A toroid has a core of non-magnetic material, inner radius r 1 , outer radius r 2 , and height h . It has N turns uniformly wound around it. What is its self-inductance?

Question

A toroid has a core of non-magnetic material, inner radius r 1 ​ , outer radius r 2 ​ , and height h . It has N turns uniformly wound around it. What is its self-inductance?

TestHub · Accepted Answer

The magnetic field inside a toroid at a radius $r$ is $B (r) = \frac{μ _{0} N I}{2 π r}$ .

Consider a differential area element $d A = h d r$ between $r$ and $r + d r$ . The magnetic flux through this area is $d Φ = B (r) d A = \frac{μ _{0} N I}{2 π r} h d r$ . The total flux linkage for $N$ turns is $Φ_{t o t a l} = N \int_{r_{1}}^{r_{2}} d Φ = N \int_{r_{1}}^{r_{2}} \frac{μ _{0} N I h}{2 π r} d r = \frac{μ _{0} N ^{2} I h}{2 π} [ln r]_{r_{1}}^{r_{2}} = \frac{μ _{0} N ^{2} I h}{2 π} ln (\frac{r _{2}}{r _{1}})$ . Since $Φ_{t o t a l} = L I$ , the self-inductance is $L = \frac{μ _{0} N ^{2} h}{2 π} ln (\frac{r _{2}}{r _{1}})$ .

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