advertisement

lec Amperes Law.notebook March 21, 2016 Ampere's Law The sum of magnetic fields around multiple wires gives us: ΣΒ∆l = µοΣΙ (l is the length of the path) As ∆l­­> 0 the sum becomes an integral around a closed path: ∮Bdl = µοΣΙ The more current that passes through the 'magnetic loop' the stronger the magnetic field at the position of the loop. Like Gauss Law, we choose regions where B is perpendicular or parallel to the chosen encompassing path. (I is the current passing through the closed path, i.e. loop) Unlike Gauss Law, we're taking the loop parallel to the magnetic field, not perpendicular! Ampere's Law The sum of magnetic fields around multiple wires gives us: ΣΒ∆l = µοΣΙ (l is the length of the path) As ∆l­­> 0 the sum becomes an integral around a closed path: ­­> ∮Bdl = µοΣΙ f Integration Path o dl integrates to the length of the closed path, just as dA (in Gauss' Law) integrates to the Gaussian surface lec Amperes Law.notebook March 21, 2016 Ampere's Law The sum of magnetic fields around multiple wires gives us: ΣΒ∆l = µοΣΙ (l is the length of the path) As ∆l­­> 0 the sum becomes an integral around a closed path: ∮Bdl = µοΣΙ For example, a circle around a single wire. B is a constant and ∆l around a circle is the circumference of the closed path. Therefore ΣΒ∆l = µοΣΙ yields Β2πr = µοΙ or Β = µοΙ / 2πr (this is our standard equation) Ampere's Law can be used to derive equations for various setups... In a current carrying wire with radius R, and a current I, what is the magnetic field at point P (at radius r<R)? . r P I R lec Amperes Law.notebook March 21, 2016 Two parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at different locations. T . . P d d Β = µοΙ / 2πr . Q Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. d 2d Β = µοΙ / 2πr 3d 4d 5d lec Amperes Law.notebook March 21, 2016 Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. d 2d Β = µοΙ / 2πr B=(µοΙ / 2πd) 3d 4d 5d Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. d 2d Β = µοΙ / 2πr B=(µοΙ / 4πd) 3d 4d 5d lec Amperes Law.notebook March 21, 2016 Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. d 2d Β = µοΙ / 2πr B=(µοΙ / 6πd) 3d 4d 5d Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. Β = µοΙ / 2πr and so on... add the fields... d 2d Bnet=B1+B2+B3+... Bnet=(µοΙ / 2π)(1/d+1/2d+...) 3d 4d 5d lec Amperes Law.notebook March 21, 2016 The Ideal Solenoid Here are many wires passing through the Ampere loop. Because of symmetry, and the fact that magnetic field lines cannot cross, they must run parallel to the solenoid. ∮Bdl = µοΣΙ I is the current in each wire, BL = µο∗N∗Ιο N is the # of turns contained in the loop. B = µο∗(N/L)∗Ιο so B = µο∗n∗Ιο Where n is the turns per unit length. The Ideal Solenoid Here are many wires passing through the Ampere loop. Because of symmetry, and the fact that magnetic field lines cannot cross, they must run parallel to the solenoid. B = µο∗n∗Ιο Note that this equation is independent of the position within the solenoid. lec Amperes Law.notebook March 21, 2016 Relate Current to Velocity... A beam of electrons has a current of 2A. There are 106 electrons per meter of beam. What is the average speed of the electrons in the beam? Relate Current to Velocity... A beam of electrons has a current of 2A. There are 1012 electrons per meter of beam. What is the average speed of the electrons in the beam? 2A = 2C/sec 1 electron = 1.6E­19 C ∴ 2C = 1.25E19 electrons If 1.25E19 electrons pass every second, and there are 1012 electrons per meter, then 1.25E6 meters worth of