1. Chapter 19 Mass Spects Review Ch. 17 - 19

2. Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! Friday, April 9, 1999 in class Ch. 17 - 19 You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you.

3. x x x x x x x x x d1d1 d2d2 Mass Spectrometer

4. Let’s first determine the distance from the entrance point that the H atom hits the photographic plate. Two singly ionized particles enter a mass spectrometer at a speed of 3 X 10 6 m/s. The strength of the magnetic field is 0.625 T. If one of the particles is H and the other particle hits the photographic plate 110 cm further away than the H atom, what chemical element is the second particle?

5. Now, let’s figure out the radius of curvature of the mystery element... We were told that the 2nd element collides with the photographic plate 110 cm further away from the entrance point than the H atom. x x x x x x x x x d1d1 d2d2 d 2 - d 1 = 110 cm d 2 = 110 + 2r 1 d 2 = 120 cm r 2 = 60 cm

6. Finally, let’s determine the mass of our mystery element...

7. PHYSICS RULES! Televisions rely upon electromagnetic fields to produce the images we see. An electron “gun” fires electrons at the television screen. When they collide with the material on the back side of the screen, colored light is emitted, producing one pixel in the image we see on the screen.

8. electromagnet television screen Changing the strength of the magnetic field changes the degree to which the electron beam is deflected.

9. We now make explicit a fact we’ve assumed implicitly in dealing with magnetic fields thus far... The superposition principle applies! x x 1 2 B tot = B 1 + B 2

11. Current Resistance Ohm’s Law Series & Parallel Circuits Kirchhoff’s Rules RC Circuits Magnetic Fields, Forces, Torques

12. E A Vd tVd t The charge carriers, each with charge q, move with an average speed v d in response to the electric field. If there are n charge carriers per volume in the conductor, then the number of charge carriers passing a surface A in a time interval  t is given by

13. V = I R R =  L / A Resisitivity and Resistance are also a function of Temperature.

14. V R 0V a b c d Power is the change in energy per unit time. As we move a charge q from a to b, it gains potential energy equal to  PE = q V Use this form for power supplied by batteries!

15. P = I 2 R V R 0V a b c d Power is the change in energy per unit time. As we move a charge q from c to d, it loses potential energy. The rate of energy loss is Use this form for power dissipated by resistors.

16. Of course, real batteries have some internal resistance, such that V R  R int I V =  - I R int load resistor internal resistance terminal voltage

17. V R  R int I What is the current in this circuit?  = I R int + I R  I = R int + R What is the power supplied by the EMF? What is the power supplied by the battery? P = I V = I 2 R So what happened to the difference? P = I  = I 2 (R int + R) Joule Heating

18. V 1 + V 2 = V I 1 = I 2 due to conservation of charge! V R2R2 + _ R1R1 I1I1 I2I2 R eq = R 1 + R 2 Resistors in series ADD.

19. V R2R2 + _ R1R1 V R1 = V R2 I I1I1 I2I2 I 1 = V / R 1 I 2 = V / R 2 V V I = I 1 + I 2 = + R 1 R 2 1 1 1 = + R eq R 1 R 2 Resistors in parallel ADD INVERSELY.

20. 1) The sum of the currents entering a junction must equal the sum of currents leaving a junction. (Conservation of Charge) 2) The sum of potential differences across all the elements on any closed loop in a circuit must be zero. (Conservation of Energy) Junction Rule Loop Rule

21. For Applying Kirchhoff’s Rules A) assign a direction to the current in each branch of the circuit. Just GUESS!! If your guess is incorrect, the current will come out as a negative number, but the magnitude will still be correct! B) when applying the loop rule, you must choose a consistent direction in which to proceed around the loop (either clockwise or counterclockwise, your choice, but stick to it).

22. 1) When you encounter a resistor in the direction of the current, the voltage drop is  V = - I R 2) When you encounter a resistor opposite the current, the voltage drop is  V = + I R R2R2 I R1R1 I1I1 I2I2 - I 1 R 1 +I2R2+I2R2

23. 3) When you encounter an emf in the direction you’re going around the loop, the voltage change is +  V 4) When you encounter an emf opposite the the direction you’re going, the voltage change is -  V - V 1 + V 2 V1V1 V2V2

24. 5) The junction rule can only be applied n-1 times in a circuit with n junctions. 6) Each new equation you write must contain a current that you haven’t yet used. 7) To solve a system of equations with k unknown quantities, you need k independent equations.

25. Find the currents in each branch of this circuit. I0I0 I1I1 I2I2 6  1  18 V 12  12 V + _ + _ 1  a f e c b d There are two junction points, so we can apply the junction rule ONCE. I 0 = I 1 + I 2 Remember, each branch has it’s own current.

26. 6  1  18 V 12  12 V + _ + _ 1  a f e c b d How many possible loops are there? b c d ae f c d a be f c b We have 3 unknowns, so we need 3 equations total. Therefore, we need use only 2 of the 3 equations provided by the loop rule. (The junction rule gave us 1 equation already!) Loops do NOT equal currents!

27. q t Q=CV Q=0.632CV  V R + _ C  = the time constant = R C Q = final charge on C Charging a capacitor

28. Discharging a capacitor R C I q t Q 0 =CV 0 Q=0.368CV 0   = the time constant = R C Q 0 = initial charge on C

29. Also as was the case with the other fields, the density of the magnetic field lines tells us the magnitude of the field: the more tightly packed the lines, the stronger the field. weaker field stronger field weaker field stronger field

30. What is the direction of the magnetic force in this case? x x x x x B into the page v +

31. Palm faces the direction of the magnetic force. Thumb points in the direction of the motion of a positive charge. Fingers point in the direction of the magnetic field.

32. This force acts in the direction perpendicular to the plane defined by the vectors v and B as indicated by the right-hand rule! WARNING: cross-product (NOT simple multiplication!) DON’T FORGET: Forces have directions!

33. But recall our definition of current in a wire... Substituting, we get the simpler expression:

34. So what happens if we put a loop of wire carrying current I in a magnetic field? B I Force on top and bottom of loop are 0! But force on left and right of the loop are equal in magnitude and opposite in direction, causing a net torque on the loop!

35. Where  is the angle between the normal to the loop and the magnetic field. Top View: B I normal  x The normal is the direction perpendicular to the plane of the loop of wire.

36. FBFB FBFB FBFB FBFB x x x x x B into the page v + x x x x x B into the page v + x x x x x B into the page v + x x x x x B into the page v +

37. So a charged particle moving in a uniform magnetic field will move in a circle! The Centripetal Force! x x x x x r

38. d We can measure d using a photographic plate (for example) to record the location of the end of the particle’s path. d = 2 r Mass Spectrometer

39. The magnetic field around a very long, current carrying wire is given by: The magnetic field is a vector quantity, so it has a direction!!!! Where the constant of proportionality is known as the permeability of free space and is found to be

40. I Wrap your hand around the wire with your thumb pointed in the direction of the current. Your fingers curl around the wire in the direction the magnetic field points.

41. I1I1 I2I2 d 1 2 If currents in same direction, wires attract. If currents in opposite directions, wires repel.

42. Where n is the number of turns per unit length of the solenoid. The magnetic field is directed parallel to the axis of the solenoid. The right hand rule for currents will help you determine toward which end the field points.