Circular Current Loop

Let us calculate the magnetic field generated by a thin circular loop of radius $a$ , lying in the $x$ -$y$ plane, centered on the origin, and carrying the steady current $I$ . Let $r$ , $\theta$ , $\varphi$ be spherical coordinates whose origin lies at the center of the loop, and whose symmetry axis is coincident with that of the loop. It follows that the distribution of current density in space is

$${\bf j}({\bf r}') = I\,\frac{\delta(r'-a)}{a}\,\delta(\cos\theta'... ...)}{a}\,\delta(\cos\theta')\,(-\sin\varphi'\,{\bf e}_x+\cos\varphi'\,{\bf e}_y).$$ (628)

Because the geometry is cylindrically symmetric, we can, without loss of generality, choose the observation point to lie in the $x$ -$z$ plane (i.e., $\varphi=0$ ). It follows from Equation (621) that

$$A_x = - \frac{\mu_0\,I}{4\pi\,a}\int \frac{\sin\varphi'\,\delta(\cos\... ...r'-a)}{\vert{\bf r}-{\bf r}'\vert _{\varphi=0}}\,r'^{\,2}\,dr'\,d{\mit\Omega}',$$ (629)

$$A_y = \frac{\mu_0\,I}{4\pi\,a}\int \frac{\cos\varphi'\,\delta(\cos\th... ...r'-a)}{\vert{\bf r}-{\bf r}'\vert _{\varphi=0}}\,r'^{\,2}\,dr'\,d{\mit\Omega}',$$ (630)

$$A_z =0,$$ (631)

where $\vert{\bf r}-{\bf r}'\vert _{\varphi=0} = [r^{\,2} + r'^{\,2}-2\,r\,r'\,(\cos\theta\,\cos\theta' + \sin\theta\,\sin\theta'\,\cos\varphi')]^{1/2}$ . It is clear that the integral for $A_x$ averages to zero. Hence, only $A_y$ , which corresponds to $A_\varphi$ , is non-zero, and we can write

$${\bf A}= A_\varphi(r,\theta)\,{\bf e}_\varphi,$$ (632)

where

$$A_\varphi(r,\theta) = \frac{\mu_0\,I}{4\pi\,a}\int \frac{\cos\var... ...ta')\,\delta(r'-a)}{\vert{\bf r}-{\bf r}'\vert}\,r'^{\,2}\,dr'\,d{\mit\Omega}',$$ (633)

which reduces to

$$A_\varphi(r,\theta) =\frac{\mu_0\,I\,a}{4\pi}\oint \frac{\cos\var... ...arphi'\,d\varphi'}{(r^{\,2}+a^{\,2} -2\,a\,r\,\sin\theta\,\cos\varphi')^{1/2}}.$$ (634)

The previous integral can be expressed in terms of complete elliptic integrals, but this is not particularly illuminating. A better approach is to make use of the expansion of the Green's function for Poisson's equation in terms of spherical harmonics given in Section 3.5:

$$\frac{1}{4\pi\,\vert{\bf r}-{\bf r}'\vert }= \sum_{l=0,\infty}\su... ...r_>^{\,l+1}}\right)Y_{l,m}^{\,\ast}(\theta',\varphi')\,Y_{l,m}(\theta,\varphi).$$ (635)

Here, $r_<$ represents the lesser of $r$ and $r'$ , whereas $r_>$ represents the greater of $r$ and $r'$ . Hence,

$$A_\varphi(r,\theta)= \mu_0\,I\,a\,\,{\rm Re}\!\sum_{l=0,\infty}\s... ...\right)\oint Y_{l,m}(\pi/2,\varphi')\,{\rm e}^{\,{\rm i}\,\varphi'}\,d\varphi',$$ (636)

where $r_<$ now represents the lesser of $r$ and $a$ , whereas $r_>$ represents the greater of $r$ and $a$ . It follows from Equation (309) that

$$\oint Y_{l,m}^{\,\ast}(\theta',\varphi')\,{\rm e}^{\,{\rm i}\,\va... ...Y_{l,m}^{\,\ast}(\theta',\varphi')\,{\rm e}^{\,{\rm i}\,\varphi'}\,\delta_{m1}.$$ (637)

Thus, Equation (637) yields

$$A_\varphi(r,\theta)= 2\pi\,\mu_0\,I\,a\,\,{\rm Re}\!\sum_{l=0,\in... ...{\,l+1}}\right)Y_{l,1}^{\,\ast}(\pi/2,\varphi')\,{\rm e}^{\,{\rm i}\,\varphi'}.$$ (638)

However, according to Equation (309),

$$Y_{l,1}(\theta,0) = \left[\frac{(2\,l+1)}{4\pi\,l\,(l+1)}\right]^{1/2}\,P_l^{\,1}(\cos\theta),$$ (639)

$$Y_{l,1}^{\,\ast}(\pi/2,\varphi')\,{\rm e}^{\,{\rm i}\,\varphi'} = \left[\frac{(2\,l+1)}{4\pi\,l\,(l+1)}\right]^{1/2}\,P_l^{\,1}(0).$$ (640)

