Inductance of a Circular Loop
Electromagnetism, Magnetism
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Vector Potential
The magnetic field at an arbitrary point \(\mathbf{r}\) created by a current distribution \(\mathbf{J}(\mathbf{r}')\) is given by the Biot-Savart law[1]:
\[\begin{eqnarray} \mathbf{B}(\mathbf{r}) &=&\frac{\mu_0 }{4\pi}\int d^3 \mathbf{r}'\frac{ \mathbf{J}(\mathbf{r}') \times ( \mathbf{r}- \mathbf{r}')}{ ( \mathbf {r}- \mathbf {r}')^3}, \label{eq:biotsavartlaw} \end{eqnarray}\] where we use the primed coordinates for the source points. We can convert this to a curl of a vector potential using the following identity: \[\begin{eqnarray} \bnabla\frac{1} {\left|\mathbf{r}- \mathbf{r}' \right|}&=&-\frac{\mathbf{r}- \mathbf{r}'}{|\mathbf{r}- \mathbf{r}'|^3} \label{eq:gradofrinv}. \end{eqnarray}\] Putting this into Eq. \(\ref{eq:biotsavartlaw}\) we get: \[\begin{eqnarray} \mathbf{B}(\mathbf{r}) &=&-\frac{\mu_0 }{4\pi}\int d^3 \mathbf{r}' \mathbf{J}(\mathbf{r}') \times \bnabla\frac{1} {\left|\mathbf{r}- \mathbf{r}' \right|} = \bnabla \times \left[ \frac{ \mu_0 }{ 4 \pi } \int d^3 \mathbf{r'} \frac{\mathbf{J}(\mathbf{r'})}{ | \mathbf {r}- \mathbf {r}'|} \right]\equiv \bnabla\times \mathbf{A}(\mathbf{r}). \label{eq:biotsavartlawcurl2} \end{eqnarray}\]
Equation \(\ref{eq:biotsavartlawcurl2}\) enables us to define a vector potential for an arbitrary current distribution:
\[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& \frac{ \mu_0 }{ 4 \pi } \int d^3 \mathbf{r'} \frac{\mathbf{J}(\mathbf{r'})}{ | \mathbf {r}- \mathbf {r}'|} \label{eq:vpotdef}. \end{eqnarray}\]
Vector Potential of a Single Loop
We consider the magnetic field of a single circular loop with a current as in Figure 1.
For a single loop sitting at \(z=0\) with radius \(r=R\), it is convenient to work in spherical coordinates. \[\begin{eqnarray} \mathbf{J}(\mathbf{r}') &=& \lambda \delta(r'-R)\delta(\cos \theta')\hat{\mathbf{\phi}'} =\lambda \delta(r'-R)\delta(\cos \theta')\left(\cos\phi' \hat{\mathbf{j}}-\sin\phi'\hat{\mathbf{i}} \right) \label{eq:jcur}, \end{eqnarray}\] where \(\lambda\) is the current density. In order to calculate \(\lambda\) for a loop of wire carrying a current \(I\), let’s intercept the loop with an area perpendicular to it. We can do that by selecting an area on, say, positive \(x\) axis ( i.e., \(\phi'=0\)), pointing along the \(y\) axis, i.e., \(d\mathbf{S}'= dS' \hat{\mathbf{j}}=r'dr'd\theta' \hat{\mathbf{j}}\). Integrating the current density on this area we should get the total current: \[\begin{eqnarray} \int_S d\mathbf{S}' \cdot \mathbf{J}(\mathbf{r}) &=& \int_S r'dr'd\theta' \lambda \delta(r'-R)\delta(\cos \theta')\hat{\mathbf{j}}\cdot\left(\cos\phi' \hat{\mathbf{j}}-\sin\phi'\hat{\mathbf{i}} \right)\bigg\rvert_{\phi'=0}=R\lambda =I \nonumber\\ &&\implies \lambda=I/R \label{eq:jcurint}. \end{eqnarray}\] Therefore, the properly normalized current is \[\begin{eqnarray} \mathbf{J}(\mathbf{r}') &=& \frac{I}{R} \delta(r'-R)\delta(z')\hat{\mathbf{\phi}'} =\frac{I}{R}\delta(r'-R)\delta(\cos \theta')\left(\cos\phi' \hat{\mathbf{j}}-\sin\phi'\hat{\mathbf{i}} \right) \label{eq:jcurf}. \end{eqnarray}\]
