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Question

(A) The magnetic field $B$ inside a toroid at a radial distance $r$ from the center is given by $B = \frac{\mu_0 n I}{2 \pi r}$.Consider an infinitesi

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$\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$

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(A) The magnetic field $B$ inside a toroid at a radial distance $r$ from the center is given by $B = \frac{\mu_0 n I}{2 \pi r}$.Consider an infinitesimal rectangular strip of width $dr$ and height $h$ at a distance $r$ from the center. The area element is $dA = h \, dr$.The magnetic flux $d\phi$ through this infinitesimal area is $d\phi = B \cdot dA = \left( \frac{\mu_0 n I}{2 \pi r} \right) (h \, dr)$.The total magnetic flux $\phi$ through the cross-section is obtained by integrating from $r = a$ to $r = b$:$\phi = \int_a^b \frac{\mu_0 n I h}{2 \pi r} dr = \frac{\mu_0 n I h}{2 \pi} \int_a^b \frac{1}{r} dr = \frac{\mu_0 n I h}{2 \pi} [\ln r]_a^b = \frac{\mu_0 n I h}{2 \pi} \ln \left( \frac{b}{a} \right)$.The self-inductance $L$ is defined as $L = \frac{n \phi}{I}$.Substituting the expression for $\phi$:$L = \frac{n}{I} \left( \frac{\mu_0 n I h}{2 \pi} \ln \left( \frac{b}{a} \right) \right) = \frac{\mu_0 n^2 h}{2 \pi} \ln \left( \frac{b}{a} \right)$.

Consider a toroid with a rectangular cross-section,of inner radius $a$,outer radius $b$,and height $h$,carrying $n$ number of turns. Then the self-inductance of the toroidal coil when current $I$ is passing through the toroid is:

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