91Ó°ÊÓ

To find the displacement current density in the sir space between the plates, according to the equations:

$$\begin{gathered} i_C=\frac{dq}{dt}\frac{d\mathrm{Ï•}E}{dt} \\ {i}_D=\frac{d\mathrm{Ï•}E}{dt} \end{gathered}$$

Therefore, we can see that the condition and the displacement currents are equal, i.e.$i_C=i_D$

Hence the displacement current density is:

$$\begin{gathered} j_D=\frac{i_D}{A}= \\ \frac{i_C}{A}=\frac{0.520A}{\mathrm{Ï€}{r}^2} \\ =\frac{0.520A}{\mathrm{Ï€}(0.0400m^2)} \\ =103A/m^2 \end{gathered}$$

Therefore, the displacement current density is$103A/m^2$

To calculate the rate at which electric field between the plates is changing .

The displacement current is:

$$\begin{gathered} i_D={\mathrm{E}}_0\frac{d\mathrm{ϕ}E}{dt} \\ =\mathrm{E}A\frac{dE}{dt}{j}_D \\ =\frac{i_D}{A}={\mathrm{E}}_0\frac{dE}{dt}\frac{dE}{dt} \\ =\frac{{\mathrm{j}}_{\mathrm{D}}}{{\mathrm{E}}_0}=\frac{103}{{\displaystyle 8.854×{10}^{-12}}} \\ =1.16×{10}^{-12}\mathrm{V}/\mathrm{m}.\mathrm{s} \end{gathered}$$

Therefore, the rate of change is$1.16×{10}^{-12}\mathrm{V}/\mathrm{m}.\mathrm{s}$

For the plates kept at a distance of from the axis.

$âˆ\circledR \stackrel{→}{B}·\stackrel{→}{dl}={\mathrm{μ}}_0(i_C+i_D)$

Since no conduction current flows through the space, and the displacement current is enclosed by the path is $i_D=j_D\mathrm{Ï€}{r}^2$.

$B\left(2\mathrm{π}r\right)={\mathrm{μ}}_0\left(j_D\mathrm{π}{r}^2\right)$

Or,

$$\begin{gathered} B=\frac{1}{2}{\mathrm{μ}}_0{j}_{D}r \\ =\frac{1}{2}(4\mathrm{π}×{10}^{-}7T·m/A)(103A/m^2)(0.0200m) \\ =1.30×{{10}^{-}}^{6}T \end{gathered}$$

For the distance of 1.00cm

$$\begin{gathered} B=\frac{1}{2}{\mathrm{μ}}_0{j}_{D}r \\ =\frac{1}{2}(4\mathrm{π}×{10}^{-}7T·m/A)(103A/m^2)(0.0100m) \\ =6.50×{{10}^{-}}^{6}T \end{gathered}$$

Therefore, the magnetic field at a plate separation of is$6.50×{{10}^{-}}^{6}T$