1. 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   v  

2. 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 F magnetic = q v B. B   F mag on e  v  

3. 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  F mag on e   v  

4. Generating Currents 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 (F el =qE) balances the Magnetic Force. B  F mag on e  v   F elec on e  

5. Generating a Voltage We now have, at equilibrium:  F = 0, or F elec on e = F mag on e, or qE = qvB, or E = vB. We know that the electric field is related to voltage by E = -  V /  s, (here  s = L, the length of the bar). Thus we have for the voltage:  V = v B L.

6. 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).

7. 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?

8. Generating Currents 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   F mag on e  v   F elec on e  

9. Generating Currents 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   F mag on e  v   F elec on e  F mag on I 

10. Generalizing We have from the previous apparatus:  V = v B L. We note that v =  w/  t where w is the width of the circuit (distance from end of rails to bar), so  V = (  w/  t) B L. Can we take the  /  t and apply it to all the variables:  V =  (B L w) /  t ? From experiment, YES!

11. Faraday’s Law We also note that wL = A (area of circuit). We can also have N number of loops, so we finally get:  V =  (N B A) /  t. 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:  V =  (N B A cos(  BA ) ] /  t.

12. Faraday’s Law If the magnetic field is not constant over the whole area, we need to break the area into pieces and then sum up the pieces. This leads to Faraday’s Law as:  V =  d   B  dA ] / dt, where we have used the dot product to represent the cos(  BA ).

13. Magnetic Flux  V =  d   B  dA ] / dt The quantity in brackets:  B  dA is called the magnetic flux with units of Webers. It is a measure of how much magnetic field cuts through a certain area. Faraday’s Law says there is a voltage produced whenever this magnetic flux changes in time. Webers = Volt-sec = T-m 2.

14. Magnetic Field  V =  d   B  dA ] / dt Because of its importance in magnetic flux, the magnetic field, B, is sometimes called the magnetic flux density. Although we already have a unit for B, the Tesla, because of B’s importance to inductance and its relation to magnetic flux density, it also has an equivalent unit called the Weber/m 2 = Tesla.

15. Lenz’s Law  V =  (N B A cos(  BA ) ] /  t or  V =  d/dt   B  dA ] The above formula is for determining the amount of voltage generated. But what is the direction of that voltage (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.

16. Example #1 Using Lenz’s Law, which direction should the induced current flow when the bar is moving to the right as indicated below? B    v  

17. Example #1 (cont.) Since the area through which the magnetic field is going is increasing, the magnetic flux is increasing. According to Lenz’s Law, the induced current will flow in a direction so as to create a magnetic field that will oppose the increase. Here it means the current will create a field going out of the loop and so using the RHR the current must flow counterclockwise! B  I  B induced  v   

18. Example #2 Consider the situation below in which the North pole of a magnet is brought closer to the center of a loop of wire. Which way will the induced current be in the wire? FRONT VIEW VIEW from LEFT S N  B B  v 

19. Example #2 (cont.) Since the magnetic field through the loop is increasing, the flux is increasing. By Lenz’s Law, the induced current will try to create a magnetic field opposing the increase, and the current direction is determined by the RHR. FRONT VIEWVIEW from LEFT  I in  I in S N  B B  v   B in B in   

20. Example #2 (cont.) If the North pole of the magnet is moved away from the coil, then the external field will decrease, and Lenz’s Law will say that the induced current should create a magnetic field to replace the decreasing external magnetic field. FRONT VIEW VIEW from LEFT  I in I in  S N  B B   v  B in B in   

21. Lenz’s Law The Computer Homework assignment, Vol. 4 #3, deals with Lenz’s Law and will give you practice with this as well.

22. Transformers and Inductors  V =  (N B A cos(  BA ) ] /  t. We have already seen how changing the area of a circuit in a magnetic field generates a voltage. We have also seen how changing the magnetic field strength through the circuit can generate a voltage - this is the basis of an inductor and a transformer.

