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A parallel plate capacitor with circular plates of radius 1m has a capacitance of 1 nF. At t=0 , It is connected for charging in series withare resistor R=1 M Q across a 2V battery (fig). 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 tau 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 `tau` =CR `=10^(-3) s` , 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)/(piepsi_(0)),A=pi (1)^(2) m^(2)=` area of the plates . Consider now a circular LOOP of radius (1/2) m parallel to theplates 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 The flux `phi_(E)=Exx` area of the loop `=Exxpixx((1)/(2))^(2)=(piE)/(4)=(q)/(4epsi_(0))` The displacement CURRENT `i_(d)=e_(0)(dphi_(E))/(dt)=(1)/(2)(dq)/(dt)=0.5xx10^(-6) "exp" (-1)` at `t=10^(-3)s` . Now , applying Ampere-Maxwell law to the loop, we get `Bxx2pixx((1)/(2))=m_(0) (i_(C)+i_(d))=m_(0)(0+i_(d))=0.5xx10^(-6)m_(0) "exp"(-1)` or `B=0.74 xx 10^(-13)T`. 
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