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Published byLilian Allison Modified over 2 years ago

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Humans are clever! With the magnetic force only pushing perpendicular to the motion, it would seem that we could not use this force to increase the speed of particles and hence generate energy. However, humans being clever, consider the following slides.

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Generating Currents Consider the following circuit: a bar moves on two rails that are connected at one end. The whole setup has a magnetic field that goes through it. B Ä Ä v ® Ä Ä

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Generating Currents If the bar is moving to the right, and the bar contains electrons that are free to move (as in metal), then the electrons are also moving to the right. There is a magnetic force on the electrons pushing them down with magnitude Fmagnetic = q v B. B Ä Ä Fmag on e ¯ v ® Ä Ä

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Generating Currents Thus, negative electrons should pile up at the bottom of the moving bar, leaving a net positive charge at the top of the bar. But this should act just like a battery! B Ä Ä Fmag on e ¯ v ® Ä - Ä

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**Generating Currents B Ä + Ä Felec on e v ® Fmag on e ¯ Ä - Ä**

To find the voltage of this “battery”, we note that as the charges pile up at the ends of the rod, an Electric field will be set up. The electrons will continue to pile up until the Electric Force (Fel =qE) balances the Magnetic Force (Fmag= qvB). B Ä Ä Felec on e v ® Fmag on e ¯ Ä - Ä

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**Generating a Voltage We now have, at equilibrium: SF = 0, or**

Felec on e = Fmag on e , or qE = qvB , or E = vB . We know that the electric field is related to voltage by E = -DV / Ds , or DV = -E*Ds (here Ds = L, the length of the bar and the minus sign indicates the electric field goes from high voltage to low voltage). Thus we have for the voltage: DV = v B L .

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Generating Power We have generated a voltage, and now to generate electric power (P=IV) we need to have that voltage drive a current. Since we have completed the circuit by connecting the ends of the rails, we will have a complete circuit - and so we will get a current depending on the resistance in the rails (V=IR).

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**Conservation of Energy**

We have made an electric generator that can generate electrical energy. But according to the Law of Conservation of Energy, we can only convert energy from one form into another. In the case of the electric generator, where does this energy come from?

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**Generating Currents B Ä ¬ I + Ä I¯ Felec on e v ® Fmag on e ¯ I**

Note that as electrons flow clockwise around the circuit, this acts the same as a current of positive charges going counterclockwise, as indicated on the diagram below. Note that a current flows up the bar. B Ä ¬ I + Ä I¯ Felec on e v ® Fmag on e ¯ I I ® Ä Ä

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**Generating Currents B Ä ¬ I + Ä I¯ Felec on e v ®**

Is there a magnetic force on this current due to its flowing through a magnetic field? YES! Note that the direction of the force on this current is to the left. This will act to slow the bar down. In effect, this apparatus converts the kinetic energy of the bar into electric energy! B Ä ¬ I + Ä I¯ Felec on e v ® Fmag on e ¯ I Fmag on I ¬ I ® Ä Ä

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**Generating Currents B Ä ¬ I + Ä I¯ Felec on e v ®**

Note that if no current flows, there will be no electric power, and there will be no magnetic force on the bar since there will be no current, and thus the bar will not slow down and no kinetic energy will be lost in the bar. Only if current flows and electric power is generated will kinetic energy be lost. B Ä ¬ I + Ä I¯ Felec on e v ® Fmag on e ¯ I Fmag on I ¬ I ® Ä Ä

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**Generalizing We have from the previous apparatus: DV = v B L .**

We note that v = Dw/Dt where w is the width of the circuit (distance from end of rails to bar). DV = (Dw/Dt) B L Can we take the D /Dt and apply it to all the variables: DV = D(w B L) / Dt ? The answer, based on experiment, is YES!

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Faraday’s Law We also note that wL = A (area of circuit). We can also have N number of loops, so we finally get: DV = D(N B A) / Dt . This is called Faraday’s Law. When we consider direction as well, we see that the magnetic field, B, has to cut through the area, A. If we assign a direction to A that is perpendicular to the surface, we get an even more general form: DV = D[ (N B A cos(qBA) ] / Dt .

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**Lenz’s Law DV = D[ (N B A cos(qBA) ] / Dt**

The above formula, Faraday’s Law, is for determining the amount of voltage generated. But what is the direction of that voltage (i.e., what direction will it try to drive a current)? The answer is Lenz’s Law: the direction of the induced voltage will tend to induce a current to oppose the change in magnetic field through the area.

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Lenz’s Law We’ll go over several cases in class. We’ll also have a lab later in the semester to play with this. The Computer Homework assignment, Vol. 4 #3, deals with Lenz’s Law and will give you practice with this as well.

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**Transformers and Inductors**

DV = D[ (N B A cos(qBA) ] / Dt . We have already seen how changing the area of a circuit in a magnetic field generates a voltage. We could also change the magnetic field strength through the circuit to generate a voltage - this is the basis of an inductor and a transformer. [We’ll consider both a little later after we look at AC voltages.]

