Module 3 : MAGNETIC FIELD
Lecture 15 : Biot- Savarts' Law
  Solution :
The current on the disk can be calculated by assuming the rotating disk to be equivalent to a collection of concentric current loops. Consider a ring of radius $r$and of width $dr$. As the disk is rotating with an angular speed $\omega$, the rotating charge on the ring essentially behaves like a current loop carrying current $\sigma\cdot 2\pi r dr\cdot\omega/2\pi = \sigma\omega r dr$.
  The field at a distance $z$due to this ring is
 
\begin{displaymath}dB = \frac{\mu_0(\sigma\omega r dr)}{2}\frac{r^2}{(r^2+z^2)^{3/2}}\end{displaymath}
  The net field is obtained by integrating the above from $r=0$to $r=R$
 
\begin{eqnarray*} B &=& \frac{\mu_0\sigma\omega}{2}\int_0^R\frac{r^3}{(r^2+z^2)^... ...0\sigma\omega}{2}\int_0^R\frac{r^2+z^2-z^2}{(r^2+z^2)^{3/2}}rdr \end{eqnarray*}

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