A parallel plate capacitor with circular plates of radius 1m has a capacitance of 1nF. At t = 0, it is connected for chargeing in series with a resistor `R=1 M Omega` across a 2V battery (Figure). 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)/(pi epsilon_(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 )=E xx` area of the loop `=E xx pi xx ((1)/(2))^(2)=(pi E)/(4)=(q)/(4epsilon_(0))`
The displacement current `i_(d)=epsilon_(0)(d phi_(E ))/(dt)=(I)/(4)(dq)/(dt)=0.5xx10^(-6)exp(-1)` at `t=10^(-3)s`.
Now, applying Ampere Maxwell law to the loop, we get `B xx 2pi 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`