electrons pass every second. So velocity = d/t = 1.25E10 m/s lec Amperes Law.notebook March 21, 2016 Relate Current to Velocity... A beam of electrons has a current of 2A. There are 1012 electrons per meter of beam. What is the average speed of the electrons in the beam? 2A = 2C/sec 1 electron = 1.6E­19 C ∴ 2C = 1.25E19 electrons If 1.25E19 electrons pass every second, and there are 1012 electrons per meter, then 1.25E6 meters worth of electrons pass every second. So velocity = d/t = 1.25E10 m/s 2A 1e 1.6E­19C 1m 1012e 1C/s 1A = 1.25E6m s What would it take to have a smaller orbital radius? lec Amperes Law.notebook March 21, 2016 What would it take to have a smaller orbital radius? F=QvB=mv2/r QBr=mv smaller r, smaller v, or bigger Q or B Biot Savart's Law This is another method of deriving magnetic fields near a current­ carrying wire. Any segment of a current­carrying wire will create a magnetic field at a distance, and this field strength decreases with distance, as expected. I These fields can overlap, creating a stronger magnetic field at given points. I lec Amperes Law.notebook March 21, 2016 Biot Savart's Law Any portion of a wire can contibute to the magnetic field at a given location, but the contribution drops off as the angle moves away from perpendicular. Thus a 'cross­product' is used. ^ / (4π r2) dB = μo I dL x r I With the cross­product, we can use trig: dB = μo I dL sinθ / (4π r2) I Biot Savart's Law dB = μo I dL sinθ / (4π r2) For a curved wire (a quarter circle) with current = I, radius = R. lec Amperes Law.notebook March 21, 2016 dB = μo I dL sinθ / (4π r2) Biot Savart's Law For a curved wire (a quarter circle) with current = I, radius = R. dB = μo I dL sinθ / (4π R2) 2 ∮dB = μo I sin90 / (4π R ) ∫dL θ = 90 ∫dL = πR/2 for quarter circle B = μoI (πR/2)/ (4π R2) B = μoIR/ (8 R2) B = μoI/ (8 R) (what would be the B in a full loop of wire?) dB = μo I dL sinθ / (4π r2) Biot Savart's Law For a long straight wire, each segment of wire contributes a little bit to the magnetic field at point p. The closer & more perpendicular, the greater the contribution. P θ I lec Amperes Law.notebook March 21, 2016 Biot Savart's Law dB = μo I dL sinθ / (4π r2) r is not independent of θ, so must be expressed in terms of θ. sinθ = y/r, where 'y' is the component of 'r' perpendicular to the wire. P θ I Biot Savart's Law dB = μo I dL sinθ / (4π r2) r = y/sinθ dB = μo I dL sinθ / (4π (y/sinθ)2) dx = y dθ / sin2θ <­ this is from a LOT of substitutions (see p781) π dB = μo I / (4π y) ∫sinθ dθ 0 π dB = μo I / (4π y) (­cosθ| 0 dB = μo I / (4π y) (­(­1) ­ (­1)) dB = μo I / (2π y) (this is our standard equation) lec Amperes Law.notebook March 21, 2016 Done. The following are post­lecture, post­handout, pre­test, practice problems. 2 parallel, current carrying wires 6A P 1 o 1m 2m 3m P 2 o 4m P 3 o 3A What is the magnetic field at each location, P1, P2, P3? Redrawing an 'end­view' might help. What is the force on the bottom wire? L = 8m lec Amperes Law.notebook March 21, 2016 2 current carrying wires 4A 1.5m P o 3m ΧΧΧΧ 4A ΧΧΧΧΧ ΧΧΧΧ What is the magnetic field location Po? Redrawing a 'top­view' might help. What is the force on the bottom wire? 2 current carrying wires 4A 1.5m P o 3m ΧΧΧΧ 4A ΧΧΧΧΧ ΧΧΧΧ If a positively charged particle of 1μC is traveling through point P, coming out of the screen, what is the force acting on it if it's travelling 5E6m/s. lec Amperes Law.notebook March 21, 2016 BP1 = 1E­6 T BP2 = 3E­7 T BP3 = 2E­7 T F = 7.2E­6 N BP = 7.5E­6 T at 45o behind board (rt) Fwire = 0N Fcharge = 2.65E­6N up