A capacitor with parallel circular plates of the radius $\mathbf{R}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{cm}$is discharging via a current of 12.0 A . Consider a loop of radiusR/3that is centered on the central axis between the plates. (a)How much displacement current is encircled by the loop? The maximum induced magnetic field has a magnitude of 12.0 mT. (b)At what radius inside and (c)outside the capacitor gap is the magnitude of the induced magnetic field 3.00 mT?

The radius of circular plates,R = 1.20 cm

Current,i = 12 A

The magnitude of maximum induced magnetic field, B = 12 mT

First, find the area enclosed and the total area of the circular plate. Substituting these values in the formula for enclosed current, calculate the displacement current encircled by the loop. Using the relations for magnetic field inside and outside find the distance r

Formulae are as follows:

$\frac{i}{A}=\frac{i_{d,enc}}{A\text{'}}$

$$\begin{gathered} B_{in}=\frac{{\mu }_{0}r}{2\pi R^2} \\ {B}_{out}=\frac{{\mu }_0{i}_{enc}}{2\pi r} \end{gathered}$$

where,i is current, r is the distance, B is the magnetic field, R is the radius.

Displacement current encircled by the loop of radius, R/3:

The current enclosed is the displacement current. Since the loop ofthegiven radius is inside, the portion of the displacement current enclosed is given by,

$\frac{i}{A}=\frac{i_{d,enc}}{A\text{'}}$

$A=\mathrm{\pi }{R}^2,A\text{'}=\mathrm{\pi }{r}^2=\mathrm{\pi }{\left(\frac{R}{3}\right)}^2$

$i_{d,enc}=\frac{i\text{A'}}{\text{A}}=i\frac{\pi {\left(\frac{R}{3}\right)}^2}{\pi R^2}$

$i_{d,enc}=\frac{i}{9}=\frac{12\text{A}}{9}=1.33\text{A}$

Therefore, the displacement current encircled by the loop of radius R/3 is 1.33 A

Radius inside the capacitor gap that has a magnitude of the magnetic field, 3 mT :

The magnetic field induced by the changing electric field is proportional to the distance r.

The maximum magnitude of induced magnetic field for radius r = R is given in the problem.

$B_{max}=12\text{mT}$

Taking the ratio of the magnetic field at distance r to R ,

$$\begin{gathered} \frac{B}{B_{max}}=\frac{r}{R} \\ r=\frac{B}{B_{max}}R \end{gathered}$$

The magnetic field inside the gap at radius r is B = 3 mT .

Therefore,

$\begin{array}{rcl}r& =& \frac{3\mathrm{mT}}{12\mathrm{mT}}\left(1.20\mathrm{cm}\right)\\ & =& 0.300\mathrm{cm}\end{array}$

Hence, the radius inside the capacitor gap that has the magnitude of an induced magnetic field 3 mT is 0.300 cm.

Radius outside the capacitor gap that has magnitude of a magnetic field 3 mT :

The magnetic field outside is inversely proportional to distance r

$B_{max}\propto \frac{1}{R}$

$B_{out}\propto \frac{1}{r}$

Taking the ratio,$\frac{B_{out}}{B_{max}}$,

$\begin{array}{rcl}\frac{B_{out}}{B_{max}}& =& \frac{R}{r}\\ r& =& \frac{B_{max}}{B_{out}}\\ & =& \frac{12\mathrm{mT}}{3\mathrm{mT}}\left(1.20\mathrm{cm}\right)\\ & =& 4.80\mathrm{cm}\end{array}$

Hence, the radius outside the capacitor gap that has the magnitude of an induced magnetic field 3 mT is 4.80 cm

Using Ampere’s law, the magnetic field inside and outside the capacitor plate can be found.