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  • A parallel plate capacitor of area 60 cm2 and separation 3 mm is charged initially to <img alt="\large 90\: \mu \: C" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D

A parallel plate capacitor of area 60 cm2 and separation 3 mm is charged initially to \large 90\: \mu \: C.

If the medium between the plate gets slightly conducting and the  plate loses the charge initially at the rate of \large 2.5\times 10^{-8}\: C/s, then what is the magnetic field between the plates.

Option: 1

\dpi{100} \large 2.5\times 10^{-8}\:T
 


Option: 2

2.0\times 10^{-7}\:T

 


Option: 3

\large 1.63\times 10^{-11}\:T


Option: 4

Zero


Answers (1)

best_answer

 Conduction current within the plates is from the positive plate to negative plate of the parallel plate capacitor.

There will also displacement current flowing between the plates and which is given as

I_{D}=\epsilon_{0} \frac{d \phi_{E}}{d t}=\epsilon_{0}A \frac{d {E}}{d t}

We know that, the electric field between the plates of the capacitor is  E= \frac{Q}{A\epsilon _o}

With the decrease in the charge on the capacitor, the electric field will decrease between the plates of the capacitor.

Therefore, the direction of I_D is opposite to that of the electric field and hence opposite to the conduction current. But the magnitude of displacement current is the same as that of conduction current. The net current between the plates is zero.

Note: Conduction current is there because the region between the plates becomes conducting.

          So total current,   I^{\prime}=I+I_{D}=0

Now Using Ampere's law,

\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I^{\prime}=0

Hence magnetic field within the plates is zero.

Posted by

Pankaj Sanodiya

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