Find the self-inductance of a toroidal coil with rectangular cross section (inner radius a, outer radius b, height h), that carries a total of N turns.
Question 7.11: Find the self-inductance of a toroidal coil with rectangular...
The magnetic field inside the toroid is (Eq. 5.60)
B(r)=\begin{cases} \frac{\mu _{0}NI}{2\pi}\hat{\phi }, for points inside the coil, \\ 0, for points outside the coil,\end{cases} (5.60)
B\frac{\mu _{0}NI}{2\pi s}.The flux through a single turn (Fig. 7.34) is
\int{B.da}=\frac{\mu _{0}NI}{2\pi}h \int_{a}^{b}{\frac{1}{s}ds }=\frac{\mu _{0}NIh}{2\pi}\ln \left(\frac{b}{a}\right) .The total flux is N times this, so the self-inductance (Eq. 7.26) is
\phi = L I. (7.26)
L=\frac{\mu _{0}N^{2}h}{2\pi}\ln \left(\frac{b}{a}\right) . (7.28)
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