2.(c) A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad\: \: s ^{-1}

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answers (1)
S Sayak

We know the Ampere - Maxwells Law,

\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})

Between the plates conduction current i_{c}=0.

For a loop of radius r smaller than the radius of the discs,

 \mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}

B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}

B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}

Since we have to find the amplitude of the magnetic field we won't use the RMS value but the maximum value of current.

\\i_{max}=\sqrt{2}\times i_{rms}\\ i_{max}=\sqrt{2}\times6.9\mu A\\ i_{max}=9.76\mu A

\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\ B_{amp}=1.63\times 10^{-11}T

The amplitude of B at a point 3.0 cm from the axis between the plates is  1.63\times 10^{-11}T.

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