A parallel plate capacitor of area 60, cm^2 a | vedclass

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(D) The magnetic field between the plates of a parallel plate capacitor is determined by the displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}

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Zero

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(D) The magnetic field between the plates of a parallel plate capacitor is determined by the displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.In this scenario,the medium between the plates is slightly conducting,which means there is a conduction current $I_c = -\frac{dq}{dt} = 2.5	imes10^{-8}\, A$ flowing between the plates.According to the modified Ampere-Maxwell law,the total current $I_{total} = I_c + I_d$ is the source of the magnetic field.For a parallel plate capacitor,the conduction current $I_c$ and the displacement current $I_d$ are equal in magnitude but opposite in direction (or effectively cancel each other out in the context of the magnetic field generated between the plates in this specific idealized setup).However,in a standard parallel plate capacitor where the plates are circular and the current is uniform,the magnetic field at a distance $r$ from the axis is given by $B = \frac{\mu_0 I r}{2\pi R^2}$.Since the problem does not specify a radial distance $r$ and asks for the magnetic field 'between the plates' generally,and given the symmetry of the conduction current flowing through the medium,the magnetic field is zero at the center or if we consider the net current enclosed by an Amperian loop to be zero due to the continuity of current,the result is zero.

$A$ parallel plate capacitor of area $60\, cm^2$ and separation $3\, mm$ is charged initially to $90\, \mu C$. If the medium between the plates becomes slightly conducting and the plate loses the charge at the rate of $2.5\times10^{-8}\, C/s$,what is the magnetic field between the plates?

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