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A parallel plate capacitor with circular plates of radius 1m has a capacitance of 0.5muF. At t=0, it is connected for charging inseries with a resistance R=2kOmega across a 4V battery. Calculate the magnetic field at a point P in between the plates and half way between the centre and periphery of the plates after 10^-3s. Fig. |
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Answer» <P> Solution :Here, `r=1m, C=0.5muF =0.5xx10^-6F,``R=2kOmega=2xx10^3Omega,V=4V,t=10^-3s`. The charge Q on the capacitor at any time t during charging is `Q=Q_0(1-e^(-t//RC))=CV(1-e^(-t//RC))` `:. (dQ)/(dt)=CVxx1/(RC) e^(-t//RC)=V/R e^(-t//tau) ( :' tau=RC)` Electric field in between the plates at time t is `E=Q/(in_0A) =Q/(in_0pir^2)` `(dE)/(dt)=1/(in_0pir^2) (dQ)/(dt) =1/(in_0pir^2)xxV/R e^(-t//tau)` Let us now consider a circular loop of radius `r'=(=1/2m)`, parallel to the plates passing through P. The magnetic field B at all points on this loop is of the smae value and acting tengentially to the loop. Applying Ampere Maxwell's Law to this loop, we get `oint vecB.vec(dl)=mu_0[I+in_0 (dphi_E)/(dt)]` or `B2pir'=mu_0(0+in_0pir' (dE)/(dt))` or `B=mu_0(in_0r')/2 (dE)/(dt) =(mu_0in_0r')/2XX1/(in_0pir^2) V/R e^(-t/tau)` `(mu_0r')/(2pir^2)xxV/R e^(-t/tau)` `=((4pixx10^-7)xx(1//2))/(2PI(1)^2)xx4/((2xx10^3))e^-1` `( :' t=10^-3s , tau=10^-3s)` `=(2xx10^-10)/2.718=7.358xx10^-11T` |
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