Question: Use Eq. 5.41 to obtain the magnetic field on the axis of the rotating disk in Prob. 5.37(a). Show that the dipole field (Eq. 5.88), with the dipole moment you found in Prob. 5.37, is a good approximation if z>> R.

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

dQ2कr $\left(\mathrm{dr}\right)$

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

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

$dt=\frac{\vartheta }{\omega }$

Here, $\omega$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.

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

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

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

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

The dipole field for the rotating disk is given as:

$B_d=\frac{{\mu }_{0}m}{4\pi r^3}(2\cos \theta +\sin \theta )$

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

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.