Question

A parallel-plate capacitor made of circular plates, each of radius $R=6.0cm$, has a capacitance $C=100pF$. The capacitor is connected to $230V$ a.c. supply with $a$ (angular) frequency of $300rad/s$. Determine the amplitude of $B$ (in $10^{-11}T$ ) at a point $3.0cm$ from the axis between the plates.

Answer 1

Answer: 1.63

We are given that, $C=100pF=100\times 10^{-12}F=10^{-10}F$, $V_{rms}=230V,\omega =300rad/s$
Capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{300\times 10^{-10}}=\frac{10^8}{3}\Omega$
If $I_{rms}$ is the rms value of the conduction current,
$I_{rms}=\frac{V_{rms}}{X_C}=\frac{230}{10^8/3}=6.9\times 10^{-6}A=6.9\mu A$
The conduction current is equal to the displacement even in case of ac.
Let $I_0$ be the peak value (amplitude) of the conduction current
(I). Clearly, $I_{rms}=\frac{I_0}{\sqrt{2}}$
Or $I_0=I_{rms}\times \sqrt{2}=(6.9\mu A)\times 1.414=9.76\mu A$
Obviously, peak value (amplitude) of $I_d$ is also the same, i.e., $9.76\mu A$.
Consider a circular path of radius $3.0cm(=0.03m)$ parallel to the plates of the capacitor and with its centre $(O)$ on the axis of the plates.
According to the Ampere's law,
$\oint \vec{B}\cdot \overrightarrow{d\ell }={\mu }_0\left(I+I_d\right)$...(i)
Now $\oint \vec{B}\cdot \overrightarrow{d\ell }=B\times$ circumference of the circle of radius
$0.03m=B(2\pi \times 0.03)$...(ii)
As there is no conduction current in the plates,
$I=0$...(iii)
Further, $I_d=\frac{9.76(\mu A)}{\pi (0.06{)}^2}\times \pi (0.03{)}^2$
$=2.44\mu A=2.44\times 10^{-6}A$....(iv)
From Eqs. (i), (ii), (iii) and (iv),
$B(2\pi \times 0.03)={\mu }_0\left(2.44\times 10^{-6}\right)$
$B=\left(\frac{{\mu }_0}{2\pi }\right)\frac{2.44\times 10^{-6}}{0.03}=\left(2\times 10^{-7}\right)\left(81.3\times 10^{-6}\right)$
$=1.63\times 10^{-11}T$