The scalar potential for the circular current loop

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 $\vec{B}$ are both zero for r > a and for r < a , and these are simply connected regions, I can write $\vec{B}$ = - $\nabla$$\Phi_{M}^{}$ , and $\Phi_{M}^{}$ satisfies Laplace's equation, is regular in both regions, and is azimuthally symmetric,

$$\Phi_{M}^{} \theta \left \{ \begin{array} {cc} \sum_n A_n r^n P_n(\cos \theta) & r < a \ \sum_n B_n r^{-n-1} P_n(\cos \theta) & r \gt a \ \end{array}\right.\,$$ (1)

Zero divergence of $\vec{B}$ 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)

which shows that B₀ = 0 as expected, and

$${\frac{n}{n+1}}$$ (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 ^$\pm$ are values slightly greater than and less than a ) gives zero if the rectangle does not go around the current, and it gives $\pm$ 4$\pi$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.

  

Figure 1: The current loop with some possible integration paths to determine the boundary conditions.

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

$$\Phi \Phi \Phi \Phi$$ (4)

which says that any discontinuity in the potential in the upper hemisphere must be the same for all points on the upper hemisphere. Similarly,

$$\Phi \Phi \Phi \Phi$$ (5)

which says any discontinuity in the potential in the lower hemisphere must be the same for all points on the lower hemisphere. I define these constant discontinuities as

$$\Phi \theta \Phi \theta \left \{ \begin{array} {cc} D_{\rm upper} & 0< \theta < \frac{\pi}{2}\ D_{\rm lower } & \frac{\pi}{2} < \theta < \pi\ \end{array}\right.\,$$ (6)

Finally, if the path encircles the current,

$$\Phi \Phi \Phi \Phi {\frac{4\pi I}{c}}$$ (7)

where I have assumed the current is in the $\hat{\phi}$ direction. Plugging in the definitions above,

$${\frac{4\pi I}{c}}$$ (8)

Notice that adding an overall constant to the potential for r < a or a different constant for r > a , does not affect the fields or these boundary conditions. In the dipole layer model, it changes the amount of dipole layer on the upper and lower hemispheres, keeping the total constant. You can choose to put the dipole layer wherever you like by choosing the value of one of these constants. However, I'll keep the arbitrary constant just to show you explicitly that it doesn't matter. The boundary condition is

$$\sum_{n}^{} \left [ -A_n a^n + B_n a^{-n-1} \right]P \theta \left \{ \begin{array} {cc} D_{\rm lower}+ \frac{4\pi I}{c} & 0 < ... ...{\pi}{2} \ D_{\rm lower} & 0 \frac{\pi}{2} < \theta < \pi\ \end{array}\right.$$ (9)

which can be written in a form more convenient for integration

$$\sum_{n}^{} \left [ -A_n a^n + B_n a^{-n-1} \right]P \theta \left \{ \begin{array} {cc} \left [D_{\rm lower}+\frac{2 \pi I}{c}... ...ght ] -\frac{2\pi I}{c} & 0 \frac{\pi}{2} < \theta < \pi\ \end{array}\right.\,$$ (10)

I now need to use the orthogonality of the Legendre polynomials, to pull out a single term on the left, and I then need to do the integral on the right. Notice that the common term is just an overall constant, while the other term is proportional to the expansion of

$$\theta \left \{ \begin{array} {cc} 1 & 0 < \theta < \frac{\pi}{2} \ -1 & \frac{\pi}{2} < \theta < \pi\ \end{array}\right.$$ (11)

whose integral with the Legendre polynomial is given in Jackson Eq. 3.23 through 3.26, with the result for n odd,

$${\frac{2\pi I}{c}} {\frac{(-1)^{(n-1)/2}}{2^{(n-1)/2}}} {\frac{(2n+1)(n-2)!!}{2 \left (\frac{n+1}{2} \right )!}}$$ (12)

while the result for n even is zero except for n = 0 where it is

$${\frac{2\pi I}{c}}$$ (13)

where I have used the previous result that B₀ = 0 . Plugging in for B_n and simplifying

$${\frac{2\pi I}{c}} {\frac{(-1)^m}{2^m}} {\frac{(2m-1)!!}{m!}}$$ A_{2m + 1} =

$${\frac{2m+1}{2m+2}}$$ B_{2m + 1} = (14)

and

$$\Phi_{M}^{} \theta {\frac{\pi I a^2}{c}} \sum_{m=0}^{\infty} {\frac{(-1)^m(2m+1)!!}{2^m(m+1)!}} \theta \left \{ \begin{array} {cc} -\frac{2m+2}{2m+1} \frac{r^{2m+1}}{a^{... ...}} & r < a\ \frac{1}{r^2} \frac{a^{2m}}{r^{2m}} & r \gt a\ \end{array}\right.$$ (15)

where an arbitrary constant can be added to the potential in either region. It is easy to take the negative gradient of this explicitly to see that it agrees with Jackson's expressions (Eqs. 5.48 and 5.49) for the $\vec{B}$ field from the vector potential.

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

$$\Phi_{M}^{} \vec{r} {\frac{I}{c}} \hat{z} \cdot \vec{\nabla} \left . \int_0^a dr' \int_0^{2\pi} d\phi' \frac{r'}{\vert\vec r - \vec r'\vert} \right\vert _ \theta$$ =

$${\frac{2\pi I}{c}} {\frac{\partial}{\partial z}} \sum_{\ell}^{} \ell \ell \theta \int_{0}^{a} {\frac{r' r_<^\ell}{r_\gt^{\ell+1}}}$$ = (16)

and the radial integrals are

$$\int_{0}^{a} {\frac{r' r_<^\ell}{r_\gt^{\ell+1}}} \left \{ \begin{array} {cc} \frac{r (2\ell+1)}{(\ell+2)(\ell-1)} -... ... r < a\\ \frac{a^{\ell+2}}{(\ell+2) r^{\ell+1}} & r \gt a \\ \end{array}\right.$$ (17)

which is the result we got by solving Laplace's equation above plus an additional term for r < a of

$$\vec{r} {\frac{\partial}{\partial z}} {\frac{2\pi I}{c}} \sum_{m=0}^{\infty} {\frac{(-1)^m (2m-1)!! (4m+1)}{2^m m! (2m+2)(2m-1)}} \theta$$ (18)

Using the completeness of the Legendre polynomials it is straightforward to calculate the expansion

$$\theta \sum_{\ell}^{} \ell \ell$$ (19)

to be

$$\ell \ell \int_{0}^{1} \ell \ell \int_{0}^{1} \ell \ell \int_{0}^{1} \ell$$ (20)

The integrals are the same ones given above (Jackson Eqs. 3.23 through 3.26), and substituting those resuls, the the extra term is seen to be

$$\vec{r} {\frac{\partial}{\partial z}} {\frac{2\pi I}{c}}$$ (21)

which gives the additional constants to move the discontinuity to the plane of the disk.