A parallel plate capacitor with circular plates of radius 1 m has a capacity `10^(-19)F`. At, t = 0 it is connected for charging in series with a resistor `R=1MOmega` across a 2 V battery. Calculate the magnetic field at a point P where, OP=0.5 m (halfway between the centre and the periphery of the plates).

Explanation

Here, radius of circular plates, r=1m
Capacitance, `C=10^(-19)F`
Resistor, `R=1MOmega=10^(5)Omega`
Potential difference, V = 2V
Time constant of RC circuit, `tau=RC`
`=10^(6)xx10^(-19)=10^(-13)s`
The charge on the capacitor at time t is
`q(t)=q_(0)[1-e^(t//tau)]=CV[1-e^(t//tau)]`
`=10^(-19)xx2[1-e^(t//tau)]`
`=2xx10^(-19)[1-e^(t//tau)]`
`=2xx10^(-19)[1-e^(t//10^(-13))]`
Electric field between the plates,
`E=(q(t))/(epsi_(A))`
[A is area of circular plates]
`=(q(t))/(epsi_(0)pi)" "[A=pir^(2)=pi(1)^(2)=pi]`
Now, we have to find magnetic field at mid point of circular loop for circular loop of radius 0.5 m Flux, `phi=Exx` Area of circular of radius 0.5 m
`=Exx[pixx((1)/(2))^(2)]`
`=(q(t))/(piepsi_(0))[pixx(1)/(4)]=(q(t))/(4epsi_(0))`
Displacement current, `i_(d)=(epsi_(0)dphi)/(dt)=(epsi_(0)d)/(dt)[(q(t))/(4epsi_(0))]`
`=(epsi_(0))/(4epsi_(0))=(d[q(t)])/(dt)`
`=(1)/(4)xx2xx10^(-19)xx(1)/(10^(-13))xxe^(t//10^(-13))`
`=(5xx10^(-7))xxe^(-t//10^(-13))`
According to Ampere.s law,
`intB.dl=mu_(0)(i_(c)+i_(d))" "(i_(c)=0)`
`Bxx2pixx(1)/(2)=mu_(0)5xx10^(-7)xxe^(-t//10^(-13))`
`B=(4pixx10^(-7))/(pi)xx5xx10^(-7)xxe^(-t//10^(-13))`
`=2xx10^(-13)e^(-10^(3)t)`