3. Announcements  Physics 2135 spreadsheets for all sections, with Exam 2 scores, will be available today on the Physics 2135 web site. You need your PIN to find your grade.  Physics 2135 Exam 2 will be returned in recitation Thursday. When you get the exam back, please check that points were added correctly. Review the course handbook and be sure to follow proper procedures before requesting a regrade. Get your regrade requests in on time! (They are due by next Thursday’s recitation.) On a separate sheet of paper, briefly explain the reason for your regrade request. This should be based on the work actually shown on paper, not what was in your head. Attach to the exam and turn it in by the end of your next Thursday’s recitation.  Preliminary exam average is about 77.7% (16 sections out of 16 reporting). Good! Scores ranged from a low of 59 to a high of 200 (21 students). I will fill in the ??’s during the “live” lecture and in its “.ppt” file.

4. Announcements  Midterm grades include your first 3 labs. If you are missing one of them, this does impact your grade. Remember, the one lowest lab score is dropped, and there are no makeups.

5. Today’s agenda: Magnetic Fields Due To A Moving Charged Particle. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart Law: Magnetic Field due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current- carrying conductor (for example: a long straight wire). Force Between Current-Carrying Conductors. You must be able to begin with starting equations and calculate forces between current- carrying conductors. We’ve been working with the effects of magnetic fields without considering where they come from. Today we learn about sources of magnetic fields.

6. + B r v It is experimentally observed that a moving point charge q gives rise to a magnetic field.  0 is a constant, and its value is  0 =4  x10 -7 T·m/A Let’s start with the magnetic field of a moving charged particle. Remember: the direction of r is always from the source point (the thing that causes the field) to the field point (the location where the field is being measured.  Magnetic Field of a Moving Charged Particle

7. cross products of unit vectors We are going to be doing lots of cross products of unit vectors. Here are some handy ways to do the cross product. The “always works if you can do math” method Example:

8. cross products of unit vectors We are going to be doing lots of cross products of unit vectors. Here are some handy ways to do the cross product. The right hand rule method http://en.wikipedia.org/wiki/Cross_product linklink to image

9. cross products of unit vectors We are going to be doing lots of cross products of unit vectors. Here are some handy ways to do the cross product. The “I learned this decades ago and forgot the name for it” method i j k

10. cross products of unit vectors i j k

11. This is example 28.1 in your text. Example: proton 1 has a speed v 0 (v 0 <<c) and is moving along the x-axis in the +x direction. Proton 2 has the same speed and is moving parallel to the x-axis in the –x direction, at a distance r directly above the x-axis. Determine the electric and magnetic forces on proton 2 at the instant the protons pass closest to each other. 1 r x y z 2 v0v0 v0v0 The electric force is E FEFE Homework Hint: this and the next 3 slides! Example: proton 1 has a speed v 0 (v 0 <<c) and is moving along the x-axis in the +x direction. Proton 2 has the same speed and is moving parallel to the x-axis in the –x direction, at a distance r directly above the x-axis. Determine the electric and magnetic forces on proton 2 at the instant the protons pass closest to each other. Skip to next slide.

12. Alternative approach to calculating electric force. This is “better” because we use the concept of field to calculate both of electric and (later) magnetic forces. 1 r x y z 2 v0v0 v0v0 At the position of proton 2 there is an electric field due to proton 1. FEFE E This electric field exerts a force on proton 2.

13. 1 r x y z 2 FEFE To calculate the magnetic force: at the position of proton 2 there is a magnetic field due to proton 1. B1B1 v0v0 v0v0

14. 1 r x y z 2 FEFE B1B1 Proton 2 “feels” a magnetic force due to the magnetic field of proton 1. FBFB v0v0 v0v0 What would proton 1 “feel?” Caution! Relativity overrules Newtonian mechanics! However, in this case, the force is “equal & opposite.”

15. 1 r x y z 2 FEFE B1B1 Both forces are in the +y direction. The ratio of their magnitudes is FBFB v0v0 v0v0 Later we will find that

16. 1 r x y z 2 FEFE B1B1 Thus FBFB v0v0 v0v0 If v 0 =10 6 m/s, then Don’t you feel sorry for the poor, weak magnetic force? What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2. If you don’t like being confused, close your eyes and cover your ears. Or see here, here, and here for a hint about how to resolve the paradox.here What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2. Another nanohuman riding on proton 2 would say “I am at rest, so there is no magnetic force on my proton, even though there is a magnetic field from proton 1.” What if you are a nanohuman, lounging on proton 1. You rightfully claim you are at rest. There is no magnetic field from your proton, and no magnetic force on 2. Another nanohuman riding on proton 2 would say “I am at rest, so there is no magnetic force on my proton, even though there is a magnetic field from proton 1.” This calculation says there is a magnetic field and force. Who is right? Take Physics 2305/107 to learn the answer.

