induction Ampere s law

Chapter 27, 28 & 29: Magnetism & Electromagnetic Induction Magnetic flux Faraday s and Lenz s law Electromagnetic Induction Ampere s law 1 Magnetic Flux and Faraday s Law of Electromagnetic Induction We