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 charge on capcitor at anytime t
`q = CE(1-e^(-t//RC)),I = (dq)/(dt) = (E)/(R )e^(-t//RC)`
`RC = 10^(6) xx 10^(-9) = 10^(-3)`
`t//RC = (10^(-3))/(10^(-3)) = 1, (E)/(R ) = (2)/(10^(6)) = 2xx10^(-6)A`
At `t = 10^(-3)sec, I = (2xx10^(-6))/(e)`
The electric field between plates
`E = (q)/(Ain_(0)) = (q)/(pi in_(0))` [since `A = pir^(2) = pi(1)^(2) = pi]`
Consider a circular loop of radius `1//2 m` parallel to plates through `P`.
flux through the loop
`phi_(E) = EA' = (q)/(pi in_(0))pi((1)/(2))^(2) = (Q)/(4 in_(0))`
Displacement current at `t = 10^(-3)sec`
`i_(d) = in_(0)(dphi_(E))/(dt) = (in_(0))/(4in_(0))(dq)/(dt) = (1)/(4)(dq)/(dt) = (1)/(4)i`
`= (1)/(4)xx(2xx10^(-6))e^(-1) = (0.5xx10^(-6))/(e)`
QAmpere-Maxwell law
`oint vec(B).dvec(l) = mu_(0)(i_(c)+i_(d))`
`B.2pi(1)/(2)= 4pixx 10^(-7) (0+(0.5 xx 10^(-6))/(e))`
`B = 0.75xx10^(-3)T` (`:'e = 2.7`)