The magnitude of the electric field between the two circular parallel plates in Fig. is$\mathit{E}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{t}\mathbf{)}$, with $\mathit{E}$in volts per meter and $\mathit{t}$in seconds. At $\mathbf{t}\mathbf{=}\mathbf{0}$, $\overrightarrow{\mathbf{E}}$is upward. The plate area is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{\hspace{0.17em}}{\mathbf{m}}^{\mathbf{2}}$. For $\mathit{t}\mathbf{\ge }\mathbf{0}$,(a) What is the magnitude and (b) What is the direction (up or down) of the displacement current between the plates, and (c) What is the direction ofthe induced magnetic field clockwise or counter-clockwise in the figure?

The magnitude of an electric field, $E=\left(4\times {10}^5\right)-\left(6\times {10}^{4}t\right)$

Plate area, $A=4\times {10}^{-2}{\text{m}}^{\text{2}}$

By substituting the electric flux into the displacement current formula and taking the derivative of an electric field with respect to time, we can get the magnitude of the displacement current. The sign of the displacement current shows its direction. Using Ampere’s law, we can find the direction of the induced magnetic field.

Formulae are as follows:

$$\begin{gathered} i_d={\epsilon }_0\left(\frac{d{\varphi }_E}{dt}\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_{enc} \end{gathered}$$

where, $i_d$is the displacement current, $\phi$is the flux, and $\overrightarrow{B}$ is the magnetic field.

The magnitude of displacement current between the plates:

The displacement current is given by,

$i_d={\epsilon }_0\left(\frac{d{\varphi }_E}{dt}\right)$

The electric flux is given by,

${\varphi }_E=EA$

Therefore, the displacement current becomes,

$i_d={\epsilon }_0\frac{d}{dt}\left(EA\right)={\epsilon }_{0}A\frac{d}{dt}\left(E\right)$

Substituting the magnitude of the electric field,

$$\begin{gathered} i_d={\epsilon }_{0}A\frac{d}{dt}\left[\left(4\times {10}^5\right)-\left(6\times {10}^{4}t\right)\right] \\ {i}_d=-{\epsilon }_{0}A\left(6\times {10}^4\frac{\text{V}}{\text{m.s}}\right) \\ {i}_d=-\left(8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{{\mathrm{Nm}}^2}\right)\left(4\times {10}^{-2}{\mathrm{m}}^2\right)\left(6\times {10}^4\frac{\text{V}}{\text{m.s}}\right) \\ {i}_d=-2.1\times {10}^{-8}\text{A} \\ \left|i_d\right|=2.1\times {10}^{-8}\text{A.} \end{gathered}$$

Hence, the magnitude of displacement current between the plates is $\left|i_d\right|=2.1\times {10}^{-8}\text{A.}$

The direction of displacement current between the plates:

The negative sign of the displacement current shows that its direction is opposite to the electric field.

Therefore, the direction of the displacement current is downward.

The direction of induced magnetic field:

Draw the Gaussian surface around the circular plates, and the enclosed current will be the displacement current.

Then, by using Ampere’s law,

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_{enc} \\ \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_d \end{gathered}$$

Since the displacement current is negative,

$\oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_d<0$

Thus,

$\oint \overrightarrow{B}.d\overrightarrow{s}<0$

Hence, this shows that the induced magnetic field must be clockwise.

Using the displacement current formula, the magnitude and direction of displacement current can be found. Using Ampere’s law, the direction of an induced magnetic field can be found.