Question: A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A . At what radius (a) inside and (b) outside the capacitor, the gap is the magnitude of the induced magnetic field equal to 75% of its maximum value? (c) What is that maximum value?

Discharged current is,$i=6.0\text{A}$ .

The magnetic field inside a capacitor is directly proportional to the loop radius The relation is written as below:

$\mathit{B}\mathbf{=}\left(\frac{{\mu }_0{i}_d}{2\pi R^2}\right)\mathit{r}$ (i)

Here, is the magnetic field, ${\mathbf{\mu }}_{\mathbf{0}}$is permeability constant, i is current, R is the inside radius, and r is the outside radius.

The magnetic field outside the capacitor is inversely proportional to the loop radius. The relation is written as below,

$\mathit{B}\mathbf{=}\mathbf{\left(}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{i}}_{\mathbf{d}}}{\mathbf{2}\mathbf{\pi }}\mathbf{\right)}$ (ii)

Here, B is the magnetic field, ${\mathbf{\mu }}_{\mathbf{0}}$is permeability constant, i is current, and r is the outside radius.

From equation 32-16, the magnetic field inside the capacitor is directly proportional to the radius so that, $B_{max}$occurs when the r =R .

$B_{max}\propto R,$ and, the given condition is $0.75B_{max}\propto r_1$.

So, to find the value of $r_1$ by taking the ratio of these two equations,

$$\begin{gathered} \frac{0.75B_{max}}{B_{max}}=\frac{r_1}{R} \\ {r}_1=0.75R \end{gathered}$$

By substituting the value, the result is,

$$\begin{gathered} r_1=0.75\times 40 \\ =30\text{mm} \end{gathered}$$

Hence, the radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is $r_1=30\text{mm}$

Similarly, for outside the capacitor, $B_{max}$ occurs, when r =R , and from equation 32- the magnetic field is inversely proportional to . So, the maximum magnetic field is at r =R . From the given condition, $0.75B_{max}$ occurs when the radius is r₂, so according to equation 32-17, write proportionality as,

$0.75B_{max}\propto \frac{1}{r_2}$

(iii)

Also, the magnetic field is inversely proportional to the inside radius, therefore,

$B_{max}\propto \frac{1}{R}$ (iv)

Taking a ratio of equations (iii) and (iv), we get

$\frac{0.75B_{max}}{B_{max}}=\frac{R}{r_2}$

Solve the above equation to calculate as,

$$\begin{gathered} r_2=\frac{R}{0.75} \\ =\frac{40\text{mm}}{0.75} \\ =53\text{mm} \end{gathered}$$

Hence, the radius outside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is$r_2=53\text{mm}$ .

$B_{max}$ can be calculated using equation (ii) as,

$$\begin{gathered} B_{max}=\frac{4\pi \times {10}^{-7}\text{T}·\text{m/A}\times 6.0\text{A}}{2\pi \times 40\times {10}^{-3}\text{m}} \\ =3.0\times {10}^{-5}\text{T} \\ {B}_{max=3.0\times {10}^{-5}\text{T}} \end{gathered}$$

Hence, the maximum value of the magnetic field $=3.0\times {10}^{-5}\text{T}$.