Simple Nature - Light and Matter

where dBzis the contribution to the total magnetic field at our point
of interest, which lies a distanceRfrom the wire.

Bz=

∫∞

y=0

∫∞

x=−∞

βkIdA

c^2 r^3

=

βkI

c^2

∫∞

y=0

∫∞

x=−∞

1

[x^2 + (R+y)^2 ]^3 /^2

dxdy

=

βkI

c^2 R^3

∫∞

y=0

∫∞

x=−∞

[(

x

R

) 2

+

(

1 +

y

R

) 2 ]−^3 /^2

dxdy

This can be simplified with the substitutionsx=Ru,y=Rv, and
dxdy=R^2 dudv:

Bz=

βkI

c^2 R

∫∞

v=0

∫∞

u=−∞

1

[u^2 + (1 +v)^2 ]^3 /^2

dudv

Theuintegral is of the form

∫∞

−∞(u

(^2) +b)− 3 / (^2) du= 2/b (^2) , so
Bz=
βkI
c^2 R

∫∞

v=0

1

(1 +v)^2

dv,

and the remainingvintegral is equals 2, so

Bz=

2 βkI

c^2 R

.

This is the field of a wire, which we already know equals 2kI/c^2 R,
so we haveβ=1. Remember, the point of this whole calculation
was not to find the field of a wire, which we already knew, but
to find the unitless constantβin the expression for the field of a
dipole. The distant field of a dipole, in its midplane, is therefore
Bz=βkIA/c^2 r^3 =kIA/c^2 r^3 , or, in terms of the dipole moment,

Bz=

km

c^2 r^3

.

The distant field of a dipole, out of its midplane

What about the field of a magnetic dipole outside of the dipole’s

midplane? Let’s compare with an electric dipole. An electric dipole,

unlike a magnetic one, can be built out of two opposite monopoles,

i.e., charges, separated by a certain distance, and it is then straight-

forward to show by vector addition that the field of an electric dipole

is

Ez=kD

(

3 cos^2 θ− 1

)

r−^3

ER=kD(3 sinθcosθ)r−^3 ,

Section 11.2 Magnetic fields by superposition 693