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Question: A parallel plate capacitor of area \(60c{{m}^{2}}\) and separation 3mm is charged initially to\(90\m...

A parallel plate capacitor of area 60cm260c{{m}^{2}} and separation 3mm is charged initially to90μC90\mu C. If the medium between the plates gets slightly conducting and the plate loses the charge initially at the rate of 2.5×10−8C/s2.5\times {{10}^{-8}}C/s then what is the magnetic field between the plates?
A. 2.5×10−8T2.5\times {{10}^{-8}}T
B. 2.0×10−7T2.0\times {{10}^{-7}}T
C. 1.63×10−11T1.63\times {{10}^{-11}}T
D. zero

Explanation

Solution

As a first step, you could note down all the given quantities from the question. Then you could find the magnitude as well as direction of the displacement current. You may then find that the displacement current and conduction current are of same magnitude but opposite direction. Then use ampere’s circuital law to find the answer.

Formula used:
Displacement current,
ID=ε0dϕEdt{{I}_{D}}={{\varepsilon }_{0}}\dfrac{d{{\phi }_{E}}}{dt}
Ampere’s circuital law,
∮B→⋅dl→=μ0I′\oint{\overrightarrow{B}}\centerdot \overrightarrow{dl}={{\mu }_{0}}I'

Complete step-by-step solution:
From our basic knowledge of electrodynamics, we know that the conduction current within the plates would be from the positive plate to the negative plate of a parallel plate capacitor.
Now, we know that the displacement current could given by the following formula,
ID=ε0dϕEdt{{I}_{D}}={{\varepsilon }_{0}}\dfrac{d{{\phi }_{E}}}{dt}
But we know that electric flux here would be,
ϕE=EA{{\phi }_{E}}=EA
So,
ID=ε0AdEdt{{I}_{D}}={{\varepsilon }_{0}}A\dfrac{dE}{dt}
Now, substituting for electric field, we get,
ID=ε0Adqε0Adt{{I}_{D}}={{\varepsilon }_{0}}A\dfrac{dq}{{{\varepsilon }_{0}}Adt}
∴ID=dqdt=1.5×10−8A\therefore {{I}_{D}}=\dfrac{dq}{dt}=1.5\times {{10}^{-8}}A
Since, the charge is decreasing with increasing time, dqdt=−ve\dfrac{dq}{dt}=-ve and hence dEdt⟨0\dfrac{dE}{dt}\langle 0
Which further implies that the direction of displacement current is opposite to that of the electric field which implies it is also opposite to the conduction current. Also, we know that the magnitude of displacement current and the conduction current is the same which would lead to the conclusion that the net current between the plates is zero. That is,
I′=I+ID=0I'=I+{{I}_{D}}=0 …………………………………. (1)
Now, we have the ampere’s law that gives,
∮B→⋅dl→=μ0I′\oint{\overrightarrow{B}}\centerdot \overrightarrow{dl}={{\mu }_{0}}I'
But from equation (1) we have that I’ = 0, thus,
∮B→⋅dl→=μ0I′=0\oint{\overrightarrow{B}}\centerdot \overrightarrow{dl}={{\mu }_{0}}I'=0
Therefore, we found the magnetic field between the two plates to be zero. Option D is correct.

Note: You may have noted that we are given a lot of parameters in the question that we haven’t used in solving the question. In the final steps, we see that these quantities, that is, the area of cross-section of the plates and the distance of separation between the plates are getting cancelled. So, we shouldn’t necessarily worry about all the quantities that are given.