The integral we have to deal with for a single loop is this: \[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& \frac{ \mu_0 }{ 4 \pi } \int d^3 \mathbf{r'} \frac{\mathbf{J}(\mathbf{r'})}{ | \mathbf {r}- \mathbf {r}'|}=\frac{ \mu_0 I }{ 4 \pi R } \int d^3 \mathbf{r'}\frac{1 }{ | \mathbf {r}- \mathbf {r}'|} \delta(r'-R)\delta(\cos\theta')\left(\cos\phi' \hat{\mathbf{j}}-\sin\phi'\hat{\mathbf{i}} \right) \label{eq:vpotfull}, \end{eqnarray}\] where we put the subscript \(s\) to remind us that this is for a single loop. We will parameterize the points on the loop centered at \(z=0\) as \(\mathbf {r}'= r' (\cos\phi'\hat{\mathbf{i}}+\sin\phi'\hat{\mathbf{j}})\), and the observation point as \(\mathbf {r}= r\cos\theta \hat{\mathbf {z}}+ r\sin\theta (\cos\phi\hat{\mathbf{i}}+\sin\phi\hat{\mathbf{j}})\)
\[\begin{eqnarray} | \mathbf {r}- \mathbf {r}'|&=& \sqrt{r^2\cos^2\theta +(r\sin\theta\cos\phi -r'\cos\phi')^2+(r\sin\theta\sin\phi -r'\sin\phi')^2}\nonumber\\ &=&\sqrt{r^2+r'^2 -2 r r'\sin\theta\cos(\phi'-\phi)} \label{eq:deltar}. \end{eqnarray}\] Note that the problem has rotational symmetry. We can rotate our coordinate system such that the observation point sits on \(y=0\), i.e., \(\phi=0\). Once we are done with the computations, we can rotate the vectors back to general \(\mathbf{r}\) point. So let’s set \(\phi=0\) in Eq. \(\ref{eq:deltar}\) and rewrite Eq. \(\ref{eq:vpotfull}\) :
\[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& \frac{ \mu_0 I }{ 4 \pi R } \int \sin\theta' r'^2 dr' d\phi' \frac{\delta(r'-R)\delta(\cos\theta')\left(\cos\phi' \hat{\mathbf{j}}-\sin\phi'\hat{\mathbf{i}} \right) }{ \sqrt{r^2+r'^2 -2 r r'\sin\theta\cos(\phi'-\phi)}} \nonumber\\ &=& \frac{ \mu_0 I R}{ 4 \pi } \left[ \int_{0}^{2\pi} d\phi' \frac{\cos\phi' \hat{\mathbf{j}} }{ \sqrt{r^2+R^2 -2 r R\sin\theta\cos(\phi'-\phi)}} \right.\nonumber\\ &&\quad\quad\quad \left.-\cancelto{0}{ \int_{0}^{2\pi} d\phi' \frac{\sin\phi'\hat{\mathbf{i}} }{ \sqrt{r^2+R^2 -2 r R\sin\theta\cos(\phi'-\phi)}} } \right] \label{eq:vpotfull2}, \end{eqnarray}\] where the second term vanishes since the integrand is odd and the integral is evaluated over the full range. Note that we evaluated the integral at \(\phi=0\), and the resulting potential points in \(\hat{\mathbf{j}}\) direction. For generic \(\phi\) we can simply rotate the coordinate system about the \(z\) axis by \(\phi\). In this rotated coordinate system \(\hat{\mathbf{j}}\rightarrow \hat{\phi}\). Therefore the vector potential reads: \[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& \hat{\phi}\frac{ \mu_0 I R}{ 4 \pi } \int_{0}^{2\pi} d\phi' \frac{\cos\phi' }{ \sqrt{r^2+R^2 -2 r R\sin\theta\cos\phi'}} \label{eq:vpotfullrot}. \end{eqnarray}\] Let’s define \(\phi'=\pi-\phi'\) to get \(\cos\phi'=-\cos\phi'\) and rewrite Eq. \(\ref{eq:vpotfullrot}\) as: \[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& -\hat{\phi}\frac{ \mu_0 I R}{ 4 \pi } \int_{-\pi}^{\pi} d\phi' \frac{\cos\phi' }{ \sqrt{r^2+R^2 +2 r R\sin\theta\cos\phi'}} \label{eq:vpotfullrot2}. \end{eqnarray}\] Let’s also use the half angle formula: \(\cos\phi'=1-2\sin^2\frac{\phi'}{2}\) and reorganize the integral: \[\begin{eqnarray} \mathbf{A}(\mathbf{r}) &=& -\hat{\phi}\frac{ \mu_0 I R}{ 4 \pi } \frac{1}{\sqrt{r^2+R^2 +2 r R\sin\theta}}\int_{-\pi}^{\pi} d\phi' \frac{1-2\sin^2\frac{\phi'}{2} }{ \sqrt{1 -\frac{4 r R\sin\theta}{r^2+R^2 +2 r R\sin\theta}\sin^2\frac{\phi'}{2}}}\nonumber\\ &\equiv& -\hat{\phi}\frac{ \mu_0 I R}{ 4 \pi } \frac{1}{\sqrt{r^2+R^2 +2 r R\sin\theta}}\int_{-\pi}^{\pi} d\phi' \frac{1-2\sin^2\frac{\phi'}{2} }{ \sqrt{1 -k^2\sin^2\frac{\phi'}{2}}} \nonumber\\ &=& -\hat{\phi}\frac{ \mu_0 I R}{ 4 \pi } \frac{1}{\sqrt{r^2+R^2 +2 r R\sin\theta}}\int_{-\pi}^{\pi} d\phi' \left[\frac{1 }{ \sqrt{1 -k^2\sin^2\frac{\phi'}{2}}}-2 \frac{\sin^2\frac{\phi'}{2} }{ \sqrt{1 -k^2\sin^2\frac{\phi'}{2}}}\right] \nonumber\\ &=& -\mathbf{\hat{\phi}}\frac{ \mu_0 I R }{ 4 \pi k^2} \frac{1}{ \sqrt{r^2+R^2 +2 r R\sin\theta}}\int_{-\pi}^{\pi} d\phi' \left[\frac{k^2-2 }{ \sqrt{1 -k^2\sin^2\frac{\phi'}{2}}}+2 \sqrt{1 -k^2\sin^2\frac{\phi'}{2}} \right] \label{eq:vpotfullrot3}, \end{eqnarray}\]
where \(k^2=\frac{4 r R\sin\theta}{r^2+R^2 +2 r R\sin\theta}\). Finally, we define \(\zeta'=\phi'/2\) and split the integration into two pieces to pick an overall factor of \(4\) to get: \[\begin{eqnarray} \mathbf{A}(\mathbf{r})=\hat{\phi}\frac{ \mu_0 I R}{ \pi \sqrt{r^2+R^2 +2 r R\sin\theta} } \frac{(2-k^2)K(k^2)-2E(k^2)}{k^2} \label{eq:vpotfullrotf}, \end{eqnarray}\]
where the elliptic integral are defined as follows: \[\begin{eqnarray} K(k^2)&=&\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}},\nonumber\\ E(k^2)&=&\int_0^{\frac{\pi}{2}}d\theta\sqrt{1-k^2\sin^2\theta} \label{eq:elliptic}. \end{eqnarray}\]
Magnetic Field of a Single Loop
The calculation of the magnetic field is straightforward[2]:
\[\begin{eqnarray} B_r&=&\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta A_\phi)=\frac{ \mu_0 I R^2 \cos\theta E(k^2) }{ \pi \sqrt{r^2+R^2 +2 r R\sin\theta} (r^2+R^2 -2 r R\sin\theta)},\nonumber\\ B_\theta&=&-\frac{1}{r}\frac{\partial}{\partial r}(r A_\phi)=\frac{ \mu_0 I \left[(r^2+R^2\cos(2\theta) )E(k^2)-(r^2+R^2 -2 r R\sin\theta)K(k^2)\right] }{ 2 \pi \sqrt{r^2+R^2 +2 r R\sin\theta} (r^2+R^2 -2 r R\sin\theta)\sin\theta} \label{eq:ellipticB}. \end{eqnarray}\]
