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 resitor R=1M `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 l=`10^(-3)`s. (The charge on the capacitor at time l is `q(l)=CV` [1-exp(-t/r)]. where the time constant r 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^(-8)[1-exp(-t//10^(-2)]`
The electric field in between the plates at time t is
`E=(q(t))/(epsilon_(0)A)=(q)/(piepsilon_(0)),A=pi(1)^(2)m^(2)=` area of the plates
Consider now a circular loop of radius `(1//2)`m parallel tot he 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
`=Exxpixx((1)/(2))^(2)=(piE)/(4)=(q)/(4epsilon_(0))`
The displacement current
`i_(d)=epsilon_(0)=(edPhi_(E))/(dt)=(1)/(4)(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))=pi_(0)(i_(c)+i_(d))=mu_(0)(0+i_(d))=0.5xx10^(-6)mu_(0)exp(-1)`
or, `B=0.74xx10^(-3)T`