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Announcements Exam 2 is one week from tomorrow. Contact me by the end of the Wednesday’s lecture if you have special circumstances different than for exam 1. Exam 2 will cover chapters 24.3-27. Physics 2135 Test Rooms, Spring 2016: Instructor Dr. Hale Dr. Kurter Dr. Madison Mr. Noble Dr. Parris Mr. Upshaw Sections G, J D, F H, N B, E K, M A, C, L Special Accommodations Exam is from 5:00-6:00 pm! Room 125 BCH 104 Physics 199 Toomey B-10 Bert. G-31 EECH G-3 Schrenk Testing Center Know the exam time! Find your room ahead of time! If at 5:00 on test day you are lost, go to 104 Physics and check the exam room schedule, then go to the appropriate room and take the exam there. We’ve been working with the effects of magnetic fields without considering where they come from. Today we learn about sources of magnetic fields. 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 currentcarrying 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 currentcarrying conductors. Magnetic Field of a Moving Charged Particle Let’s start with the magnetic field of a moving charged particle. r It is experimentally observed that a moving point charge q gives rise to a magnetic field. B r̂ + v μ0 qv rˆ B= . 2 4π r 0 is a constant, and its value is 0=4x10-7 T·m/A 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. 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 C = A B Example: ˆi C = det A x B x ˆj Ay By ˆi ˆj ˆk ˆj = det 0 0 0 -1 kˆ Av z B z kˆ ˆ 1 = i 0 - -1 = ˆi 0 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 link to image http://en.wikipedia.org/wiki/Cross_product 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 i j k ˆi ˆj =kˆ cross products of unit vectors i j k i j k ˆj ˆi = -kˆ i j k i j k ˆi kˆ = -jˆ i j k i j k ˆj kˆ = - -iˆ = ˆi Example: proton 1 has a speed v0 (v0<<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. y This is example 28.1 in your text. FE Skip to next slide. The electric force is v0 1 q1q2 ˆ FE = r 2 4 r E r 1 2 1 e ˆ FE = j 2 4 r Homework Hint: this and the next 3 slides! 2 z r̂ v0 x 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. At the position of proton 2 there is an electric field due to proton 1. y 1 q1 ˆ 1 eˆ E1 = r= j 2 2 4 r 4 r FE E v0 This electric field exerts a force on proton 2. 1 eˆ 1 e2 ˆ FE = qE1 = e j= j 2 2 4 r 4 r 2 r 1 z r̂ v0 x To calculate the magnetic force: at the position of proton 2 there is a magnetic field due to proton 1. q1 v1 rˆ B1 = 4 r 2 ev 0 ˆi ˆj B1 = 4 r2 y FE v0 ev 0 ˆ B1 = k 2 4 r 2 B1 r 1 z r̂ v0 x Proton 2 “feels” a magnetic force due to the magnetic field of proton 1. FB = q2 v2 B1 FB = ev 0 ev 0 ˆ ˆ i k 2 4 r y FE v0 e2 v 02 ˆ FB = j 2 4 r 2 FB B1 What would proton 1 “feel?” r 1 Caution! Relativity overrules Newtonian mechanics! However, in this case, the force is “equal & opposite.” z r̂ v0 x Both forces are in the +y direction. The ratio of their magnitudes is e 2 v 02 2 4 r FB = FE 1 e 2 2 4 r y FE FB = v 20 FE v0 2 FB B1 Later we will find that r 1 1 = 2 c z r̂ v0 x FB v 20 Thus = 2 FE c If v0=106 m/s, then 10 FB -5 = 1.11 10 FE 3 108 2 6 2 y FE Don’t you feel sorry for the poor, weak magnetic force? v0 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. B1 If you don’t like being confused, Another nanohuman riding on proton 2 would say “I am close your eyes at rest, so there is no magnetic force on my proton, and iscover your even though there a magnetic fieldears. from proton 1.” This calculation says there is a magnetic field and force. Who is here, right? Take Physics 2305/107 totolearn Or see here, and here for a hint about how resolvethe the answer. paradox. 2 FB r 1 z r̂ v0 x 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 currentcarrying 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 currentcarrying conductors. Biot-Savart Law: magnetic field of a current element 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 dl of wire gives rise to a little bit of magnetic field. r dB r̂ dl I μ 0 I d rˆ dB = 4π r 2 The Biot-Savart Law 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 dl because the script l does not display very well. You may see the equation written using r = r rˆ . Applying the Biot-Savart Law P dB r μ0 I ds rˆ r ˆ dB = where r = 2 4π r r ds rˆ = ds rˆ sin = ds sin because rˆ =1 r̂ ds I dB = μ 0 I ds sin θ 4π r2 B = dB Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate. 