Use Eq. $\mathbf{5}\mathbf{.}\mathbf{41}$to obtain the magnetic field on the axis of the rotating disk in Prob. $\mathbf{5}\mathbf{.}\mathbf{37}\mathbf{\left(}\mathbf{a}\mathbf{\right)}$. Show that the dipole field (Eq. $\mathbf{5}\mathbf{.}\mathbf{88}$), with the dipole moment you found in Prob. $\mathbf{5}\mathbf{.}\mathbf{37}$, is a good approximation if$\mathbf{z}\mathbf{>}\mathbf{>}\mathbf{}\mathbf{R}\mathbf{.}$

The total charge on the small element of ring is given as:

$\mathbf{}\mathbf{dC}\mathbf{2}\mathbf{r}\mathbf{\left(}\mathbf{dr}\mathbf{\right)}$

Here, $\mathrm{\sigma }$is the surface charge density for the rotating disk.

The time period for the revolution of disk is given as:

role="math" localid="1658118995711" $\mathbf{dt}\mathbf{=}\frac{\mathbf{2}\mathbf{r}}{\mathbf{\omega }}$

Here, $\mathrm{\omega }$is the angular velocity of rotating disk.

The current in the small element of the disk is given as:

$\mathrm{l}=\frac{\mathrm{dQ}}{\mathrm{dt}}$

$\mathrm{l}=\frac{\mathrm{\sigma }\left(2\mathrm{\pi r}\right)\mathrm{dr}}{\left(\frac{2\mathrm{\pi }}{\mathrm{\omega }}\right)}$

role="math" localid="1658119084316" $\mathrm{I}=\mathrm{\sigma }\left(\mathrm{r\omega }\right)\mathrm{dr}$

The magnetic field on the axis of rotating disk due to small element of disk is given as:

$\mathrm{dB}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2}\left(\frac{{\mathrm{r}}^2}{{\left({\mathrm{r}}^2+{\mathrm{z}}^2\right)}^{3/2}}\right)$

$\mathrm{dB}=\frac{{\mathrm{\mu }}_0\left(\mathrm{\sigma }\right(\mathrm{r\omega }\left)\mathrm{dr}\right)}{2}\left(\frac{{\mathrm{r}}^2}{{\left({\mathrm{r}}^2+{\mathrm{z}}^2\right)}^{3/2}}\right)$

The total magnetic field on the axis of rotating disk is given as:

$\mathrm{B}={\int }_0^{\mathrm{R}} \mathrm{dB}$

$\mathrm{B}={\int }_0^{\mathrm{R}} \left(\frac{{\mathrm{\mu }}_0\left(\mathrm{\sigma }\right(\mathrm{r\omega }\left)\mathrm{dr}\right)}{2}\left(\frac{{\mathrm{r}}^2}{{\left({\mathrm{r}}^2+{\mathrm{z}}^2\right)}^{3/2}}\right)\right)$

$\mathrm{B}=\frac{{\mathrm{\mu t}}_0\mathrm{\sigma \omega }}{2}\left(\frac{{\mathrm{R}}^2+2{\mathrm{z}}^2}{\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}}-2\mathrm{z}\right)$

Apply the approximation $\mathrm{z}>>\mathrm{R}$in the above expression.

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\sigma \omega }}{2}\left(\frac{{\mathrm{R}}^2+2{\mathrm{z}}^2}{\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}}-2\mathrm{z}\right)$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\sigma \omega }}{2}\left[\frac{2{\mathrm{z}}^2\left(1+\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}\right)}{\mathrm{z}{\left(1+\frac{\mathrm{R}}{\mathrm{z}}\right)}^{-1/2}}-2\mathrm{z}\right]$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\sigma \omega }}{2}\left[2\mathrm{z}\left(1+\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}\right)\left(1-\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}+\frac{3}{8}\left(\frac{{\mathrm{R}}^4}{{\mathrm{z}}^4}\right)\right)-1\right]$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\sigma \omega R}}^4}{2{\mathrm{z}}^3}$

${\mathrm{B}}_{\mathrm{d}}=\frac{{\mathrm{\mu }}_0{\mathrm{\sigma \omega R}}^4}{8{\mathrm{z}}^3}$The dipole moment for the rotating disk by equation$5.37$ is given as:

$\mathrm{m}=\frac{{\mathrm{\pi \sigma \omega R}}^4}{4}$

The dipole field for the rotating disk is given as:

${\mathrm{B}}_{\mathrm{d}}=\frac{{\mathrm{\mu }}_0\mathrm{m}}{4{\mathrm{\pi r}}^3}(2\cos \mathrm{\theta }+\sin \mathrm{\theta })$

The points on the z axis $\mathrm{z}=\mathrm{r}$and $\mathrm{\theta }=0$.

Substitute all the values in the above equation.

${\mathrm{B}}_{\mathrm{\sigma }}=\frac{{\mathrm{\mu }}_0\left(\frac{{\mathrm{\pi \sigma \omega R}}^4}{4}\right)}{4\mathrm{\pi }(\mathrm{z}{)}^3}(2\cos (0)+\sin (0\left)\right)$

Therefore, it is clear that to obtain magnetic field on the axis of rotating disk the approximation in dipole field for very small distance compared to radius of disk.