If the frequency of the voltage signal impressed on the parallel plate capacitor is doubled from f to 2f, the amplitude of the magnetic field Bo at r = a at the mid-plane of the gap between the plates of this capacitor will: Increase by a factor of two. Remain the same. Decrease by a factor of four. Decrease by a factor of two. Increase by a factor of four.

Increase by a factor of four

Explanation:

Given data

capacitor = f to 2f

r = a

to find out

capacitor will be

solution

first we calculate the current

displacement current will be = ɛ * electric flux

displacement current will be = ɛ * electric flux

electric flex = A * electric field

electric field = voltage / d

electric field = V (o) sin ωt / d

so electric flex

electric flex = A * V (o) sin ωt / d

and displacement current will be

displacement current = ɛ * A * V (o) sin ωt / d

so the magnetic filed will be here

magnetic filed = μ * current + μ * ɛ d∅/dt

so we can say that frequency will be double when magnetic field increase

so Increase by a factor of four