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Electric Generators Finally, we could change the direction of the area in relation to the field - this is the basis for the most common kind of generator. This generator looks just like the electric motor, except we put in rotational motion and get current instead of putting in current and getting rotational motion!

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**Electric Generator (two complete rings rather than one split ring) N S**

crank (turn at frequency, f)

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**Electric Generators DV = D[ (N B A cos(qBA) ] / Dt .**

When we change the angle, qBA, with respect to time (qBA=wt) , the calculus gives us the following relation: DV = N B A w sin(qBA) , or VAC = Vm sin(wt) where Vm=NBAw, is the max voltage, and where w = DqBA/Dt = 2pf. This kind of voltage is an alternating voltage (AC voltage) since the sine function alternates between positive and negative.

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**AC Circuits VAC = Vm sin(wt)**

For this kind of AC voltage, we can determine the amplitude (max) of the voltage (Vm= NBAw) . But since the average of sine is zero, how do we treat the average? What is usually important is the power delivered by the electric circuit. From P=IV we see that both the current and the voltage are important.

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AC Circuits From Ohm’s Law, we have V = IR, where R is a constant that depends on the material and geometry of the materials used to conduct the current (R=rL/A). Thus, I = V/R = (NBAw/R) sin(wt) = Im sin(wt) , where Im = NBAw/R = Vm/R . The alternating voltage creates an alternating current! From this we see that the electric power is: P = I V = ImVm sin2(wt) .

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**AC Power P = I V = ImVm sin2(wt)**

Note that the Power involves the square of the sine function, and so the Power oscillates but is always positive. But what we are usually interested in is the average power. From the calculus, we find that the average of sin2(q) = 1/2. Thus: Pavg = (1/2)ImVm .

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**RMS Voltage and Current**

In order to work with AC circuits just as we did with DC circuits, we create a voltage and current called rms (root mean square). Vrms = Vm (1/2)1/2 and Irms = Im (1/2)1/2 so that we have Pavg = Irms Vrms and Vrms = Irms R . Note that the power formula and Ohm’s Law are the same for DC and for AC-rms, but NOT for instantaneous AC.

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**Transformers Transformers work this way: An AC voltage:**

VAC-1 = V1 sin(wt) , is used to generate a current: IAC-1 = I1sin(wt) in coil #1 which generates an oscillating magnetic field in coil #1 since B is proportional to I: BAC-1 = B1sin(wt); a second coil (#2) is inside coil #1; this #2 coil then, by Faraday’s Law, DV = D[ (N B A cos(qBA) ] / Dt has a voltage, VAC-2, induced in it. By adjusting the number of loops in both coils, the induced (AC) voltage in #2 can be different than that in coil #1. We can adjust (or transform) the voltage up or down!

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Transformers For safety, we like to have rather low voltages in the house. For economy, since Plost = I2R, we like to have low currents (which means high voltages) in our transmission lines. We can have both if we use transformers located in our neighborhoods!

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Inductors A coil of wire can create a magnetic field if a current is run through it. If that current changes (as in the AC case), the magnetic field created by the coil will change. Will this changing magnetic field due to the changing current through the coil cause a voltage to be created across the coil? YES! This is called self-inductance and is the basis behind the circuit element called the inductor.

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**B = (mo/4p) I DL sin(qIr) / r2 (DL=length)**

Inductors Since the voltage created depends in this case on the changing magnetic field, DV = D[ (N B A cos(qBA) ] / Dt . and the field depends on the changing current, B = (mo/4p) I DL sin(qIr) / r2 (DL=length) we have: Vinductor = -L DI / Dt (L=inductance) where the L (called the inductance) depends on the shape and material (just like capacitance and resistance).

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**Inductors Vinductor = -L DI / Dt**

Here the minus sign means that when the current is increasing, the voltage across the inductor will tend to oppose the increase, and it also means when the current is decreasing, the voltage across the inductor will tend to oppose the decrease.

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**Units: Henry From Vinductor = -L dI /dt**

L has units of Volt / [Amp/sec] which is called a Henry: 1 Henry = Volt-sec / Amp . For a solenoid: L = mpN2R2 / Length (A Henry is a rather large amount of inductance.)

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**Energy Stored in an Inductor**

We start from the definition of voltage: V = PE/q (or PE = qV). But since the voltage across an inductor is related to the current change, we might express q in terms of I: I = dq/dt, or dq = I dt. Therefore, we have: Estored = S qi Vi = V dq = V I dt and now we use VL = L dI/dt to get: Estored = (L dI/dt) I dt = L I dI = (1/2)LI2.

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**Review of Energy in Circuits**

There is energy stored in a capacitor (that has Electric Field): Estored = (1/2)CV2 . Recall that V is related to E (Electric Field). There is energy stored in an inductor (that has Magnetic Field): Estored = (1/2)LI2 . Recall the I is related to B (Magnetic Field). There is power dissipated (as heat) in a resistor: Plost = RI2 .

