Physical Chemistry Third Edition

24.1 Magnetic Fields and Magnetic Dipoles 1005

b.If the dipole is east of the wire and is rotated to point horizontally toward the north, find

its energy.

Solution

a.Since the dipole and the magnetic field are perpendicular to each other,Emag0.

b.

B

μ 0 I

2 πr



(4π× 10 −^7 TmA−^1 )(15.0A)

2 π(0.200 m)

 1. 5 × 10 −^5 T

Emag−(0.0025 J T−^1 )(1. 5 × 10 −^5 T)cos(0)− 3. 75 × 10 −^8 J

If a particle of chargeQis moving with a constant speed in a circular orbit as in

Figure 24.1b, its motion is equivalent to an average electric current with magnitude

I

Q

torbit



Qv

2 πr

(24.1-7)

wheretorbitis the time required to make one circuit of the orbit, whereris the radius

of the particle’s orbit, and wherevis its speed. Equations (24.1-5) and (24.1-7) give

μ|μ|

πr^2 Qv

2 πr



Qvr

2

(24.1-8)

This can be restated in terms of the angular momentum, which for a circular orbit is

given by Eq. (E-19) of Appendix E:

L|L|mvr (circular orbit) (24.1-9)

so that

μ|μ|

Q

2 m

|L| (circular orbit) (24.1-10)

The same relation holds for the vector quantities as for their magnitudes, even for orbits

that are not circular (although we do not prove this fact):

μ

Q

2 m

L (general relation) (24.1-11)

If the orbiting particle is positively charged, the magnetic moment is in the same direc-

tion as the angular momentum, and if it is negatively charged, the magnetic moment is

in the opposite direction.

EXAMPLE24.5

An electron is moving in a circular orbit of radius 0.529 Å (52.9 pm) at a speed of

2. 187 × 106 ms−^1 (the average speed of an electron in a 1sstate of a hydrogen atom).

Find the magnitude of the magnetic dipole.