Module 3 : MAGNETIC FIELD
Lecture 19 : Mutual Inductance
  Example 24 :
  Obtain an expression for the self inductance in a toroid of inner radius $a$, outer radius $b$ and height $h$.
Solution :
  We have seen that the field inside the toroid at a distance $r$ from the axis of the toroid is given by $ B(r) = \mu_0NI/2\pi r$.
  The flux through one turn of the coil is the integral of this field over the cross section of the coil
 
\begin{displaymath}\Phi {\rm (one\ turn)}\ = h\int_a^b \frac{\mu_0NI}{2\pi r} dr = \frac{\mu_0INh}{2\pi}\ln(\frac{b}{a})\end{displaymath}
  The flux threading $N$ turns is
 
\begin{displaymath}\Phi = \frac{\mu_0IN^2h}{2\pi}\ln(\frac{b}{a})\end{displaymath}

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