Self-Inductance of a Wire
Electromagnetism, Magnetism
Magnetic Field of a Wire Segment
Figure 1 shows a wire segment of length \(L\) carrying a current \(I\) and located at the origin.
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]:
\[\begin{eqnarray} \mathbf{B}(\mathbf{r}) &=&\frac{\mu_0 }{4\pi}\int d^3 \mathbf{r}'\frac{ \mathbf{J}(\mathbf{r}') \times ( \mathbf{r}- \mathbf{r}')}{ ( \mathbf {r}- \mathbf {r}')^3}, \label{eq:biotsavartlaw} \end{eqnarray}\]
\[\begin{eqnarray} \mathbf{J}(\mathbf{r}') &=&\frac{I}{\pi \rho_0^2} \Theta(\rho'-\rho_0) \hat{\mathbf{x}} \label{eq:jcur}, \end{eqnarray}\]
Due to the rotational symmetry of the set up, we can compute the magnetic field at \(y=0\) and then rotate it to the desired angle. \[\begin{eqnarray} \mathbf{r}-\mathbf{r}' &=& z \, \hat{\mathbf{z}} +(x-x')\, \hat{\mathbf{x}} \label{eq:deltar}, \end{eqnarray}\]
Let us first consider the magnetic field outside the wire., i.e., \(z> \rho_0\). \[\begin{eqnarray} \mathbf{B}(x,z) &=&\frac{\mu_0 }{4\pi} I \int_{-\frac{L}{2}}^{\frac{L}{2}} dx' \frac{ \hat{\mathbf{x}} \times ( z \, \hat{\mathbf{z}} +(x-x')\, \hat{\mathbf{x}} )}{ ( z^2+(x-x')^2)^{3/2}} =z \, \hat{\mathbf{y}} \frac{\mu_0 }{4\pi} I \int_{-\frac{L}{2}}^{\frac{L}{2}} dx' \frac{ 1 }{( z^2+(x-x')^2)^{3/2}}\nonumber\\ &=& \, \hat{\mathbf{y}} \frac{\mu_0 }{4\pi z} I \int_{-\frac{L/2-x}{z}}^{\frac{L/2+x}{z}} du \frac{ 1 }{( 1+u^2)^{3/2}} = \, \hat{\mathbf{y}} \frac{\mu_0 }{4\pi z} I \left[ \frac{u}{\sqrt{1+u^2}} \right]_{-\frac{L/2-x}{z}}^{\frac{L/2+x}{z}}\nonumber\\ &=& \, \hat{\mathbf{y}} \frac{\mu_0 }{4\pi z} I \left[ \frac{x+L/2}{\sqrt{z^2+\left(x+L/2 \right)^2}} -\frac{x- L/2}{\sqrt{z^2+\left(x- L/2 \right)^2}} \right]. \label{eq:biotsavartlaw2} \end{eqnarray}\]
The Flux Linkage
Now we have to deal with a bit of ambiguity. We will want to integrate over \(x\) to compute the linked-flux, however, we can’t really integrate from \(-\infty\) to \(\infty\) because this will enclose other wires supplying current to the wire we are considering. The exact definition of the linked-flux is heavily dependent on how the circuit is closed. We will assume the boundaries \([-L/2,L/2]\) and integrate over this finite interval. This methodology is consistent with the other method of computing the self-inductance of a wire segment as proposed by Neumann. [2] [3] [4]
The differential flux going through a strip of width \(dx\) at \(z\) is given by \[\begin{eqnarray} d\Phi(z)&=&d z \int_{-L/2}^{L/2} dx \mathbf{B}(x,z) =dz \frac{\mu_0 I}{4\pi z} \left[\sqrt{z^2+\left(x+L/2 \right)^2}-\sqrt{z^2+\left(x- L/2 \right)^2}\right]_{-L/2}^{L/2} =dz \frac{\mu_0 I}{2\pi z} \left[\sqrt{z^2+L^2}-z\right]. \label{eq:dphi} \end{eqnarray}\]
