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Magnetic Force on moving charges

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1 Magnetic Force on moving charges
Physics 212 Lecture 12 Today's Concept: Magnetic Force on moving charges

2 Tribute to New Orleans and Mardi Gras
Music Who is the Artist? The Meters The Neville Brothers Trombone Shorty Michael Franti Radiators Tribute to New Orleans and Mardi Gras

3 Your Comments “How can a magnetic field point in a direction? What about the wire example in the prelecture where is shown to move in a counter-clockwise/clockwise direction?” “This stuff is really neat... It is fun to actually see the calculations for magnetism. However, since this is the first time I’ve really seen it, it is still a bit confusing. If you could go through different examples and go over the actual concepts more, that would be great.” A little at a time! Good questions! But very deep! “Magnets. How do they work?” “Please explain what causes magnetic fields, and why they exist. Unless we learn that later or not at all, I just don’t understand.” “What are we supposed to do if we have two left hands?” “Could you go over this right hand rule again? There are so many of them...” RHR confused many “We most definitely need examples. Also, every time you post an awful pun, a puppy dies. Just so you know.” “I refuse to make an attempt at a bad pun.” 05 3

4 Key Concepts: Today’s Plan:
The force on moving charges due to a magnetic field. The cross product. Today’s Plan: Review of magnetism Review of cross product Example problem 05

5 Magnetic Observations
Bar Magnets Compass Needles N S N S S N N S N S You may be familiar with the behavior of bar magnets like the ones shown, and their ability to exert a force without touching. Each magnet has a north and south pole. If you bring two magnets near each other, like poles repel and opposite poles attract. In many ways the bar magnets behave just as a rod would behave if it had positive charge at one end and negative charge at the other (e.g. an electric dipole). N S Magnetic Charge? N S cut in half 07

6 Magnetic Observations
Compass needle deflected by electric current I Magnetic fields created by electric currents Direction? Right thumb along I, fingers curl in direction of B Magnetic fields exert forces on electric currents (charges in motion) V I F V I F Indeed, one can describe the force on one magnet due to the other in terms of the magnetic field created by one magnet, producing a force on the other. Similar to the electric field, the magnetic field is a vector with both a magnitude and direction, commonly denoted by the letter B. One can “map” the magnetic field due to the bar magnet but looking at its interaction with a “test” magnet. The interpretation of these magnetic field lines is analogous to electric field lines, the arrows give the direction, and the density is proportional to the strength. Notice that the magnetic field produced by the North and South poles of the magnetic look identical to the electric field produced by an electric dipole. There is however one very important difference. If you “break” a magnet, you do not get one piece that is the south pole and one piece that is the North pole. Instead you get two smaller magnets, each with a north and south pole. So there does not appear to be a magnetic charge that can be isolated. That means unlike the electric field lines that start and stop on “charges”, magnetic field lines are always continuous, and we picture the lines running through the bar magnet from north to south. 12

7 Magnetic Observations
P P I (out of the screen) I (into of the screen) Case I Case II The magnetic field at P points A. Case I: left, Case II: right B. Case I: left, Case II: left C. Case I: right, Case II: left D. Case I: right, Case II: right WHY? Direction of B: right thumb in direction of I, fingers curl in the direction of B

8 Magnetism & Moving Charges
All observations are explained by two equations: Today Next Week In addition to producing forces on other magnets, magnetic fields can also produce forces on charges. Although this force is proportional to the magnitude of the charge and the magnetic field, the magnetic force on a charge has a very different nature from the gravitational and Coulomb force we have studied thus far. First, the magnitude of the force is proportional to the velocity of the charge perpendicular to the magnetic field. Second the direction of the force is perpendicular to both the magnetic field, and the velocity of the charge. This can be expressed mathematically using the cross product. So the magnetic force is q v cross B. The units for the magnetic field are Tesla such that a 1 Coulomb charge moving 1 m/s perpendicular to a magnetic field of 1 Tesla will experience a force of 1 Newton. 14

9 Cross Product Review Cross Product different from Dot Product
A●B is a scalar; A x B is a vector A●B proportional to the component of B parallel to A A x B proportional to the component of B perpendicular to A Definition of A x B Magnitude: ABsinq Direction: perpendicular to plane defined by A and B with sense given by right-hand-rule Cross products show up a lot when considering magnetic forces, so lets take a minute to review their use. The magnitude of the cross product is obtained by multiplying the magnitude of the vectors, and the sin of the angle between them. The direction can be obtained using the right hand rule. There are several versions of this rule, here we will use the flat hand rule. Hold you right hand flat. Now orient your thumb so it points in the direction of the first vector (v), your fingers should point in the direction of the second vector B, then your palm points in the resulting directions (F). Lets illustrate this with an example. Consider a proton (q=1.6e-19C) moving (v=300 m/s) in a uniform magnetic field B=0.75T pointing to the right. The proton is traveling at an angle of 30 degrees w.r.t. the horizontal as shown. The magnitude of the force is just the product q x v x B x sin(30)= The direction is found using the RHR. Thumb in direction of v, fingers B, palm gives F into the page. Note that if this was an electron (negatively charge) instead of a proton, the magnitude of the force would be the same, however, the direction would be opposite (out of the page). It will often be useful to represent magnetic field lines going into and out-of the page. We will use the notation that an x represents a field line into the page, and a dot represents a field line coming out of the page. 16

10 Remembering Directions: The Right Hand Rule
y x F B qv z 17

11 Checkpoint 1a Three points are arranged in a uniform magnetic field. The B field points into the screen. 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is A. right B. left C. into the screen D. out of the screen E. zero “all are into the screen” “Right hand rule says its pointing out of the screen” “qvXB=0 if v=0.” 18

