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## Electromagnetism

by: Miss Lucienne Hamill

12

0

3

# Electromagnetism PHY 712

Miss Lucienne Hamill
WFU
GPA 3.91

Natalie Holzwarth

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COURSE
PROF.
Natalie Holzwarth
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Physics 2

This 3 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 712 at Wake Forest University taught by Natalie Holzwarth in Fall. Since its upload, it has received 12 views. For similar materials see /class/230730/phy-712-wake-forest-university in Physics 2 at Wake Forest University.

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Date Created: 10/28/15
February 267 2008 Notes for Lecture 17 Derivation of the hyper ne interaction Magnetic dipole eld These notes are very similar to the notes on the electric dipole eld The magnetic dipole moment is de ned by 1 m idsrr gtlt Jr 1 with the corresponding potential 0 m gtlt r A 7 2 r 4W 2 lt gt and magnetost atic eld Bmr 7 8 m63r 3 The rst terms come form evaluating V gtlt A in Eq 2 The last term of the eld expression follows from the following derivation We note that Eq 3 is poorly de ned as r a 07 and consider the value of a small integral of Br about zero For this purpose7 we are supposing that the dipole m is located at r 0 In this case we will approximate Br 0 m We Brd3rgt 631 4 First we note that Brd3r R2 7 i gtlt Ar do 5 792 77R This result follows from the divergence theorm v Vd3r VdA 6 vol surface In our case7 this theorem can be used to prove Eq 5 for each cartesian coordinate of V gtlt A since VXAEEV gtlt ASV gtltA22V gtltA Notethat V gtlt A 7V 2 gtlt A and that we can use the Divergence theorem with V E 2 gtlt AI39 for the x7 component for example 3 i V01Vx gtlt Ad r A x gtlt A rdA A gtlt r di 7 Therefore7 3 i i 2 LBW gtlt Ad r 7 LR A gtlt r xx yy zzdA R LR r gtlt AdQ s which is identical to Eq We can use the identity as in Lecture Notes 157 f39 47139 rlt d i if 9 lr 7 r l 3 T r Now7 expressing the vector potential in terms of the current density E 3 Jr A1 A 47Td r ri W 10 the integral over 9 in Eq 5 becomes 47TR2 no r 2 7 7 3 RTRrxAdQ 3 4WdrrirXJr 11 If the sphere R contains the entire current distribution7 then rgt R and rlt r so that 11 becomes dgr I39 gtlt Jr E sl m 12 3 47139 3 47139 which thus justi es the delta function contribution in Eq 3 and results so called Fermi contact77 contribution in the hyper ne interaction R2 f gtlt Ad 2 7 Magnetic eld due to electrons in the Vicinity of a nucleus In Lecture Notes 167 we showed that the current density associated with an electron in a bound state of an atom as described by a quantum mechanical wavefunction bnlmxr can be written A iehmla JOquot 7 Wan 112 13 mar sint9 In the following7 it will be convenient to represent the azimuthal unit vector 3 in terms of cartesian coordinates 1 A z gtlt r 1 7s1n xcos y 14 7quot sin 039 The vector potential for this current density can be written 0 ch 2 gtlt Iquot lwnlm 1quot12 Altrgt 77 ml d3Tlr 7 r l T Z sin20 15 We want to evaluate the magnetic eld B V gtlt A in the vicinity of the nucleus 1 a 0 Taking the curl of the Eq 157 we obtain 0 6h 1quot i Iquot X 2 X 1quot lwnlmxr B0 if d3 16 139 4W me m1 7 139 7 IJ g Tz Sinz 6 Evaluating this expression with 1 a 07 we obtain 2 0 6h 3 A X Z X 1quot lzbnzmxr B 0 777 m d r 17 47139 mg l r 3 r 2 sin2 0 Expanding the cross product and expressing the result in spherical polar coordinates7 we obtain in the numerator fquot gtlt 2 gtlt fquot 21 7cos2 0 7 cos 0 sin 0 cos 737 cos 0 sin 0 sin b In evaluating the integration over the azimuthal variable 15 the x and 3 components vanish which reduces to h 2 A 26 nm 2 1300 ampL ml d3TZ 1 WM 1quotl 18 47139 m5 T Z sin2 0 and Moehmli 1 1300 Wdgr lwnlmllz W 05 19 4me r Hyper ne interaction The so called hyper ne77 interaction results from the magnetic dipole moment of a nucleus MN responding to the magnetic eld formed by the magnetic dipole of the electron spin Me as well as the electron orbital current contribution HHF MN BME 1300 20 77 3MN 12Me 12 MN Me 81 3 L MN HHF 4W 73 3 MN Me5 1quot me T3 21

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