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[l-exp(-t//tau)]`
`= 2 xx 10^(-9) [1-exp (-t//10^(-3)]`
The electric field in between the plates at time t is `A = pi(l)^(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. Then flux
`E xxpixx1/2^(2)=(piE)/(4)=(q )/(4epsilon_(0))`
The displacement current
Now applying amere maxwell law to the loop we get
or `B=0.74 xx10^(-13)` T