lect16 - Physics 227 Lecture 16 Ampere’s Law • Lecture...

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Unformatted text preview: Physics 227: Lecture 16 Ampere’s Law • Lecture 15 review: • • Magnetic field magnitudes for charged particle or current. Ratio of magnetic to electric force for two charged particles. • • Long straight wire: B = μ0I/2πr FM 2 = µ0 ￿0 v FE Parallel wires: F/L = μ0I2/2πr. µ0 q￿ × r vˆ ￿r B (￿) = 4π r 2 vˆ µ0 Id￿ × r lˆ µ0 q￿ × r ￿ (￿) = nAdl = dB r 2 4π r 4π r 2 Monday, October 31, 2011 Physics 227: Exam 2 Information • Note: exam 2: 16 questions covering chapters 25 - 28 • • • • Monday, October 31, 2011 Thursday, Nov 17, 2011, 9:40 PM - 11:00 PM Room assignments: • A-I Arc 103 • J-M SEC 111 • N-R PLH • S-Z Beck Auditorium, Livingston Campus!!! (NOT Hill 114) Anyone with a conflict should contact Prof. Cizewski ASAP Bring pencils, 1 formula sheet w/ anything you want, NO calculators or other electronics needed or allowed! Ampere’s Law vs. Ampere’s Law involves a line integral over a closed path of the magnetic field. It relates the integral to the current through the surface defined by the closed path. Like with Gauss’s Law in electrostatics, we will use it in situations with a symmetry that allows us to calculate the field. Monday, October 31, 2011 • • • Ampere’s Law Let’s do a simple case we know first - a long straight wire w ith magnetic field B = μ0I/2πr. B is constant for fixed r. Thus the path is a circle centered on the wire, in a plane perpendicular to it. B is tangential to the circle. We find the integral is ∫B.dl = BC = (μ0I/2πr)(2πr) = μ0I. • • • Monday, October 31, 2011 If we reverse the direction of the current, the B field changes direction, and the sign of the integral changes. If we change the direction we go around the circle, B.dl and the integral change sign. Thus, we find the integral is ±μ0I, depending on how we choose directions. iClicker: Ampere’s Law with no Current Enclosed What is the magnitude of ∫B.dl around the quartercircle path shown? A. 0. B. μ0I. C. μ0I (r1/r2). D. (π/2) μ0I (r1/r2)2. E. (π/2) μ0I (r2-r1). Numerical integral done in lecture. Also, since no current is inside the loop, the integral has to be 0. Monday, October 31, 2011 What if There are Several Currents? Add them up. Note the direction to use - if the thumb points in the direction of the positive current, the direction around is the way the RH fingers curl. In this case, the current is >0 and the integral will be +μ0I. Positive current out of clock face: integrate around CCW. Positive current into clock face, integrate around CW. Monday, October 31, 2011 A Note on Conservative Forces The electric force is conservative. When a charge moves along some path and returns to a previous position, the energy is the same. The total work done by the electric force vanishes (=0). Mathematically: W= ￿ ￿l q E · d￿ → ￿ ￿l E · d￿ = 0 The magnetic force is always perpendicular to the direction of motion, so necessarily: W= ￿ ￿ Fm · d￿ = 0 l While there is often a symmetry between how we handle electric and magnetic fields, the line integrals of the fields over a closed loop are unrelated, with very different meanings. Monday, October 31, 2011 iClicker: Ampere’s Law for Uniform Current Density What if we have uniform current density inside a wire? How will B vary with r? A. B is 0 inside the wire. B. B is constant inside the w ire. C. B ≈ r. D . B ≈ r 2. E. B ≈ 1/r. Answer derived on next slide. Monday, October 31, 2011 Use Ampere’s Law to Determine the Field of a Uniform Current Density Wire • • • • • Current density in the w ire is J = I/πR2. For a circle of radius r: ∫B.dl = B 2πr = μ0I = μ0 Jπr2 ➪ B = μ0Jr/2 = μ0Ir/2πR2. The field inside the w ire varies with r. Outside the wire, we again have B = μ0I/2πr. Note the units: B = μ0 x current / length Monday, October 31, 2011 Field of a Uniform Current Density Wire Monday, October 31, 2011 What is the Magnetic Field of a Solenoid? Solenoid: a group of parallel, coaxial coils, with a current that flows through all of them. You can see from the drawing that as you get more coils, the field inside the solenoid grows, while the field outside decreases. In the limit of an infinitely long solenoid, the field inside is constant while the field outside vanishes. Apply Ampere’s Law as shown above to the right... Monday, October 31, 2011 Is it ``Obvious’’ that the Field Outside an Infinitely Long Solenoid Vanishes? The ``upper’’ wires lead to a field in the plane of the board that is to the right above the wires to the left below the wires The ``lower’’ wires lead to a field in the plane of the board that is to the left above the wires to the right below the wires The contributions cancel above and below the solenoid, but add inside it. Because the field vanishes outside, the field inside cannot diverge - it must remain parallel and constant. Monday, October 31, 2011 iClicker: What is the Magnetic Field...? ... inside an infinitely long solenoid? Use n loops/m, current I A, radius r m, an Amperian loop on length L, and ∫B.dl = μ0I. A. It depends on where points a&b are inside the solenoid. B. B = μ0nI. C. B = μ0nIL. D. B = μ0nI/L. E. B = μ0nI/L2. Monday, October 31, 2011 Previous slides indicated the field is constant. It cannot depend on the length of the loop L that you draw on the infinite solenoid. Also recall that the units of B are μ0 x current / length. Since n has units of 1/length, only answer B has the right units. What is the Magnetic Field of a ``Short’’ Solenoid? The field is not constant, so an integration is needed, rather than Ampere’s Law. Monday, October 31, 2011 What is the Magnetic Field for a toroid? The field is constant. Use Ampere’s Law with n loops/m, current I A, radius r m: ∫B.dl = μ0I ➪ B 2πr = μ0 In2πri ➪ B = μ0nIri/r. We can use N = 2πrin to obtain B = μ0NI/2πr. In the limit that ri ≈ ro, we also have ri ≈ r, and B = μ0nI, the same as for an infinitely long solenoid. Monday, October 31, 2011 Thank you. See you Thursday. Monday, October 31, 2011 ...
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This document was uploaded on 11/03/2011 for the course PHSYICS 227 at Rutgers.

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