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A parallel plate capacitor with circular plates of radius 1m has a capacitance of 1nF. At t = 0, it is connected for chargeing in series with a resistor R=1 M Omega across a 2V battery (Figure). 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 tauis 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))/(epsilon_(0)A)=(q)/(pi epsilon_(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 )=E xx` area of the loop `=E xx pi xx ((1)/(2))^(2)=(pi E)/(4)=(q)/(4epsilon_(0))` The displacement current `i_(d)=epsilon_(0)(d phi_(E ))/(dt)=(I)/(4)(dq)/(dt)=0.5xx10^(-6)exp(-1)` at`t=10^(-3)s`. Now, APPLYING Ampere Maxwell LAW to the loop, we get `B xx 2pi 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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