Q63P - Fundamentals Of Physics

Question

A parallel-plate capacitor with circular plates of radius   is being charged. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{\%}$ of its maximum value?

Step-by-Step Solution

Answer

(a) The radius inside the capacitor plate is $27.5 \text{ mm}$.

(b) The radius outside the capacitor plate is $\text{110 mm}$.

Listing the given quantities:

Radius of the circular plates,  $R= 55.\text{0 mm}$

Understanding the concepts of magnetic field:

Use the formula of the magnitude of the magnetic field at the radius inside and outside the capacitor. Then compare these magnetic fields with maximum magnetic field to find the radius inside and outside the capacitor gap at which the magnitude of the induced magnetic field is equal to   of its maximum value.

Formula: 

Magnitude of the induced magnetic field inside the capacitor is,

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

Here, ${\mu }_0$ is the permittivity of free space,  $i_d$ is the displacement current, $R$ is the radius of the circular plate, and $r$ is the radius of the capacitor plate.

(a) Calculations of the radius inside the capacitor gap:

The magnitude of the induced magnetic field inside the capacitor as,

$B=\left(\frac{{\mu }_0{i}_d}{2\pi R^2}\right)r$                                                                                                     ..... (1)

The induced magnetic field will be maximum when  $r=R$. 

Therefore,

$B_{max}=\left(\frac{{\mu }_0{i}_d}{2\pi R^2}\right)R$

$B_{max}=\left(\frac{{\mu }_0{i}_d}{2\pi R}\right)$                                                                                                 ..... (2)

Now, according to the given condition, 

$B=\frac{50}{100} B_{max}$

Substitute equation (1) and (2) in the above equation.

$$\begin{gathered} \left(\frac{{\mu }_0{i}_d}{2\pi R^2}\right)r=0.50\left(\frac{{\mu }_0{i}_d}{2\pi R}\right) \\ \frac{r}{R}=0.50 \end{gathered}$$

$\begin{array}{rcl}r& =& 0.50\left(55.0\text{ mm}\right)\\ & =& 27.5\text{ mm}\end{array}$

Hence, the radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to $50\%$ of its maximum value is  $\begin{array}{rcl}& & 27.5\text{ mm}\end{array}$.

(b) Calculations of the radius outside the capacitor gap:

The magnitude of the induced magnetic field outside the capacitor as,

$B=\left(\frac{{\mu }_0{i}_d}{2\pi r}\right)$                                                                                                                  ..... (3)

The induced magnetic field will be maximum when $r=R$.

Therefore, 

$B_{max}=\left(\frac{{\mu }_0{i}_d}{2\pi R}\right)$

Now, according to the given condition, 

$\begin{array}{l}B=\frac{50}{100} B_{max}\\ \left(\frac{{\mu }_0{i}_d}{2\pi r}\right)=0.50\left(\frac{{\mu }_0{i}_d}{2\pi R}\right)\\ \frac{R}{r}=0.50\end{array}$

$\begin{array}{c}r=\frac{55.0\text{ mm}}{0.5}\\ =110\text{ mm}\end{array}$

Hence, the radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to $50\%$ of its maximum value is $110\text{ mm}$.