17. Today’s agenda: Magnetic Fields Due To A Moving Charged Particle. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart Law: Magnetic Field due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current- carrying conductor (for example: a long straight wire). Force Between Current-Carrying Conductors. You must be able to begin with starting equations and calculate forces between current- carrying conductors.

18. From the equation for the magnetic field of a moving charged particle, it is “easy” to show that a current I in a little length d l of wire gives rise to a little bit of magnetic field. dB r dldl The Biot-Savart Law I You may see the equation written using If you like to be more precise in your language, substitute “an infinitesimal” for “a little length” and “a little bit of.” I often use ds instead of d l because the script l does not display very well. Biot-Savart Law: magnetic field of a current element

19. Applying the Biot-Savart Law  I ds r  dB Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate.  P

20. Example: calculate the magnetic field at point P due to a thin straight wire of length L carrying a current I. (P is on the perpendicular bisector of the wire at distance a.) ds is an infinitesimal quantity in the direction of dx, so I y r x  dB P ds  x z  a L 

21. I y r x  dB P ds  x z  a L

22. I y r x  dB P ds  x z  a look integral up in tables, use the web,or use trig substitutions web L

23. I y r x  dB P ds  x z  a L

24. I y r x  dB P ds  x z  a When L , or The r in this equation has a different meaning than the r in the diagram! L

25. I y r x  dB P ds  x z  a When L , or The r in this equation has a different meaning than the r in the diagram! L

26. I B r Magnetic Field of a Long Straight Wire We’ve just derived the equation for the magnetic field around a long, straight* wire… …with a direction given by a “new” right- hand rule. *Don’t use this equation unless you have a long, straight wire! linklink to image source

27. Looking “down” along the wire:  I B The magnetic field is not constant (and not uniform). At a fixed distance r from the wire, the magnitude of the magnetic field is constant (but the vector magnetic field is not uniform). The magnetic field direction is always tangent to the imaginary circles drawn around the wire, and perpendicular to the radius “connecting” the wire and the point where the field is being calculated.

28. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . I  A´ C A R Thanks to Dr. Waddill for the use of the diagram. C´ O Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . I see three “parts” to the wire. As usual, break the problem up into simpler parts. A’ to A A to C C to C’ http://autosystempro.com/tag/generators/

29. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . I  A´ C A R C´ O Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . For segment A’ to A: ds

30. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .  A´ C A R C´ O Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . For segment C to C’: ds I

31. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle .  A´ C A R C´ Important technique, handy for homework and exams: The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths. O ds

32. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . For segment A to C:  A´ C A R C´ O I ds

33.  A´ C A R C´ O I ds The integral of ds is just the arc length; just use that if you already know it. We still need to provide the direction of the magnetic field.

34.  A´ C A R C´ O I ds If we use the standard xyz axes, the direction is Cross into. The direction is “into” the page, or . x y z

35.  A´ C A R C´ O I ds Important technique, handy for exams: Along path AC, ds is perpendicular to.

36. Today’s agenda: Magnetic Fields Due To A Moving Charged Particle. You must be able to calculate the magnetic field due to a moving charged particle. Biot-Savart Law: Magnetic Field due to a Current Element. You must be able to use the Biot-Savart Law to calculate the magnetic field of a current- carrying conductor (for example: a long straight wire). Force Between Current-Carrying Conductors. You must be able to begin with starting equations and calculate forces between current- carrying conductors.

37. Magnetic Field of a Current-Carrying Wire “Knowledge advances when Theory does battle with Real Data.”—Ian Redmount It is experimentally observed that parallel wires exert forces on each other when current flows. I1I1 I2I2  F 12 F 21 I1I1 I2I2  F 12 F 21

38. Example: use the expression for B due to a current-carrying wire to calculate the force between two current-carrying wires. I1I1 I2I2  F 12 d L L2L2 L1L1 B2B2  The force per unit length of wire is Homework Hint: use this technique for any homework problems with long parallel wires! x y z

39. I1I1 I2I2  F 12 F 21 d L L2L2 L1L1 B1B1  The force per unit length of wire is B x y z

40. Be able to show that if the currents in the wires are in the opposite direction, the force is repulsive. I1I1 I2I2  F 12 F 21 d L L2L2 L1L1 The official definition of the Ampere: 1 A is the current that produces a force of 2x10 -7 N per meter of length between two long parallel wires placed 1 meter apart in empty space.