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(a) Fig shows a capacitor made of two circular plates each of radius 12cm and separated by 5.0mm. The capacitor is being charged by an external source (not shownin the figure). The charging current is constant and equal to 0.15A. Use Ampere's law (modified to include displacement current as given in the text) and the symmetryin the problem to calculate magnetic field between the plates at a point (i) on the axis (ii) 6.5 cm from the axis (iii) 15cm from the axis. (b) At what distance from the axis is the magnetic field due to displacement current greatest? Obtain the maximum value of the field. |
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Answer» Solution :Here, `R=0.12m,I=0.15A` `:.` Area of the plate, `A=piR^2=pixx(0.12)^2m^2` (a) Consider a loop of RADIUS r between the two circular plates, PLACED coaxially with them. Then area of the loop, `A'=pir^2` By SYMMETRY magnetic FIELD induction `vecB` is equal in magnitude and is tangentially to the CIRCLE at every point. In this. only displacement current `I_D` will cross the loop. Therefore, using Ampere's Maxwell law, we have `ointvecB.vec(dl)=mu_0I_D` `2pirB=mu_0xx`(current passing through the area A') `=mu_0I_D (pir^2)/(piR^2)` for `r lt R` `=mu_0I_D` for` r gt R` Thus, `B=(mu_0I_Dr^2)/(R^2 2pir)=(mu_0I_Dr)/(2piR^2)`....(i) (If` r lt R)` and `B=(mu_0I_D)/(2pir)`....(ii) (If `r gt R)` (i) On the axis, `r=0` Using (i), we get, `B=0` (ii) For a point 6.5cm from the axis, `r=6.5cm=6.5xx10^-2m`. Using (i), wehave, `B=(4pixx10^-7xx0.15xx6.5xx10^-2)/(2pixx(12xx10^-2)^2)` `=1.35xx10^-7T` (iii) For a point 15 cm from the axis, `r=15 cm=0.15m`. Using (ii), we have `B=(4pixx10^-7xx0.15)/(2pixx0.15)=2xx10^-7T` (b) From equation (i) and (ii) we note that B is maximum if `r=R=12cm=0.12m` `:. B_(max)=(mu_0I_D)/(2piR)=(4pixx10^-7xx0.15)/(2pixx0.12)` `=2.5xx10^-7T` |
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