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A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Omega across a 2V battery (Fig. 8.3). 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) = CV [1 – exp (–t// tau)], where the time constant tau is equal to CR.) |
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Answer» Solution :The time constant of the CR circuit is `tau = CR = 10^(-3) s`. Then, we have `q(t)=CV [1-"exp"(-t//tau)]` `=2xx10^(-9)[1-"exp "(-t//10^(-3))]` The electric field in between the plates at time t is `E=(q(t))/(epsi_(0)A)=(q)/(pi epsi_(0)), A= pi (1)^(2) m^(2)`= area of the plates. Consider now a circular loop of radius (1/2) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same value. The flux `Phi_(E)` through this loop is `Phi_(E)=Exx` area of the loop `=E xx pi xx((1)/(2))^(2)=(pi E)/(4) =(q)/(4 epsi_(0))` The displacement CURRENT `i_(d) =epsi_(0)"" (d Phi_(E))/(dt) =(1)/(4)""(dq)/(dt) =0.5 xx10^(-6)" exp "(-1)` at `t=10^(-3)s`. Now, applying Ampere-Maxwell law to the loop, we get `B xx 2 pi xx((1)/(2))= mu_(0) (i_(c)+i_(d))=mu_(0)(0+i_(d))=0.5xx10^(-6) mu_(0)" exp"(-1)` or, `B=0.74xx10^(-13) T` |
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