Hence, we obtain

$$A_\varphi(r,\theta) =\frac{1}{2}\,\mu_0\,I\,a \sum_{l=1,3,5,\cdot... ...P_l^1(0)}{l\,(l+1)}\left(\frac{r_<^{\,l}}{r_>^{\,l+1}}\right)P_l^1(\cos\theta),$$ (641)

where we have made use of the fact that $P_l^1(0)=0$ when $l$ is even.To be more exact,

$$A_\varphi(r,\theta) = \frac{1}{2}\,\mu_0\,I\sum_{l=1,3,5,\cdots}\frac{P_l^1(0)}{l\,(l+1)}\left(\frac{r}{a}\right)^lP_l^1(\cos\theta)$$ (642)

for $r<a$ , and

$$A_\varphi(r,\theta) = \frac{1}{2}\,\mu_0\,I\sum_{l=1,3,5,\cdots}\frac{P_l^1(0)}{l\,(l+1)}\left(\frac{a}{r}\right)^{l+1}P_l^1(\cos\theta)$$ (643)

for $r>a$ .

Now, according to Equations (620) and (633),

$$B_r = \frac{1}{r\,\sin\theta}\,\frac{\partial}{\partial\theta}\,(\sin\theta\,A_\varphi),$$ (644)

$$B_\theta = -\frac{1}{r}\,\frac{\partial}{\partial r}\,(r\,A_\varphi),$$ (645)

$$B_\varphi =0.$$ (646)

Given that

$$\frac{1}{\sin\theta}\,\frac{d}{d\theta}\left[\sin\theta\,P_l^1(\cos\theta)\right]=-l\,(l+1)\,P_l(\cos\theta),$$ (647)

we find that

$$B_r(r,\theta) = -\frac{\mu_0\,I}{2\,a}\sum_{l=1,3,5,\cdots}P_l^1(0)\left(\frac{r}{a}\right)^{l-1}P_l(\cos\theta),$$ (648)

$$B_\theta(r,\theta) = -\frac{\mu_0\,I}{2\,a}\sum_{l=1,3,5,\cdots}\frac{P_l^1(0)}{l}\left(\frac{r}{a}\right)^{l-1}P_l^1(\cos\theta)$$ (649)

in the region $r<a$ . In particular, because $P_l(x)=-x$ and $P_1^1(x)=-(1-x^2)^{1/2}$ , we obtain

$$B_r({\bf0}) = \frac{\mu_0\,I}{2\,a}\,\cos\theta,$$ (650)

$$B_\theta({\bf0}) = -\frac{\mu_0\,I}{2\,a}\,\sin\theta.$$ (651)

The previous two equations can be combined to give

$${\bf B}({\bf0}) = \frac{\mu_0\,I}{2\,a}\,{\bf e}_z.$$ (652)

Of course, this result can be obtained in a more straightforward fashion via the direct application of the Biot-Savart law. We also have

$$B_r(r,\theta) = -\frac{\mu_0\,I}{2\,a}\sum_{l=1,3,5,\cdots}P_l^1(0)\left(\frac{a}{r}\right)^{l+2}P_l(\cos\theta),$$ (653)

$$B_\theta(r,\theta) = \frac{\mu_0\,I}{2\,a}\sum_{l=1,3,5,\cdots}\frac{P_l^1(0)}{l+1}\left(\frac{a}{r}\right)^{l+2}P_l^1(\cos\theta)$$ (654)

in the region $r>a$ . A long way from the current loop (i.e., $r/a\rightarrow\infty$ ), we obtain

$$A_\varphi(r,\theta) = \frac{\mu_0}{4\pi}\,m\,\frac{\sin\theta}{r^{\,2}},$$ (655)

$$B_r(r,\theta) = \frac{\mu_0}{4\pi}\,m\,\frac{2\,\cos\theta}{r^{\,3}},$$ (656)

$$B_\theta(r,\theta) = \frac{\mu_0}{4\pi}\,m\,\frac{\sin\theta}{r^{\,3}},$$ (657)

where $m=I\,\pi\,a^{\,2}$ .

Now, a small planar current loop of area $A$ , carrying a current $I$ , constitutes a magnetic dipole of moment

$${\bf m}= I\,A\,{\bf n}.$$ (658)

Here, ${\bf n}$ is a unit normal to the loop in the sense determined by the right-hand circulation rule (with the current determining the sense of circulation). It follows that in the limit $a\rightarrow 0$ and $I\,\pi\,a^{\,2}\rightarrow m$ the current loop considered previously constitutes a magnetic dipole of moment ${\bf m} = m\,{\bf e}_z$ . Moreover, Equations (656)-(658) specify the non-zero components of the vector potential and the magnetic field generated by the dipole. It is easily seen from Equation (656) that

$${\bf A} = \frac{\mu_0}{4\pi}\,\frac{{\bf m}\times {\bf r}}{r^{\,3}}.$$ (659)

Taking the curl of this expression, we obtain

$${\bf B} =\frac{\mu_0}{4\pi}\left[\frac{3\,({\bf m}\cdot{\bf r})\,{\bf r} -r^{\,2}\,{\bf m}}{r^{\,5}}\right],$$ (660)

which is consistent with Equations (657) and (658).