**K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ U.S.A.
**

The vector potential and magnetic induction and of the circular current loop,
of radius *a* in the *x* - *y* plane carrying a current *I* ,
are calculated by Jackson. Let's calculate the magnetic scalar potential.

First let's use an expansion in Legendre polynomials. Since the curl
and divergence of are both zero for *r* > *a* and for *r* < *a* ,
and these are simply connected regions, I can write
= - ,
and satisfies Laplace's equation, is regular in both regions, and
is azimuthally symmetric,

(r,) = .
| (1) |

Zero divergence of means that its normal component is continuous,
and using the orthogonality of the *P*_{n} ,

na^{ n - 1}A_{n} = - (n + 1)a^{ - n - 2}B_{n}
| (2) |

B_{n} = - a^{ 2n + 1}A_{n}
| (3) |

The curl equation says that the line integral around the usual
little rectangle with half the loop at *r* = *a*^{ +}, and the other half
at *r* = *a*^{ -}, ( *a*^{ } are values slightly greater than and less than *a* )
gives zero if the rectangle does not go around the current, and it
gives
4*I*/*c* if it does, where the sign is determined by the
integration direction around the rectangle and the direction of the current
in the usual way. The integral of the negative gradient of the potential
is just the difference of the potential at the end points. In figure 1
I show the current loop, the sphere at *r* = *a* , and 3 different rectangular
paths. You are of course supposed to imagine that these have been shrunk
in size and possibly changed shape
so the length of the parts of the path that cross the sphere
are arbitrarily small.

If I integrate around the rectangle starting from point 1 to point 2 to point 3 to point 4, I get

(1) - (4) = (2) - (3) | (4) |

(9) - (12) = (10) - (11) | (5) |

(a^{ +},) - (a^{ -},) = .
| (6) |

Finally, if the path encircles the current,

(5) - (6) + (7) - (8) = - | (7) |

D_{upper} - D_{lower} =
| (8) |

_{n}(cos ) =
| (9) |

_{n}(cos ) = .
| (10) |

f (cos ) =
| (11) |

- A_{n}a^{ n} + B_{n}a^{ - n - 1} =
| (12) |

- A_{0} = D_{lower} + = arbitrary constant
| (13) |

A_{2m + 1}
| = |
- a^{ - 2m - 1}
| |

B_{2m + 1}
| = |
- A_{2m + 1}a^{ 4m + 3}
| (14) |

(r,) = P_{2m + 1}(cos )
| (15) |

As a final note, you could take the result above, and divide it into
3 regions. The region *r* > *a* , and the upper and lower halves
of the sphere for *r* < *a* . If you add a constant to the upper half of
the sphere to make the potential continuous at *r* = *a* , and add a different
constant to the lower half sphere to again make the potential continuous
at *r* = *a* , you have moved the discontinuity and therefore the dipole
layer to the plane of the loop.

You should be able to see now how to move the dipole layer to any arbitrary surface bounded by the current loop by appropriately dividing up space and making the potential continuous except when crossing the layer.

You can also find the scalar potential by integrating the charge density
of the dipole layer with Coulomb's law. If you do that using a layers of
the surface of the sphere you will get the result above. If you use the
layer over the plane of the disk, the integral is

() | = | - cos =0 | |

= |
- P_{}(0)P_{}(cos )dr'
| (16) |

dr' =
| (17) |

F() = rP_{2m}(cos ) .
| (18) |

|cos | = D_{}P_{}
| (19) |

D_{} = (2 + 1)dxP_{}(x)x = ( + 1)P_{ + 1}(x) + P_{ - 1}(x)dx
| (20) |

F() = |z|
| (21) |