Q72P - Fundamentals Of Physics

Question

Two plates (as in Fig. 32-7) are being discharged by a constant current. Each plate has a radius of  $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{cm}$. During the discharging, at a point between the plates at radial distance  $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{cm}$ from the central axis, the magnetic field has a magnitude of  $\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{nT}$ . (a) What is the magnitude of the magnetic field at radial distance $\mathbf{6}\mathbf{.}\mathbf{00}$  ? (b) What is the current in the wires attached to the plates?

Step-by-Step Solution

Answer

1. The magnitude of the magnetic field at radial distance $6.00\text{ cm}$   is $16.7\mathrm{nT}$ .

Step 1: Listing the given quantities:

Radius of plates  $\mathrm{R}=0.04 \mathrm{m}$

Radial distance,  ${\mathrm{r}}_1=0.02\mathrm{m}$

Radial distance, ${\mathrm{r}}_2=0.06\mathrm{m}$           

Magnetic field,  ${\mathrm{B}}_1=12.5\times {10}^{-9}\mathrm{T}$

Step 2: Understanding the concepts of magnetic field:

This magnetic field increases from the center of the capacitor to a maximum value at the edges of the capacitor and then begins to decrease again as we move away from the capacitor.

 

Formulas:                                          

The magnetic field inside the plates: ${\mathrm{B}}_1=\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2{\mathrm{\pi R}}^2}\right){\mathrm{r}}_1$

The magnetic field outside the plates: ${\mathrm{B}}_2=\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2{\mathrm{\pi r}}_2}\right)$

Here, ${\mathrm{B}}_1$  and  ${\mathrm{B}}_2$ are the magnetic field inside and outside the plates respectively,  $i_d$ is the fictitious current, $\mathrm{R}$  is the radius of the plate, ${\mathrm{r}}_1$  and $r_2$  are the radial distance, and ${\mathrm{\mu }}_0$  is the permeability of free space having a value of  $4\mathrm{\pi }\times {10}^{-7}\text{ N}/{\text{A}}^2$.

Step 3: (a) Calculations of the magnitude of magnetic field at radial distance 6 . 00  cm :

The point at the radial distance of  ${\mathrm{r}}_1$ is between the plates; the magnetic field at the point is given by,

  $\begin{array}{c}{\mathrm{B}}_1=\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2{\mathrm{\pi R}}^2}\right){\mathrm{r}}_1\\ 12.5\times {10}^{-9}\mathrm{T}=\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2{\mathrm{\pi R}}^2}\right){\mathrm{r}}_1\end{array}$

 

Rearrange the above equation as below.

 $\begin{array}{c}\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2\mathrm{\pi }}\right)=\frac{(12.5\times {10}^{-9}\text{ T})\times {\mathrm{R}}^2}{{\mathrm{r}}_1}\\ =\frac{(12.5\times {10}^{-9}\text{ T})\times {(0.04\text{ m})}^2}{0.02\text{ m}}\end{array}$

$\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2\mathrm{\pi }}\right)={10}^{-9}\text{ T}\cdot \text{m}$

                                                                                                      ….. (1)

 

Now the point at a radial distance of  ${\mathrm{r}}_2$ is outside the plates. So the magnetic field at that point is,

 $\begin{array}{c}{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}}{2{\mathrm{\pi r}}_2}\\ =\frac{1}{0.06\text{ m}}\times ({10}^{-9}\text{ T}\cdot \text{m})\\ =16.7\times {10}^{-9}\mathrm{T}\\ =16.7\mathrm{nT}\end{array}$

 

Hence, the magnitude of magnetic field at radial distance $6.00\text{ cm}$  is  $16.7\mathrm{nT}$.