Physical Chemistry Third Edition

1006 24 Magnetic Resonance Spectroscopy

Solution

μ

|Q|vr

2



(1. 6022 × 10 −^19 C)(2. 187 × 106 ms−^1 )(5. 29 × 10 −^11 m)

2

 9. 27 × 10 −^24 Am^2  9. 27 × 10 −^24 JT−^1

PROBLEMS

Section 24.1: Magnetic Fields and Magnetic Dipoles

24.1 Find the current in amperes necessary to give a magnetic
field of 1.00 gauss (10−^4 T) at a distance of 1.00 m from
a long wire. Express this current in electrons per second.

24.2 Assume that a long straight wire carries a current of 1.00 A.
Find the distance from the wire such that the magnetic field
is equal to 0.50 gauss, the value of the earth’s magnetic
field in some parts of North America.

24.3 The magnetic field at the center of a circular loop carrying

a currentIis given by

B

μ 0 I

2 r

whereris the radius of the loop. IfB 10 .0 gauss and

r 0 .500 cm, find the value ofI. If the current is

counterclockwise, what is the direction of the magnetic

field?

24.2 Electronic and Nuclear Magnetic Dipoles

Equation (24.1-11) can be used to construct the operator for a magnetic dipole due to

an orbiting electron:

̂μ−

e

2 me

̂L (24.2-1)

wheremeis the mass of the electron, 9. 10939 × 10 −^31 kg,−eis the charge on the

electron, and̂Lis the operator for the orbital angular momentum.

EXAMPLE24.6

Find the magnitude of the magnetic dipole due to the orbital angular momentum of a hydrogen

atom in a 3dstate.

Solution

〈μ^2 〉

(

e

2 me

) 2

〈L^2 〉

(

e

2 me

) 2

h ̄^2 (2)(3)



(

1. 60218 × 10 −^19 C

2(9. 10939 × 10 −^31 kg)

) 2 (

6. 62608 × 10 −^34 Js

2 π

) 2

(6)

 5. 16 × 10 −^46 C^2 kg−^2 J^2 s^2  5. 16 × 10 −^46 J^2 T−^2

|μ| 2. 27166 × 10 −^23 CkgJs 2. 27166 × 10 −^23 JT−^1