12 Checkpoint 1a Three points are arranged in a uniform magnetic field. The B field points into the screen. 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is A. right B. left C. into the screen D. out of the screen E. zero The particle’s velocity is zero. There can be no magnetic force. 18

13 Checkpoint 1b Three points are arranged in a uniform magnetic field. The B field points into the screen. 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: The positive charge moves from A toward B. The direction of the magnetic force on the particle is A. right B. left C. into the screen D. out of the screen E. zero “it should be diagonally outward” “v is up, b is into screen, right hand rule says it should be left.” “Because the magnetic force goes into the screen, so will the direction” 21

14 Checkpoint 1b Three points are arranged in a uniform magnetic field. The B field points into the screen. 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: The positive charge moves from A toward B. The direction of the magnetic force on the particle is A. right B. left C. into the screen D. out of the screen E. zero X qv B F 21

15 Cross Product Practice
protons (positive charge) coming out of screen (+z direction) Magnetic field pointing down What is direction of force on POSITIVE charge? A) Left B) Right C) UP D) Down E) Zero - x + x + y - y y x B screen z 24

16 Motion of Charge q in Uniform B Field
Force is perpendicular to v Speed does not change Uniform Circular Motion x x x x x x x q v F Solve for R: R Consider a positively charged particle moving to the right, in a uniform magnetic field pointing out of the page. Since the magnetic force on this particle is perpendicular to its velocity, the magnetic force will NOT change the particles speed. It will however change the particles direction. When the particle is moving to the right, the force will be down, as the particle starts moving down, the direction of the force will change always remaining perpendicular to the particles velocity. The result is the particle will undergo uniform circular motion, with constant magnitude of acceleration v^2/r pointing to the center of the circle. Using Newton’s second law, we can set the magnetic force equal to the acceleration, to determine the radius of the orbit. r = mv/qB. Note that the radius is proportional to both the mass and the velocity, and inversely proportional to the charge and magnetic field. An important feature is the time required to complete one revolution T = d/v = 2 pi m / qB is independent of the velocity of the particle. Uniform B into page 30

17 LHC 17 miles diameter

18 Checkpoint 2a The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. What is the direction of the magnetic field in chamber 1? A. up B. down C. into the page D. out of the page “Since the direction of the magnetic field has to be perpendicular to the charged particle’s direction, it will point up..” “Having a magnetic field going into the screen indicates counterclockwise motion.” “Thumb points to the left because particle is deflected to the left and fingers point up because that is velocity vector. Fingers curl out of the screen.” 34

19 Checkpoint 2a The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. What is the direction of the magnetic field in chamber 1? A. up B. down C. into the page D. out of the page qv F B . 34

20 Checkpoint 2b The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the magnetic field in chamber 2 A. |B1| > |B2| B. |B1| = |B2| C. |B1| < |B2| “because the curve is much sharper inside box 1” “It equals out.” “Because the particle changed direction more slowly in the second chamber.” 36

21 Checkpoint 2b The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the magnetic field in chamber 2 A. |B1| > |B2| B. |B1| = |B2| C. |B1| < |B2| Observation: R2 > R1 36

22 Calculation Conceptual Analysis
exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Conceptual Analysis What do we need to know to solve this problem? (A) Lorentz Force Law (B) E field definition (C) V definition (D) Conservation of Energy/Newton’s Laws (E) All of the above The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Absolutely ! We need to use the definitions of V and E and either conservation of energy or Newton’s Laws to understand the motion of the particle before it enters the B field. We need to use the Lorentz Force Law (and Newton’s Laws) to determine what happens in the magnetic field.

23 Calculation Let’s Do It !! Strategic Analysis
exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Strategic Analysis Calculate v, the velocity of the particle as it enters the magnetic field Use Lorentz Force equation to determine the path in the field as a function of B Apply the entrance-exit information to determine B The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Let’s Do It !!

24 Calculation exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B What is v0, the speed of the particle as it enters the magnetic field ? (A) (B) (C) (D) (E) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Why?? Conservation of Energy Initial: Energy = U = qV = qEd Final: Energy = KE = ½ mv02 Newton’s Laws a = F/m = qE/m v02 = 2ad

25 Calculation exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B X X X X X X X X X What is the path of the particle as it moves through the magnetic field? (A) (B) (C) Why?? Path is circle ! Force is perpendicular to the velocity Force produces centripetal acceleration Particle moves with uniform circular motion The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class.

26 Calculation What is the radius of path of particle? Why??
exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B What is the radius of path of particle? (A) (B) (C) (D) (E) Why?? May want to point out that we don’t know B in answer D X X X X X X X X X R xo/2

27 Calculation exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B (A) (B) (C) (D) (E) B Why?? The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class.

28 Follow-Up exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? (A) (B) (C) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. 1. q changes  v changes No slam dunk.. As Expected ! Several things going on here 2. q & v change  F changes 3. v & F change  R changes

29 Follow-Up exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? How does v, the new velocity at the entrance, compare to the original velocity v0? (A) (B) (C) (D) (E) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Why??

30 Follow-Up exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? How does F, the magnitude of the new force at the entrance, compare to F0, the magnitude of the original force? The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. (A) (B) (C) (D) (E) Why??

31 Follow-Up exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E x0 d enters here B B Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? How does R, the radius of curvature of the path, compare to R0, the radius of curvature of the original path? The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. (A) (B) (C) (D) (E) Why??

32 Follow-Up exits here A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? x0/2 X X X X X X X X X q,m E E x0 d enters here B B Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. (A) (B) (C) A Check: (Exercise for Student) Given our result for B (above), can you show:


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