lec24_030507_-_Amperes_Law

# lec24_030507_-_Amperes_Law - Exam II Results y d An...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exam II Results y d An infinite wire: r I Ampere's Law o I 2x I A I-carrying loop: d y P B B= x R r B P x An infinite solenoid: = IA = IR 2 B= 4 x 2 + R 2 ( o 2 ) 3/ 2 If you do not have a score in LON-CAPA, something is amiss LONwith your ID#! Please come see me to fix it. Monday, 5 March 2007 1 Bx = o nI Magnetic flux: Monday, 5 March 2007 B = B dA B dA = 0 2 d I R Ampere's Law Ampere' B d = o I thru Example: Long Solenoid 3 B d = o I encl 2 B By symmetry, the magnetic field lines will be circular around the wire, as is the d : B d = B d = o I thru By symmetry, the magnetic field will have the same strength all along the path: B d = B d = B d = o I thru 4 1 d 1. Choose a path. 2. Left hand side of Ampere's law. Ampere' Bd = Bd + Bd + Bd + Bd 1 2 3 4 B d = o I thru Monday, 5 March 2007 B(2R ) = o I thru B= o I thru 2R 3 Monday, 5 March 2007 4 1 Example: Long Solenoid Bd = Bd + Bd + Bd + Bd 1 3 Example: Long Solenoid Bd = Bd + Bd + Bd + Bd 1 3 2 3 4 2 3 4 Bd = 0 1 2 Bd = Bd 3 2 Bin 4 1 d Bout Bin 1. Choose a path. Bfar out = 0 Bd = 2 out Bd + Bd Bout d in 4 1 d = Bin d 3 2. Left hand side of Ampere's law. Ampere' =0 Bin d 1. Choose a path. 2. Left hand side of Ampere's law. Ampere' 3. Right hand side of Ampere's law. Ampere' = Bin d 3 Bd = Bd + Bd 4 Monday, 5 March 2007 = Bin d number of turns/length in out 5 =0 o I encl = o I total = o NI = o (nd )I Monday, 5 March 2007 number of turns 6 Example: Long Solenoid o I encl = o (nd )I B d = B d = Bin d 3 3 Example: Toroid d Amperian loop! B d = o I encl 2 Bin 4 Bin d = o (nd )I Bin = o nI B & d are always parallel : B d = o I encl 1 d 1. Choose a path. 2. Left hand side of Ampere's law. Ampere' 3. Right hand side of Ampere's law. Ampere' 4. Ampere's law! Ampere' Monday, 5 March 2007 Bd = Bd B is constant everywhere on a path : 7 Monday, 5 March 2007 =B d = B(2r ) 8 2 Example: Toroid B d = o I encl R2 R1 Example: Toroid B d = o I encl R2 R1 B d = B(2r ) B d = B(2r ) Path 1: r < R1 B inside = 0 B d = o I encl B(2r ) = o I encl B outside Path 1: r < R1 B d = o I encl B(2r ) = o I encl I encl = 0 9 Path 3: r > R2 I encl = 0 10 B Monday, 5 March 2007 inside =0 Monday, 5 March 2007 =0 Example: Toroid B d = o I encl R2 R1 J =D Example: Non-uniform Current Density Nonr R 3 B d = o I encl B d Path 2: B d = o I encl B(2r ) = o I encl Bin between B d = B(2r ) Path 1: r < R1 B inside = 0 Path 3: r > R2 B outside = 0 I encl = NI Bsolenoid = o nI 11 R r d B B & d are always parallel : Bd = Bd =B d = B(2r ) R1 < r < R2 B is constant everywhere on a path : Monday, 5 March 2007 NI = o 2r Monday, 5 March 2007 12 3 Example: Non-uniform Current Density Nonr J =D R 3 Example: Non-uniform Current Density Nonr J =D R 3 B d = o I encl B d B d = o I encl B d B d = B(2r ) R R B d d r r d B dr I encl = dI = JdA = d r dr d r Monday, 5 March 2007 D 0 r3 R3 (2rdr ) 13 r d Monday, 5 March 2007 = 2D r 5 R3 5 Inside: r < R 2D r 5 B(2r ) = o 3 R 5 r r3 Binside = o D 3 r = o J 5 R 5 Outside: r > R R2 2D R 5 = o 2D o I encl = o 3 = o I total R 5 5 B(2r ) = o I total Boutside = o I total 2r Infinite Wire!! 14 B d = B(2r ) 2D r 5 o I encl = o 3 R 5 Magnetic Materials I: Ferromagnetism A ferromagnetic material is one in which the electronic spins are aligned in relatively large domains (~ 100 m). Examples of these material include iron and steel. Magnetic Materials II: Paramagnetism Paramagnetic materials also have net electronic spins (and thus magnetic dipole moments). They are, however, randomly oriented on an individual basis. When placed in a large external magnetic field, the spins, and magnetic dipole magnetic moments, of the domains all line up with the applied field. As a result, their alignment in a strong external magnetic field does not produce as large an increase in the overall magnetic field in the material (B ~ 1.00002Bo). Examples include Aluminum, Magnesium, and Tungsten. This alignment can result in large increases in the magnetic field in the field material (B ~ 1000Bo to 100000Bo). Permanent magnets are made from ferromagnetic materials whose spins have spins all been fixed in one direction. Monday, 5 March 2007 15 Monday, 5 March 2007 16 4 Magnetic Materials III: Diamagnetism Diamagnetic materials have no net electronic spin (and thus no magnetic dipole moments). Therefore, you would think that there is nothing to align to an external magnetic field. However, in the presence of a large external magnetic field, a magnetic dipole moment will be induced, or created, in such a induced, way as to oppose the external field. The net result is to reduce the magnetic field in the diamagnetic diamagnetic material by a small amount. Usually, in materials like Copper, Diamond, and Gold, B ~ 0.99995 Bo. In a perfectly diamagnetic material, like a superconductor, the magnetic field inside is ZERO! Monday, 5 March 2007 17 Recap on Magnetics You are responsible for all of Chapter 27 Magnetic Force & Magnetic Field We did not, and will not, cover Section 28.1 You are responsible for the rest of Chapter 28 Source of Magnetic Field Note: The previous 3 slides are all you need to know for Section 28.8 Monday, 5 March 2007 18 5 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online