A parallel plate capacitor with circular plates of radius `1m` has a capacitor of `1nF`. At `t = 0`, it is connected for charging in series with a resistor `R = 1MOmega` across a `2V` battery. Calculate the magnetic field at a point `P`, halfway between the cnetre and the periphery of the plates, after `t = 10^(-3)sec`.

Explanation

The time constant of the CR circuit is `r=CR=10^(-3)s`. Then. We have
`q(l)=CV[1-"exp"(-l//r)]`
`=2xx10^(-9)[1-"exp"(-t//10^(-3))]`
the electric field in between the plates at time t is
`E=(q(t))/(epsiolon_(0)A))=(q)/(piepsilon_(@)):A=pi(1)^(2)m^(2)=` area of the plates.
Consider no a circular loop 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
`=Exxpixx((1)/(2))^(2)=(piE)/(4)=(q)/(4epsilon_(0))`
the displacement current
`i_(d)=epsilon_(0)(dphi_(E))/(dl)=(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`