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.) y dB P r r̂ ds μ 0 I ds rˆ dB = 4π r 2 a z x x I ds rˆ = ds sinθ kˆ dB = L μ 0 I ds sinθ 4π r2 ds is an infinitesimal quantity in the direction of dx, so μ 0 I dx sinθ dB = 4π r2 sinθ = a r 2 r r̂ ds x L μ 0 I dx sinθ 4π r2 μ 0 I dx a μ 0 I dx a dB = = 3 4π r 4π x 2 + a2 3/2 a z dB = r = x +a y dB P 2 x I μ 0 I dx a -L/2 4π 2 2 3/2 x +a B= L/2 μ 0I a L/2 dx B= 4π -L/2 x 2 + a2 3/2 μ 0I a L/2 dx B= 4π -L/2 x 2 + a2 3/2 y dB P r r̂ ds x look integral up in tables, use the web,or use trig substitutions a x z I L dx x 2 2 3/2 +a = x a x +a 2 L/2 μ 0I a x B= 4π a2 x 2 + a2 1/2 -L/2 μ 0I a L/2 = 4π a2 L/2 2 +a2 1/2 -L/2 1/2 2 2 2 a -L/2 +a 2 2 1/2 y dB P r r̂ ds x μ 0I a 2L/2 B= 1/2 4π a2 L2 /4 +a2 a z L x I B= μ 0I L 1 4πa L2 /4 + a2 1/2 μ 0I L 1 B= 2πa L2 + 4a2 μ 0I B= 2πa 1 4a2 1+ 2 L y dB P r r̂ ds x a z x μ 0I B= 2πa 1 4a2 1+ 2 L I L When L, B = μ 0I . 2 πa μ 0I or B = 2 πr The r in this equation has a different meaning than the r in the diagram! y dB P r r̂ ds x a z x μ 0I B= 2πa 1 4a2 1+ 2 L I L When L, B = μ 0I . 2 πa μ 0I or B = 2 πr The r in this equation has a different meaning than the r in the diagram! Magnetic Field of a Long Straight Wire We’ve just derived the equation for the magnetic field around a long, straight* wire… μ0 I B= 2 πr I B r …with a direction given by a “new” righthand rule. link to image source *Don’t use this equation unless you have a long, straight wire! Looking “down” along the wire: The magnetic field is not constant (and not uniform). I B 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. 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´ I see three “parts” to the wire. A’ to A A A to C O C I C´ R Thanks to Dr. Waddill for the use of the diagram. C to C’ As usual, break the problem up into simpler parts. http://autosystempro.com/tag/generators/ 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 . μ 0 I ds rˆ dB = 4π r 2 A´ I ds r̂ A For segment A’ to A: C´ O ds rˆ = ds rˆ sin 0 = 0 C R dB A'A = 0 B A'A = 0 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 . μ 0 I ds rˆ dB = 4π r 2 A´ A r̂ ds O C R I C´ For segment C to C’: ds rˆ = ds rˆ sin 180 = 0 dB CC' = 0 B CC' = 0 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´ r̂ A Important technique, handy for homework and exams: ds r̂ O C R ds C´ The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to r̂ along those paths. 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 . μ 0 I ds rˆ dB = 4π r 2 A´ A For segment A to C: ds r̂ O C I C´ ds rˆ = ds rˆ sin 90 = 0 = ds 1 1 R = ds dB AC μ 0 I ds = dB AC = 4π R 2 A´ A μ 0 I ds dB AC = 4π R 2 r̂ O μ 0I ds B AC = dB AC = 2 4π R arc arc ds C I C´ μ 0I B AC = 4πR 2 R μ 0I B AC = 4πR 2 ds arc The integral of ds is just the arc length; just use that if you already know it. R dθ arc μ 0I B AC = dθ 4πR arc μ 0I B AC = θ 4πR We still need to provide the direction of the magnetic field. A´ A Cross ds into r̂ . The direction is “into” the page, or . ds r̂ O C I C´ If we use the standard xyz axes, ˆ the direction is -k. y R z B = B A'A + B AC + B CC' = - μ 0 Iθ ˆ k 4πR x Important technique, handy for exams: A´ A Along path AC, ds is perpendicular to r̂ . ds r̂ O C R I C´ ds rˆ = ds rˆ sin 90 ds rˆ = ds 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 currentcarrying 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 currentcarrying conductors. Magnetic Field of a Current-Carrying Wire It is experimentally observed that parallel wires exert forces on each other when current flows. I1 I2 F12 F21 I1 I2 F12 F21 “Knowledge advances when Theory does battle with Real Data.”—Ian Redmount Example: use the expression for B due to a current-carrying wire to calculate the force between two current-carrying wires. d F12 = I1L1 B2 I1 L ˆB = μ 0 I2 kˆ 2 2 πd The force per unit length of wire is Homework Hint: use this technique for any homework problems with long parallel wires! F12 B2 μ I F12 = I1Ljˆ 0 2 kˆ 2 πd μ 0 I1I2L ˆ F12 = i 2 πd I2 L1 y z x F12 μ 0 I1I2 ˆ = i. L 2πd L2 B d F21 =I2L2 B1 I1 L μ 0 I1 ˆ B1 = k 2 πd The force per unit length of wire is F12 F21 B1 μ 0 I1 ˆ ˆ F21 = I2Lj k 2πd μ 0 I1I2L ˆ F21 = i 2 πd I2 L1 y z x μ II F21 = - 0 1 2 ˆi. L 2 πd L2 Be able to show that if the currents in the wires are in the opposite direction, the force is repulsive. F12 = F21 = μ 0 I1I2L 2πd d L I1 I2 F12 F21 4 π 10-7 I1I2L L -7 F12 =F21 = = 2 10 I1I2 2πd d 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. L1 L2