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Considering the case of a parallel plate capacitor being charged, show how one is required to generalise Ampere.s circuital law to include the term due to displacement current. |
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Answer» Solution :Consider charging of a parallel plate capacitor by a time-varying current iw Let us FIND the magnetic field at a point P in a region outside the capacitor. For this we consider a plane CIRCULAR loop of radius r whose plane is perpendicular to the direction of the current carrying wire and which is centred symmetrically with respect to the wire [Fig]. Using symmetry condition and applying Ampere.s circuital law, we have `B(2pir)=mu_(0)i_((t))"....(i)"` However, if we consider a different surface, having the same boundary [either as SHOWN in Fig. or Fir. and apply Ampere.s circuital law as before, we find `B(2pir)=mu_(0)(0)=0"....(ii)"` It is because nocurrent passes through the surface of Figs. and 8.04. It causes a contradiction because we get a finite value of magnetic field B by doing calculation in one way and zero value of B by doing calculation in another way. To remove this contradiction Maxwell introduced the CONCEPT of displacement current  `i_(D)=in_(0)(dphi_(E))/(DT),` where `(dphi_(E))/(dt)` is the rate of change of electric flux between the plates of given capacitor. Now, the Ampere - Maxwell.s circuital law is expressed as `oint vecB.vecdl=mu_(0)[i_((t))+i_(D)]` Thus, Ampere - Maxwell law successfully explains the flow the current through a capacitor when it is being charged (or discharged) by a battery. |
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