Question

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?

Short answer

Answer

1. The radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is $r_1=30\text{mm}$
2. The radius outside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is $r_2=53\text{mm}$
3. The maximum value of magnetic field $B_{max}=3.0\times {10}^{-5}\text{T}$

Step-by-step solution

Step 1: Given information

The radius of a plate of parallel plate capacitor is,$r=40\text{mm}$ ,

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

Understanding the concept  

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.

(a) Determining at what radius inside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value

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 $r_1=30\text{mm}$

(b) Determining at what radius outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value 

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$r_2=53\text{mm}$ .

(c) Determining the maximum value of the magnetic field 

$B_{max}$ 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 $=3.0\times {10}^{-5}\text{T}$.