91影视

Consider that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

Write the formula of electric field between the plates, as a function of.

$\mathbf{E}=\frac{蟽\left(\mathbf{t}\right)}{{\mathbf{蔚}}_{\mathbf{0}}}$ 鈥︹�� (1)

Here, $蟽\left(\mathbf{t}\right)$ is charge on the plate and ${\mathbf{蔚}}_{\mathbf{0}}$is relative permittivity.

Write the formula of magnetic field at a distance s from the axis.

$\mathbf{B}=\frac{{渭}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{d}}}{\mathbf{2}蟺s}$ 鈥︹�� (2)

Here, ${渭}_{\mathbf{0}}$is permeability, $I_d$is displacement current and s is the surface of the plates.

Write the formula of magnetic field in the same direction as the case b.

$\mathbf{B}=\frac{{渭}_{\mathbf{0}}\mathbf{I}}{2蟺\mathbf{s}}\left[\mathbf{1}-\frac{\mathbf{I}\left(\mathbf{s}\right)}{\mathbf{I}}\right]$ 鈥︹�� (3)

Here, ${渭}_{\mathbf{0}}$is permeability, I is current, I_S is current passing through curve.

Determine the charge on the plate is:

$$\begin{gathered} Q\left(t\right)=蟽\left(t\right)S \\ =It \end{gathered}$$

Since the charge density grows linearly with time and is homogeneous throughout the plate. The electric field is thus:

Determine the electric field between the plates, as a function of t.

Substitute $\frac{Q\left(t\right)}{S}$for$蟽\left(t\right)$into equation (1).

$$\begin{gathered} E=\frac{Q\left(t\right)}{{蔚}_{0}S} \\ =\frac{It}{S{蔚}_0} \end{gathered}$$

Here, S is the surface of the plates.

The magnetic field will curve around the current if it flows down the z-axis, using Ampere's equation with the displacement current term.

Determine the displacement current.

$I_d=EdS$

Determine the magnetic field at a distance from the axis.

Substitute EDS for I_d into equation (2).

$$\begin{gathered} B=\frac{{渭}_0{蔚}_0}{2蟺s}\frac{d}{dt}{鈭�}_{S}EdS \\ =\frac{{渭}_0\left(s^2蟺\right)}{2蟺s}\frac{I}{蟺a^2} \\ =\frac{{渭}_{0}I}{2}\frac{s}{A}\widehat{蠒} \end{gathered}$$

Whereas, the integration loop was positioned between the capacitor plates as a circle whose radius was coaxial with the plates.

Using a new bounding surface but the same Amperian loop once more, the magnetic field is:

$2蟺sB={渭}_0{I}_{enc}$

There will be two contributions from "regular" current but there is contribution from the displacement current. Given that the surface's normal points inwards, one is on surface S₁ (positive), and the other is at curve C₁ (negative).

$$\begin{gathered} 2蟺sB={渭}_0{I}_{enc} \\ 2蟺sB={渭}_{0}I-{渭}_{0}I\left(s\right) \\ B=\frac{{渭}_{0}I}{2蟺s}\left[1-\frac{I\left(s\right)}{I}\right] \end{gathered}$$

The charge density must only be a function of time for the current I(s) to travel through curve. Let's say that all we search for is the charge density on the circle enclosed by curve:

$$\begin{gathered} 蟽\left(t\right)=\frac{Q\left(s,t\right)}{s^2蟺} \\ =\frac{1}{s^2蟺}\left[I-I\left(s\right)\right]t \end{gathered}$$

For to only be a function of time the term $I-I\left(s\right)$ needs to be proportional to;

$I-I\left(s\right)=C{s}^2$

Substitute 0 for l(s) into above equation.

$C=\frac{I}{s^2}$

Now substitute $\frac{1}{s^2}$for C into above equation.

$I\left(s\right)=I-I{\left(\frac{s}{a}\right)}^2$

Determine the magnetic field in the same direction as the case b.

$$\begin{gathered} B=\frac{{渭}_{0}I}{2蟺s}\left[1-1+{\left(\frac{s}{a}\right)}^2\right] \\ =\frac{{渭}_{0}I}{2蟺}\frac{s}{a^2} \end{gathered}$$

Therefore, the value of magnetic field in the same direction as the case b is .

$\stackrel{鈫�}{B}=\frac{{渭}_{0}I}{2蟺}\frac{s}{a^2}$