Question

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

Answer 1

We are given that, $C=100\, pF =100 \times 10^{-12} F=10^{-10}\,F$, $V_{ rms }=230\, V , \omega=300 \,rad / s$
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 $3.0\, cm (=0.03\, m )$ 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
$0.03 \,m =B(2 \pi \times 0.03)\,\,\,$...(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) $
$B=\left(\frac{\mu_{0}}{2 \pi}\right) \frac{2.44 \times 10^{-6}}{0.03}=\left(2 \times 10^{-7}\right)\left(81.3 \times 10^{-6}\right)$
$=1.63 \times 10^{-11} \,T$