91Ó°ÊÓ

The radius of the disk is$R$, total charge on the disk is$Q$and the angular velocity f the disk is$\mathrm{Ó\neg }$.

The relation between the linear velocity $\left(v\right)$and the angular velocity $\left(\mathrm{Ó\neg }\right)$is given by

$v=\mathrm{Ó\neg }R......................\left(1\right)$

The surface charge density $\left(\mathrm{σ}\right)$of the disk is given by

role="math" localid="1649565802454" $\mathrm{σ}=\frac{Q}{\mathrm{π}{R}^2}.....................\left(2\right)$

Let's consider an infinitesimal ring of thickness $dr$at a distance $r$from the center of the disk.

The amount of charge $\left(dq\right)$contained within this infinitesimal ring is given by

$dq=2\mathrm{π}r\mathrm{σ}dr......................\left(3\right)$

According to Biot-Savart's law, the magnetic field$\left(dB\right)$ due to this infinitesimal charge is given by

role="math" localid="1649565540422" $dB=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{dqv}{r^2}...................\left(5\right)$

Here, ${\mathrm{μ}}_0$is the permeability of free space.

Substitute the expression for $dq$from equation (3) and the expression for $v$from equation (1) and simplify to obtain the infinitesimal magnetic field.

role="math" localid="1649565738953" $\begin{array}{rcl}dB& =& \frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\left(2\mathrm{Ï€}r\mathrm{σ}dr\right)\left(\mathrm{Ó\neg }r\right)}{r^2}\\ & =& \frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }}{2}dr............................\left(6\right)\end{array}$

The formula to calculate the net magnetic field $\left(B\right)$is given by

$B={∫}_0^{R}dB..........................\left(7\right)$

Substitute the expression for $dB$from equation (6) into equation (7) and simplify to obtain the magnetic field.

role="math" localid="1649565846430" $\begin{array}{rcl}B& =& {∫}_0^R\frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }}{2}dr\\ & =& \frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }}{2}R...........................\left(8\right)\end{array}$

Substitute role="math" localid="1649565878042" $\left(\frac{Q}{\mathrm{π}{R}^2}\right)$for $\mathrm{σ}$into equation (8) to obtain the required magnetic field.

role="math" localid="1649565929821" $\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_0\left(\frac{Q}{\mathrm{Ï€}{R}^2}\right)\mathrm{Ó\neg }}{2}R\\ & =& \frac{{\mathrm{μ}}_0\mathrm{Ó\neg }Q}{2\mathrm{Ï€}R}\end{array}$

The required magnetic field is given by$B=\begin{array}{rcl}& & \frac{{\mathrm{μ}}_0\mathrm{Ó\neg }Q}{2\mathrm{Ï€}R}\end{array}$.