91影视

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

诲蚕2啶时 $\left(\mathrm{dr}\right)$

Here, $蟽$is the surface charge density for the rotating disk.

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

$dt=\frac{蠎}{蝇}$

Here, $蝇$is the angular velocity of rotating disk.

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

$\backslash begin\left\{aligned\right\}\&I=\backslash frac\{dQ\}\{dt\}\backslash \backslash \&I=\backslash frac\{\backslash sigma(2\backslash pir)dr\}\{\backslash left(\backslash frac\{2\backslash pi\}\{\backslash omega\}\backslash right\left)\right\}\backslash \backslash \&I=\backslash sigma(r\backslash omega)dr\backslash end\left\{aligned\right\}$

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

$\backslash begin\left\{aligned\right\}\&dB=\backslash frac\{\backslash mu\_\{0\}l\}\left\{2\right\}\backslash left(\backslash frac\{r^\left\{2\right\}\left\}\right\{\backslash left(r^\{2\}+z^\{2\}\backslash right)^\{3/2\}\}\backslash right)\backslash \backslash \&dB=\backslash frac\{\backslash mu\_\{0\left\}\right(\backslash sigma(r\backslash omega)dr\left)\right\}\left\{2\right\}\backslash left(\backslash frac\{r^\left\{2\right\}\left\}\right\{\backslash left(r^\{2\}+z^\{2\}\backslash right)^\{3/2\}\}\backslash right)\backslash end\left\{aligned\right\}$

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

$\backslash begin\left\{aligned\right\}B\&=\backslash int\_\left\{0\right\}^\left\{R\right\}dB\backslash \backslash B\&=\backslash int\_\left\{0\right\}^\left\{R\right\}\backslash left(\backslash frac\{\backslash mu\_\left\{0\right\}(\backslash sigma(r\backslash omega)dr)\left\}\right\{2\}\backslash left(\backslash frac\{r^\{2\left\}\right\}\{\backslash left(r^\left\{2\right\}+z^\left\{2\right\}\backslash right)^\{3/2\left\}\right\}\backslash right)\backslash right)\backslash \backslash B\&=\backslash frac\{\backslash mu\_\{0\}\backslash sigma\backslash omega\}\left\{2\right\}\backslash left(\backslash frac\{R^\left\{2\right\}+2z^\left\{2\right\}\left\}\right\{\backslash sqrt\{R^\{2\}+z^\{2\left\}\right\}\}-2z\backslash right)\backslash end\left\{aligned\right\}$

Apply the approximation $z>>R$in the above expression.

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

$\backslash begin\left\{aligned\right\}\&B=\backslash frac\{\backslash mu\_\{0\}\backslash sigma\backslash omega\}\left\{2\right\}\backslash left[2z\backslash left(1+\backslash frac\{R^\{2\left\}\right\}\{2z^\{2\left\}\right\}\backslash right)\backslash left(1-\backslash frac\{R^\{2\left\}\right\}\{2z^\{2\left\}\right\}+\backslash frac\left\{3\right\}\left\{8\right\}\backslash left(\backslash frac\{R^\left\{4\right\}\left\}\right\{z^\left\{4\right\}\}\backslash right)\backslash right)-1\backslash right]\backslash \backslash \&B=\backslash frac\{\backslash mu\_\{0\}\backslash sigma\backslash omegaR^\{4\left\}\right\}\{2z^\{3\left\}\right\}\backslash end\left\{aligned\right\}$

The dipole moment for the rotating disk by equation 5.37 is given as:

$m=\frac{蟺蟽蝇R^4}{4}$

The dipole field for the rotating disk is given as:

$B_d=\frac{{渭}_{0}m}{4蟺r^3}(2\cos 胃+\sin 胃)$

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

Substitute all the values in the above equation.

$\backslash begin\left\{aligned\right\}\&B\_\left\{d\right\}=\backslash frac\{\backslash mu\_\{0\}\backslash left(\backslash frac\{\backslash pi\backslash sigma\backslash omegaR^\{4\left\}\right\}\left\{4\right\}\backslash right\left)\right\}\{4\backslash pi(z)^\{3\left\}\right\}(2\backslash \cos (0)+\backslash \sin (0\left)\right)\backslash \backslash \&B\_\left\{d\right\}=\backslash frac\{\backslash mu\_\{0\}\backslash sigma\backslash omegaR^\{4\left\}\right\}\{8z^\{3\left\}\right\}\backslash end\left\{aligned\right\}$

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.