Page 171 - Electromagnetics

Figure 3.18: Circular loop of wire.

The differential equations (3.131)and (3.134)are vector versions of Poisson’s equation,
and may be solved quite easily for unbounded space by decomposing the vector source
into rectangular components. For instance, dotting (3.131)with ˆ x we find that

2
∇ A x =−µJ x .
This scalar version of Poisson’s equation has solution

µ J x (r )
A x (r) = dV
4π V |r − r |

in unbounded space. Repeating this for each component and assembling the results, we
obtain the solution for the vector potential in an unbounded homogeneous medium:

µ J(r )

A(r) = dV . (3.135)
4π V |r − r |

Any surface sources can be easily included through a surface integral:
µ J(r ) µ J s (r )

A(r) = dV + dS . (3.136)
4π V |r − r | 4π S |r − r |

In unbounded free space containing materials represented by M,wehave

µ 0 J(r ) + J M (r ) µ 0 J s (r ) + J Ms (r )

A(r) = dV + dV (3.137)

4π V |r − r | 4π S |r − r |

where J Ms =−ˆ n × M is the surface density of magnetization current as described in
(3.153). It may be verified directly from (3.137) that ∇· A = 0.
Field of a circular loop. Consider a circular loop of line current of radius a in
ˆ
unbounded space (Figure 3.18). Using J(r ) = I φ δ(z )δ(ρ − a ) and noting that r =

ρ ˆρ + zˆ z and r = a ˆρ , we can write (3.136)as
µI 2π adφ
A(r) = ˆ .
4π 0 φ ρ + a + z − 2aρ cos(φ − φ ) 1/2
2
2
2

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