Finally, we integrate over \(z\) to get the total flux: \[\begin{eqnarray} \Phi&=& \int_{\rho_0}^\infty d\Phi(z)= \frac{\mu_0 I}{2\pi} \int_{\rho_0}^\infty \frac{dz}{z} \left[\sqrt{z^2+L^2}-z\right] = \frac{\mu_0 I}{2\pi} \int_{\rho_0}^\infty dz \left[\frac{\sqrt{z^2+L^2}}{z}-1\right] \label{eq:flux} \end{eqnarray}\] Let’s do the harder integral first with the substitution \(z=L \sinh t\): \[\begin{eqnarray} I&=& \int dz \frac{\sqrt{z^2+L^2}}{z} =L \int dt \frac{\cosh^2 t}{\sinh t}=L \int dt\left[ \frac{1}{\sinh t}+\sinh t\right]=L \int dt \frac{2}{ e^t-e^{-t}}+L\cosh t\nonumber\\ &=&L \int d(e^t) \frac{2}{ e^{2t}-1}+L\cosh t=L \left[ \ln \left( \frac{e^t+1}{e^t-1} \right) + \sinh t \right]=L \left[ \ln \left( \tanh \frac{t}{2} \right) + \cosh t \right] \label{eq:theintegral} \end{eqnarray}\] Now use \(\tanh \frac{t}{2}=\frac{\sinh t}{1+\cosh t}\) to get:
\[\begin{eqnarray} I&=& L \ln \left( \frac{z}{L+\sqrt{z^2+L^2}} \right) + \sqrt{z^2+L^2} \label{eq:theintegral2}. \end{eqnarray}\] Putting it all together we get: \[\begin{eqnarray} \Phi&=& \frac{\mu_0 I}{2\pi} \left[L \ln \left( \frac{z}{L+\sqrt{z^2+L^2}} \right) + \sqrt{z^2+L^2} -z\right]_{\rho_0}^\infty\nonumber\\ &=& \frac{\mu_0 I}{2\pi} \left[-L \ln \left( \frac{\rho_0}{L+\sqrt{\rho_0^2+L^2}} \right) - \sqrt{\rho_0^2+L^2} +\rho_0\right] \simeq \frac{\mu_0 I L}{2\pi} \left[\ln \left( \frac{2L}{\rho_0} \right) -1\right] \label{eq:fluxfinal}, \end{eqnarray}\] where we assumed \(\rho_0 \ll L\). Using the relation \(\mathcal{L}=I\Phi\) we get: \[\begin{eqnarray} \mathcal{L}&=&\frac{\mu_0 L}{2\pi} \left[\ln \left( \frac{2L}{\rho_0} \right) -1\right] \label{eq:lout}. \end{eqnarray}\]
The Energy in the wire
If the current is distributed uniformly over the wire, the magnetic field inside the wire is given by \[\begin{eqnarray} B(r)&=&\frac{\mu I}{2\pi} \frac{r}{\rho_0^2} \label{eq:bin}, \end{eqnarray}\] where \(\mu\) is the permeability of the wire. The corresponding energy is given by The corresponding energy is given by \[\begin{eqnarray} U&=&\int d^3 \mathbf{r} \frac{B^2}{2\mu} =2\pi \int_0^{\rho_0} dr r \frac{B^2}{2\mu} =\frac{\mu I^2}{4\pi} \int_0^{\rho_0} dr \frac{r^3}{\rho_0^4} =\frac{1}{2}\left[\frac{\mu}{8\pi} \right] I^2\equiv \frac{1}{2} \mathcal{L} I^2 \label{eq:u}, \end{eqnarray}\] which results in \[\begin{eqnarray} \mathcal{L}&=&\frac{\mu}{8\pi} \label{eq:lin}. \end{eqnarray}\]
Putting it all together we get: \[\begin{eqnarray} \mathcal{L}&=&\frac{\mu_0 L}{2\pi} \left[\ln \left( \frac{2L}{\rho_0} \right) -1 +\frac{\mu}{4 \mu_0}\right] \label{eq:lfinal}. \end{eqnarray}\]