23. Transformer We’ll consider an inductor in a little bit. But now consider the situation in the figure below. Note that circuit 1 is not electrically connected to circuit 2. But it is connected magnetically, since the magnetic field will “flow” through the iron square ring. I 1  B 1   B 1 12

24. Transformer (cont.) If I 1 changes in circuit 1, then B 1 changes. But since the iron ring carries B 1 through circuit 2, a voltage and current will be induced in circuit 2. The current in circuit 2 depends on how the current in circuit 1 changes (as well as the number of loops in both circuit 1 and circuit 2). I 1  B 1   B 1 12

25. Transformer (cont.) The voltage in circuit 2 depends on the change in the current in circuit 1: V 2 = M 12 dI 1 /dt M 12 is called the mutual inductance, and is measured in Henries, where henry = volt / [amp/sec] = volt-sec / amp. I 1  B 1   B 1 12

26. Mutual Inductance Like capacitance and resistance, mutual inductance depends NOT on voltage and/or current, but on the geometry and materials. In this case, the mutual inductance will depend on the number of turns in both circuits 1 and 2 as well as the geometries (solenoids) used in circuits 1 and 2.

27. 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!

28. Electric Generator If we use the crank to turn the area inside the magnetic field, we are changing the magnetic flux through the area. By Lens’ Law this should generate a voltage and current! crankr NB   S L w

29. Electric Generators  V =  d/dt  (N B A cos(  BA ) ]. When we change the angle,  BA, with respect to time:  BA =  t, we get the following relation:  V =  N B A  sin(  BA ), or V AC = V o sin(  t) where V o = NBA   and where  d    dt  f. This kind of voltage is an alternating voltage (AC voltage) since the sine function alternates between positive and negative.

30. Electric Generators You should be able to DESIGN your own generator based on the voltage (V) and frequency (  ) you want out of the generator.

31. AC Circuits V AC =  V o  sin(  t) For this kind of AC voltage, we can determine the amplitude of the voltage (V o = NBA . 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.

32. 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. Thus, I = V/R = (NBA  R) sin(  t) = I o sin(  t), where I o = NBA  /R = V o /R. From this we see that the electric power is: P(t) = I(t)*V(t) = I o V o sin 2 (  t).

33. AC Power P(t) = I(t)*V(t) = I o V o sin 2 (  t) 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 sin 2 (  ) = 1/2. Thus: P avg = (1/2)I o V o.

34. Average of sin 2 (  Avg of sin 2 (  ) =  sin 2 (  i ) /  (1) This is approximate, so we could extend the summation to an integral: Avg of sin 2 (  ) =  sin 2 (  ) d  /  (1) d , where the limits of integration go from 0 to 2  for both the numerator and the denominator.

35. Another way A tricky way to get this without integrating is: Note now that sin 2 (  ) + cos 2 (  ) = 1 for all  ’s, so the average of [sin 2 (  ) + cos 2 (  )] = 1. But the average of [sin 2 (  ) + cos 2 (  )] = 1 = average of sin 2 (  ) + average of cos 2 (  ). Sine and cosine differ only by the starting point, so the average of sin 2 (  ) = average of cos 2 (  ) over a complete oscillation. Therefore, average of sin 2 (  ) = 1/2 !

36. 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). V rms = V o (1/2) 1/2 and I rms = I o (1/2) 1/2 so that we have P avg = I rms V rms = (1/2) I o V o, and V rms = I rms R.

37. Eddy Currents Another consequence of Faraday’s Law is the existence of eddy currents. We’ll look at a demonstration in class to see these eddy currents at work.

38. Extension of Ampere’s Law Recall from Part 3: Ampere’s Law:  closed loop B  dL =  o I encircled Consider a circuit with a capacitor that is being charged. There is a current flowing to the capacitor, and so there will be a magnetic field that encircles the wire. However, there is no current flowing across the capacitor. Will there be a magnetic field encircling the charging capacitor?

39. Ampere’s Law (extended) Gauss’ Law for Electric Fields:  E  dA = Q enclosed /  o Recall also that I = dq/dt. These two can be combined to give for a charging capacitor: I effective =  o d/dt[  E  dA]. With this, Ampere’s Law can be extended:  closed loop B  dL =  o I encircled +  o  o d/dt [  E  dA]

40. Maxwell’s Equations We now have the four main laws for electric and magnetic fields. Together they are called Maxwell’s Equations. The four laws are:

41. Maxwell’s Equations Gauss’ Law for Electric fields  closed area E  dA = Q enclosed /  o Gauss’ Law for Magnetic fields  closed area B  dA = 0 Ampere’s Law  closed loop B  dL =  o I encircled +  o  o d/dt [  E  dA] Faraday’s Law  closed loop E  dL =  V =  d/dt   B  dA ].