Chap28_Notes

# Chap28_Notes - Physics 184 Physics for Scientists Engineers...

This preview shows pages 1–8. Sign up to view the full content.

10/20/2011 Physics for Scientists & Engineers 2 1 Physics 184 Physics for Scientists & Engineers 2 Fall Semester 2011 Chapter 28: Magnetic Fields due to moving charges

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

View Full Document
10/20/2011 Physics for Scientists & Engineers 2 2 Electric Field Magnetic Field Let’s address the problem of calculating magnetic fields generated by moving charges We calculated the electric field in terms of the electric charge using the form where dq is a charge element and is a unit vector in the radial direction 2 0 1 ˆ 4 dq dE r r  ˆ r
10/20/2011 Physics for Scientists & Engineers 2 3 Magnetic Fields The magnetic field has an added complication because the current that produces the magnetic field has a direction As opposed to charge which is a scalar We can write the magnetic field produced by a current element ids as This formula is the Biot-Savart Law This is the same as for the electric field, except 0 is the magnetic permeability of free space whose value is 7 0 Tm 4 10 A   2 0 3 0 ˆ 4 4 r r s id r r s B d 0 0 / 1 , ˆ ˆ  r s r dq

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

View Full Document
10/20/2011 Physics for Scientists & Engineers 2 4 Magnetic Fields (2) The direction of the magnetic field produced by the current element is perpendicular to both the radial direction and to the current element The magnitude of the magnetic field is given by Where is the angle between the radial direction and the current element We will now calculate the magnetic field for various current element distributions 2 0 sin 4 r ids dB
10/20/2011 Physics for Scientists & Engineers 2 5 Magnetic Field from a Long, Straight Wire First application: calculate the magnetic field from an infinitely long straight wire Demo We calculate the magnetic field dB at a point P at a distance r from the wire as illustrated below The magnitude of the magnetic field will be given by the Biot-Savart Law and the direction will be out of the page

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

View Full Document
10/20/2011 Physics for Scientists & Engineers 2 6 Magnetic Field from a Long, Straight Wire (2) We will calculate the magnetic field from the right half of the wire and multiply by two to get the magnetic field from the whole wire The magnitude of the magnetic field from the right side of the wire is given by Recall that for the electric field from a line of charge: 00 22 0 0 0 sin sin 42 i ids ds B dB rr    r E r dx dE E a a a 0 0 2 0 0 2 lim cos 2 2 
10/20/2011 Physics for Scientists & Engineers 2 7 Magnetic Field from a Long, Straight Wire (3) We can relate r, s, and by Substituting we get use integral identity Carrying out this integral we get Magnetic field at a perpendicular distance from a long, straight wire carrying a current is   0 3/ 2 0 22 2 i r ds B sr   0 1/ 2 2 0 1 2 i rs B r    B ( r ) 0 i 2 r   0 1/ 2 0 2 s i s r   

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

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

## This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.

### Page1 / 65

Chap28_Notes - Physics 184 Physics for Scientists Engineers...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online