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**Review of Circuit Elements**

Resistor: VR = R I where I = Dq/Dt Capacitor: VC = (1/C)q (from C = q/V) Inductor: VL = -L DI/Dt We can make an analogy with mechanics: q is like x; V is like F; t is like t; L is like m; I = Dq/Dt is like v = Dx/Dt; C is like 1/k (spring); DI/Dt is like a = Dv/Dt; R is like air resistance.

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RL Circuit - DC What happens when we have a resistor in series with an inductor in a circuit with a battery? From Conservation of Energy, the voltage changes around the circuit must equal zero: Vbattery – Vresistor – Vinductor = 0 , or Vbattery = I*R + L*dI/dt . If we replace I with v, and dI/dt with a, and R with b (air resistance), and L with m, and identify Vbattery = constant with the constant force of gravity, mg, we have an equivalent equation: mg = bv + ma, or mg – bv = ma (with down being the positive direction for a falling object).

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RL Circuit - DC From the mechanical analogy, this should be like having a mass with air resistance. If we have a constant force (like gravity), the object will accelerate up to a terminal speed (due to force of air resistance increasing up to the point where it balances the gravity). SF=ma -bv + mg = ma, or m dv/dt + bv = mg

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**Mechanical Analogy: mass falling with air resistance**

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RL Circuit - DC (cont.) If we connect the resistor and the inductor to a battery and then turn the switch on, from the mechanical analogy we would expect the current (which is like velocity) to begin to increase until it reaches a constant amount. From conservation of energy: Vbattery = Vresistor + Vinductor where Vbattery = constant; VR = I*R, and VL = L*dI/dt .

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**LR Circuit - qualitative look**

From the circuit point of view, initially we have zero current so there is no VR =I*R (voltage drop across the resistor). Thus the full voltage of the battery is trying to change the current, hence VL = Vbattery, and so dI/dt = Vbattery /L. However, as the current increases, there is more voltage drop across the resistor, VR = I*R, which reduces the voltage across the inductor (VL = Vbattery - VR), and hence reduces the rate of change of the current, dI/dt = VL/L !

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**RL Circuit - DC: I(t) versus t**

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**Review of Circuit Elements**

Resistor: VR = R I where I = Dq/Dt Capacitor: VC = (1/C)q (from C = q/V) Inductor: VL = -L DI/Dt We can make an analogy with mechanics: q is like x; V is like F; t is like t; L is like m; I = Dq/Dt is like v = Dx/Dt; C is like 1/k (spring); DI/Dt is like a = Dv/Dt; R is like air resistance.

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**Inductive Reactance - AC**

Since VL = -L I/t, for an AC current we will have a voltage induced that will oppose the changing current. This opposition will tend to limit the current in the circuit and behave in some sense like a resistance. We call this action Reactance. For an inductor, since VL = -L I/t, VL will be big if L is big and/or if is big (causing I/t to be big). A more detailed calculation for reactance (or resistance to an AC circuit) gives: XL = L and is used in the Ohm’s Law-like relation: VL = XLI (the bigger XL, the smaller the I or the bigger the VL)

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**Capacitive Reactance - AC**

There is a similar effect for a capacitor in an AC circuit. For a capacitor, since VC = q/C, VC will be small if C is big and/or if is big (causing little q to accumulate in the short time). A more detailed calculation for reactance gives XC = 1/(C), and it is used in the Ohm’s Law-like relation: VC = XCI (the bigger XC, the smaller the I or the bigger the VC)

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**AC Circuits Z = [R2 + (wL - 1/wC)2]1/2.**

A resistor obviously limits the current in a circuit. But, as we just saw, a capacitor and an inductor also limit the current in an AC circuit. However, the reactances do not just add together. Using the fundamental relations and the calculus, we come up with the concept of impedance, Z: V = IZ where Z takes into account all three reactances: XR=R, XL=L and XC= 1/C: Z = [R2 + (wL - 1/wC)2]1/2. Power, however, is still: Pavg = I2R (not P=I2Z).

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Oscillations Newton’s Second Law: SF = ma can be written as: SF - ma = 0 . This is like Conservation of Energy: SV = 0 . If we put an inductor with a capacitor with an AC voltage, we have the analogy with a mass connected to a spring that has an oscillating applied force. In each of these cases, we get resonance. We’ll demonstrate this in class with a mass and spring. This is the basis of tuning a radio!

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**AC Circuits V = IZ where Z = [R2 + (wL - 1/wC)2]1/2 .**

Note that when (wL - 1/wC) = 0, Z is smallest and so I is biggest! This is the condition for resonance. Thus when w = [1/LC]1/2, we have resonance. This is equivalent to the resonance of a spring when w = [k/m]1/2 . Computer Homework Vol. 4, #4, gives practice with problems involving inductance (L) and impedance (Z).

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