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))/(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 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
`Phi_(E)=Exx` area of the loop
`=Exxtauxx((1)/(2))^(2)=(piE)/(4)=(q)/(4epsilon_(0))`
The displacement current
`i=epsilon_(0)(dPhi_(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))=mu_(0)(i_(c)+i_(d))`
`=mu_(0)(0+i_(d))=0.5xx10^(-6)mu_(0)exp(-1)`
or, `B=0.74xx10^(-13)T`