241-28_Lec-15 - Physics 241 Lecture 15 Y. E. Kim October...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 241 Lecture 15 Y. E. Kim October 19, 2010 Chapter 28, Sections 3, 4, 6 October 18, 2010 University Physics, Chapter 25 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
October 18, 2010 University Physics, Chapter 28 2 Ampere’s Law is Where the integral is carried out around an Amperian loop and i enc is the current enclosed by the loop As an example of Ampere’s Law , consider the five currents shown below The currents are perpendicular to the plane shown An Amperian loop is represented by the red line 0 enc B ds i   0 1 2 3 B ds i i i 0 3 4 ids r dB r Biot-Savart Law Ampere’s Law (1)   1 2 3 enc i i i i
Background image of page 2
October 18, 2010 University Physics, Chapter 28 3 Ampere’s Law (2) This loop encloses i 1 , i 2 , and i 3 and excludes i 4 and i 5 A direction of integration is shown above along with the resulting magnetic field The sign of the contributing currents can be determined using a right hand rule by pointing your fingers along the direction of integration Your thumb will indicate the positive direction of the contributing currents   enc i 0 enc B ds i   enc i   enc i ds
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
October 18, 2010 University Physics, Chapter 28 4 Magnetic Field inside Long Wire (1) Consider a current i flowing out of the page in a wire with a circular cross section of radius R This current i is uniformly distributed over the cross sectional area of the wire To find the magnetic field , we use an Amperian loop with radius r represented by the red circle The magnetic field is tangential to this Amperian loop so we can write the left side of Ampere’s Law as 2 B ds B ds B r   0 enc B ds i B B
Background image of page 4
October 18, 2010 University Physics, Chapter 28 5 Magnetic Field inside Long Wire (2) The right hand side of Ampere’s Law contains the enclosed current which can be calculated from the ratio of the area of the Amperian loop to the cross sectional area of the wire Equating the left and right sides using Ampere’s law we get or and i enc i A loop A wire i r 2 R 2 B 2 r   0 i r 2 R 2 0 2 ( ) , 2 i B r r r R R     enc i 0 enc B ds i 0 ( ) , 2 i B r r R r  
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
October 18, 2010 University Physics, Chapter 28 6 Magnetic Field inside Long Wire (3) 0 2 ( ) , 2 i B r r r R R     0 ( ) , 2 i B r r
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 25

241-28_Lec-15 - Physics 241 Lecture 15 Y. E. Kim October...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online