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?

Short answer

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

Step-by-step solution

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

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 $4\mathrm{\pi }\times {10}^{-7}\text{N}/{\text{A}}^2$.

(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 $16.7\mathrm{nT}$.