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?

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

Use the formula of the magnitude of the magnetic field attheradius 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 role="math" localid="1663087015119" $r$is the radius of the capacitor plate.

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.

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

role="math" localid="1663087305362" $\begin{array}{rcl}r& =& 0.50\left(55.0\text{mm}\right)\\ & =& 27.5\text{mm}\end{array}$

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 $\begin{array}{rcl}& & 27.5\text{mm}\end{array}$.

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,

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

role="math" localid="1663087671454" $\begin{array}{c}r=\frac{55.0\text{mm}}{0.5}\\ =110\text{mm}\end{array}$

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 $110\text{mm}$.