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A parallel plate capacitor with circular plates of radius \(1\; m\) has a capacitance of \(1\;\mathit{nF}\) . At \(t=0\) , it is connected for charging in series with a resistor \(R=1\;\mathit{M{\Omega}}\)  across a \(2\;V\)  battery . Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after \(t = 10^{–3} s\). \((\)The charge on the capacitor at time t is \(q(t)=\mathit{CV}[1\text{-}e^{\text{-}\frac t{\tau }}]\) , where the time constant τ is equal to \(CR).\)

Answer

Here the time constant is given by \(\mathit{CR}=10^{-3}\mathit{sec}\)
And the charge on the capacitor at any time t is given by \(q(t)=\mathit{CV}[1-\exp \left(\frac{-t}{\tau }\right)]\)
= \(2\times 10^{-9}[1-\exp \left(\frac{-t}{\tau }\right)]\)
We also know that
Electric field in between the plates
\(E=\dfrac{q}{A \epsilon_0}\)
Here area= \(\pi r^2,\) \(r=1\)
E= \(\frac q{\mathit{\pi \epsilon_0}}\)
Consider a loop of radius ½ parallel to the plated of the capacitor. So by gauss law
Electric flux \(=E\times A\)
= \(\frac{\mathit{E\pi }} 4=\frac q{4\mathit{\upsilon o}}\)
Maxwell developed displacement current which is
Id= \(\frac{\mathit{\epsilon_0 d\varphi }}{\mathit{dt}}=0.5\times 10^{-5}\exp \left(-1\right)\)
A t\(=10^{-3}\;\text{sec}\)
Applying the ampere maxwell law to calculate the magnetic field
\(B\times 2\pi \times \frac 1 2=\mathit{\mu o}\left(\mathit{ic}+\mathit{id}\right)\)
so we get \(B=0.74\times 10^{-13}T\)
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