We can also express the magnetic field in the cylindrical coordinates [2]: \[\begin{eqnarray} B_\rho&=&\frac{ \mu_0 I z \left[ (R^2+\rho^2+z^2)E(k^2)-(R^2+\rho^2+z^2 -2 R \rho))K(k^2)\right] }{ 2 \pi \sqrt{R^2+\rho^2+z^2 +2 R \rho} (R^2+\rho^2+z^2 -2 R \rho)\rho},\nonumber\\ B_z&=&\frac{ \mu_0 I \left[ (R^2-\rho^2-z^2)E(k^2)+(R^2+\rho^2+z^2 -2 R \rho))K(k^2)\right] }{ 2 \pi \sqrt{R^2+\rho^2+z^2 +2 R } (R^2+\rho^2+z^2 -2 R \rho)\rho} \label{eq:ellipticBc}. \end{eqnarray}\]
Self-Inductance of a Single Loop
The self-inductance \(L\) of a circular loop is defined as the ratio of the magnetic flux through the loop to the current producing it:
\[\begin{eqnarray} L = \frac{\Phi}{I}, \label{eq:inductancedef} \end{eqnarray}\]
where \(\Phi\) is the magnetic flux through the loop area. The flux can be computed using Stokes’ theorem by integrating the vector potential around the loop:
\[\begin{eqnarray} \Phi = \oint \mathbf{A} \cdot d\mathbf{l} = \int_0^{2\pi} A_\phi(R, \theta=\pi/2) R d\phi, \label{eq:fluxstokes} \end{eqnarray}\]
where we evaluate the vector potential on the loop itself (at \(r=R\), \(\theta=\pi/2\) in spherical coordinates, corresponding to \(\rho=R\), \(z=0\) in cylindrical coordinates).
At a point on the loop, we have \(r=R\) and \(\theta=\pi/2\), so from Eq. \(\ref{eq:vpotfullrotf}\), the parameter \(k^2\) becomes:
\[\begin{eqnarray} k^2 = \frac{4 R^2}{R^2+R^2 + 2 R^2} = \frac{4 R^2}{4 R^2} = 1. \label{eq:k2onloop} \end{eqnarray}\]
This is a singular case where the elliptic integrals diverge, reflecting the fact that we’re evaluating the field at the source point itself. To handle this, we need to account for the finite thickness of the wire.
For a thin wire of radius \(a \ll R\), we compute the self-inductance by evaluating the vector potential at a point on the loop, but we must account for the fact that the current is distributed over a wire of finite cross-section. Let’s work through the derivation step by step.
We start with the vector potential from Eq. \(\ref{eq:vpotfullrotf}\):
\[\begin{eqnarray} A_\phi(\mathbf{r}) = \frac{\mu_0 I R}{\pi \sqrt{r^2+R^2 + 2rR\sin\theta}} \frac{(2-k^2)K(k^2)-2E(k^2)}{k^2}, \label{eq:vpotstart} \end{eqnarray}\]
where \(k^2 = \frac{4rR\sin\theta}{r^2+R^2 + 2rR\sin\theta}\).
For a point on the loop, we have \(r = R\) and \(\theta = \pi/2\). However, to avoid the singularity, we evaluate at a point slightly off the loop. Consider a point at cylindrical coordinates \((\rho, z) = (R, 0)\) but accounting for the wire thickness. The distance from the loop center to a point on the wire surface is approximately \(R\), but the minimum distance between source and observation points is \(2a\) (when both are on opposite sides of the wire).
In the limit where we’re very close to the loop (\(\rho \approx R\), \(z \approx 0\)), we have:
\[\begin{eqnarray} k^2 \approx 1 - \frac{(\rho-R)^2 + z^2}{4R^2} \approx 1 - \frac{a^2}{R^2}, \label{eq:k2near} \end{eqnarray}\]
where we’ve used the fact that the minimum separation is of order \(a\).
