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 center and the periphery of the plates, after `t = 10^(-3)sec`.

Explanation

Time constant of C-R circuit is given
by
`tau=CR=10^-9xx10^6=10^-3s`
Change on capacitor at any time t during
charging is given by
`q=CV(1-e^(-t//tau))=(CV)/tau e^(-t//tau)`
The electric field in between the plates at time
t is,
`E=q/(in_0A)=q/(in_0pir^2)`
Consider a circular loop of radius `r'=1//2m`
parallel to the plates passing through P. The
magnetic field at all points on this loop is of same
value and acting tangentially to the loop.
The electric flux through this loop is
`phi_E=Epir'^2=q/(in_0pir^2)xxpir'^2=q/(in_0) (r'^2)/(r^2)`
The displacement current, `i_D=in_0(dphi_E)/(dt)`
`in_0 d/(dt) (q/(in_0) (r'^2)/(r^2)) =(r'^2)/(r^2) (dq)/(dt) =(r'^2)/(r^2) (CV)/tau e^(-t/tau)`
When `t=10^-3s`,
`I_D=((1//2)^2)/(1^2)xx(10^-9xx2)/(10^-3) e^(-10^3//10^-3)`
`=(0.5xx10^-6)/eA`
Applying Ampere Maxwell's law to the loop,
we have,
`ointvecB.vec(dl)=mu_0(i+i_D)=mu_0(0+i_D)=mu_0i_D`
or `B2pir'=(mu_0xx0.5xx10^-6)/e`
or `B=(mu_0)/(2pi(1//2))xx(0.5xx10^-6)/e`
`=(2xx10^-7)/((1//2))xx(0.5xx10^-6)/2.718=(2xx10^-13)/2.718`
`=0.74xx10^-13T`