Physics 227: Lecture 16 Ampere s Law

Transcription

1 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 = μ 0I/2πr Parallel wires: F/L = μ 0I 2 /2πr. F M F E = µ 0 0 v 2 B(r )= µ 0 4π qv ˆr r 2 d B(r )=nadl µ 0 4π qv ˆr r 2 = µ 0 4π Id l ˆr r 2

2 Physics 227: Exam 2 Information Note: exam 2: 16 questions covering chapters 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!

3 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.

4 Ampere s Law Let s do a simple case we know first - a long straight wire with magnetic field B = μ 0 I/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 = (μ 0 I/2πr)(2πr) = μ 0 I. 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 ±μ 0 I, depending on how we choose directions.

5 iclicker: Ampere s Law with no Current Enclosed What is the magnitude of B. dl around the quartercircle path shown? A. 0. B. μ 0 I. C. μ 0 I (r 1 /r 2 ). D. (π/2) μ 0 I (r 1 /r 2 ) 2. E. (π/2) μ 0 I (r 2 -r 1 ). Numerical integral done in lecture. Also, since no current is inside the loop, the integral has to be 0.

6 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 +μ 0 I. Positive current out of clock face: integrate around CCW. Positive current into clock face, integrate around CW.

7 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 = q E d l E d l =0 The magnetic force is always perpendicular to the direction of motion, so necessarily: W = F m d l =0 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.

8 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 wire. C. B r. D. B r 2. E. B 1/r. Answer derived on next slide.

9 Use Ampere s Law to Determine the Field of a Uniform Current Density Wire Current density in the wire is J = I/πR 2. For a circle of radius r: B. dl = B 2πr = μ 0 I = μ 0 Jπr 2 B = μ 0 Jr/2 = μ 0 Ir/2πR 2. The field inside the wire varies with r. Outside the wire, we again have B = μ 0 I/2πr. Note the units: B = μ 0 x current / length

10 Field of a Uniform Current Density Wire

11 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...

12 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.

13 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 = μ 0 I. A. It depends on where points a&b are inside the solenoid. B. B = μ 0 ni. C. B = μ 0 nil. D. B = μ 0 ni/l. E. B = μ 0 ni/l 2. 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 μ 0 x current / length. Since n has units of 1/length, only answer B has the right units.

14 What is the Magnetic Field of a ``Short Solenoid? The field is not constant, so an integration is needed, rather than Ampere s Law.

15 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 = μ 0 I B 2πr = μ 0 In2πr i B = μ 0 nir i /r. We can use N = 2πr i n to obtain B = μ 0 NI/2πr. In the limit that r i ro, we also have r i r, and B = μ 0 ni, the same as for an infinitely long solenoid.

16 Thank you. See you Thursday.