91影视

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

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

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

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

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

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

$\mathrm{l}=\frac{\mathrm{蟽}\left(2\mathrm{蟺谤}\right)\mathrm{dr}}{\left(\frac{2\mathrm{蟺}}{\mathrm{蝇}}\right)}$

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

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

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

$\mathrm{dB}=\frac{{\mathrm{渭}}_0\left(\mathrm{蟽}\right(\mathrm{r蝇}\left)\mathrm{dr}\right)}{2}\left(\frac{{\mathrm{r}}^2}{{\left({\mathrm{r}}^2+{\mathrm{z}}^2\right)}^{3/2}}\right)$

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

$\mathrm{B}={鈭�}_0^{\mathrm{R}}鈥�\mathrm{dB}$

$\mathrm{B}={鈭�}_0^{\mathrm{R}}鈥�\left(\frac{{\mathrm{渭}}_0\left(\mathrm{蟽}\right(\mathrm{r蝇}\left)\mathrm{dr}\right)}{2}\left(\frac{{\mathrm{r}}^2}{{\left({\mathrm{r}}^2+{\mathrm{z}}^2\right)}^{3/2}}\right)\right)$

$\mathrm{B}=\frac{{\mathrm{渭迟}}_0\mathrm{蟽蝇}}{2}\left(\frac{{\mathrm{R}}^2+2{\mathrm{z}}^2}{\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}}鈭�2\mathrm{z}\right)$

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

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{蟽蝇}}{2}\left(\frac{{\mathrm{R}}^2+2{\mathrm{z}}^2}{\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}}鈭�2\mathrm{z}\right)$

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{蟽蝇}}{2}\left[\frac{2{\mathrm{z}}^2\left(1+\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}\right)}{\mathrm{z}{\left(1+\frac{\mathrm{R}}{\mathrm{z}}\right)}^{鈭�1/2}}鈭�2\mathrm{z}\right]$

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{蟽蝇}}{2}\left[2\mathrm{z}\left(1+\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}\right)\left(1鈭�\frac{{\mathrm{R}}^2}{2{\mathrm{z}}^2}+\frac{3}{8}\left(\frac{{\mathrm{R}}^4}{{\mathrm{z}}^4}\right)\right)鈭�1\right]$

$\mathrm{B}=\frac{{\mathrm{渭}}_0{\mathrm{蟽蝇搁}}^4}{2{\mathrm{z}}^3}$

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

$\mathrm{m}=\frac{{\mathrm{蟺蟽蝇搁}}^4}{4}$

The dipole field for the rotating disk is given as:

${\mathrm{B}}_{\mathrm{d}}=\frac{{\mathrm{渭}}_0\mathrm{m}}{4{\mathrm{蟺谤}}^3}(2\cos 鈦�\mathrm{胃}+\sin 鈦�\mathrm{胃})$

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

Substitute all the values in the above equation.

${\mathrm{B}}_{\mathrm{蟽}}=\frac{{\mathrm{渭}}_0\left(\frac{{\mathrm{蟺蟽蝇搁}}^4}{4}\right)}{4\mathrm{蟺}(\mathrm{z}{)}^3}(2\cos 鈦�(0)+\sin 鈦�(0\left)\right)$

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.