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.)

Explanation

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`