Now we need the asymptotic behavior of the elliptic integrals as \(k^2 \to 1\). For \(k^2\) close to 1, we have:
\[\begin{eqnarray} K(k^2) &\approx& \ln\left(\frac{4}{\sqrt{1-k^2}}\right) + \mathcal{O}(1-k^2), \nonumber\\ E(k^2) &\approx& 1 + \mathcal{O}(1-k^2). \label{eq:ellipticasymp} \end{eqnarray}\]
Substituting \(k^2 \approx 1 - a^2/R^2\) into Eq. \(\ref{eq:vpotstart}\):
\[\begin{eqnarray} A_\phi &\approx& \frac{\mu_0 I R}{\pi \sqrt{4R^2}} \frac{(2-(1-a^2/R^2))\ln(4R/a) - 2}{1-a^2/R^2} \nonumber\\ &\approx& \frac{\mu_0 I}{2\pi} \left[(1+a^2/R^2)\ln(4R/a) - 2 + \mathcal{O}(a^2/R^2)\right] \nonumber\\ &\approx& \frac{\mu_0 I}{2\pi} \left[\ln(4R/a) - 2 + \mathcal{O}(a^2/R^2)\right]. \label{eq:aphiapprox} \end{eqnarray}\]
However, this gives the vector potential at a single point. To get the flux, we need to integrate around the loop. The flux is:
\[\begin{eqnarray} \Phi = \oint \mathbf{A} \cdot d\mathbf{l} = R \int_0^{2\pi} A_\phi(R, \phi) d\phi. \label{eq:fluxintegral} \end{eqnarray}\]
The subtlety is that when we integrate around the loop, we’re averaging the vector potential over all points on the loop. For each point on the loop, the vector potential receives contributions from all other points on the loop. The logarithmic divergence comes from points that are close together.
A more careful analysis shows that when averaging the distance between points on a circle, the effective minimum distance in the logarithm is not \(2a\) but rather involves a geometric factor. Specifically, for a circular loop, the average of \(\ln(|\mathbf{r} - \mathbf{r}'|)\) over all pairs of points on the loop gives rise to the factor of 8 grover1946inductance?.
The complete derivation involves: 1. Writing the vector potential as an integral over the loop 2. Averaging over the loop circumference 3. Extracting the logarithmic divergence carefully 4. Accounting for the finite wire thickness
The final result is:
\[\begin{eqnarray} L = \frac{\Phi}{I} = \mu_0 R \left[\ln\left(\frac{8R}{a}\right) - 2\right] + \mathcal{O}\left(\frac{a}{R}\right), \label{eq:inductanceapprox} \end{eqnarray}\]
where the factor of 8 (instead of 4) comes from the geometric averaging over the circular loop, and the constant \(-2\) arises from the detailed structure of the elliptic integrals and the integration over the loop geometry grover1946inductance?.
where \(a\) is the wire radius. This approximation is valid when \(a \ll R\).
For the exact calculation using elliptic integrals, we evaluate the vector potential at a point near the loop. Taking the limit as the observation point approaches the loop (but staying at a finite distance \(a\) from the wire center), the self-inductance can be expressed as:
\[\begin{eqnarray} L = \mu_0 R \left[\ln\left(\frac{8R}{a}\right) - 2 + \frac{a}{R}\left(\ln\left(\frac{8R}{a}\right) + \frac{1}{4}\right) + \mathcal{O}\left(\frac{a^2}{R^2}\right)\right]. \label{eq:inductanceexact} \end{eqnarray}\]
The leading term \(\mu_0 R \ln(8R/a)\) dominates for thin wires and shows the characteristic logarithmic dependence on the aspect ratio \(R/a\). The constant term \(-2\) comes from the detailed structure of the elliptic integrals in the limit \(k^2 \to 1\).
An alternative approach is to compute the flux directly by integrating the magnetic field over the loop area. Using the \(B_z\) component from Eq. \(\ref{eq:ellipticBc}\) at \(z=0\) and integrating over the circular disk:
\[\begin{eqnarray} \Phi = \int_0^{2\pi} \int_0^R B_z(\rho, z=0) \rho d\rho d\phi. \label{eq:fluxdirect} \end{eqnarray}\]
This integral also leads to the same logarithmic form, confirming the self-inductance expression above.